How to read and parse a multipart response as a stream?

Question

I am creating a java process to download WebEx recordings using their NBR API's NBRFileOpenService call. It returns a multipart response with the recording file contents attached. I have it somewhat working with the code below. However, when the recording file is large enough, I get OutOfMemoryError exception.

It is quite common for the recordings to be large and if the API only returned the file alone, I could just stream the download, however I'm not so sure how I can safely handle the multipart response. So I'm wondering if there is any way to read the file metadata as well as save the binary content to a file without holding the entire response in memory.

API Response Format:

------=_Part_674_458057647.1593732813745
Content-Type: text/xml; charset=UTF-8
Content-Transfer-Encoding: binary
Content-Id: <AD79B5747EFC01CDDA9A281BA8CDEF0C>

[SOAP RESPONSE]
------=_Part_674_458057647.1593732813745
Content-Type: application/octet-stream
Content-Transfer-Encoding: binary
Content-Id: <C498AB4664B57130F869695A1C5B584E>

[FILE METADATA]
------=_Part_674_458057647.1593732813745
Content-Type: application/octet-stream
Content-Transfer-Encoding: binary
Content-Id: <003D9EBA1E491CE2E9E5903C996EFD4C>

[BINARY FILE CONTENT]
------=_Part_674_458057647.1593732813745--

My Code:

public void retrieveRecordingFile(String uri, String recordId, String serviceType) throws Exception {
    HttpClient httpClient = generateHttpClient();
    
    HttpPost httpPost = new HttpPost(uri);
    httpPost.addHeader("Content-Type", ContentType.APPLICATION_XML.getMimeType());
    httpPost.addHeader("SOAPAction", "NBRFileOpenService");
    
    String requestXml = buildNBRDownloadFileXml(recordId, serviceType);
    HttpEntity httpEntity = new ByteArrayEntity(requestXml.getBytes(Charset.forName("UTF-8")));
    
    httpPost.setEntity(httpEntity);
    HttpResponse httpResponse = httpClient.execute(httpPost);
    if (httpResponse.getStatusLine().getStatusCode() == 200) {
        MimeMultipart mimeMultipart = new MimeMultipart(new ByteArrayDataSource(httpResponse.getEntity().getContent(), "multipart/form-data"));
        String filename = null;
        File targetFile = null;
        for (int i = 0; i < mimeMultipart.getCount(); i++) {
            if (i == 1) {
                filename = retrieveFileName(mimeMultipart.getBodyPart(i).getInputStream());
            } else if (i == 2) {
                targetFile = new File(DOWNLOAD_DIR + filename);
                FileUtils.copyInputStreamToFile(mimeMultipart.getBodyPart(i).getInputStream(), targetFile);
            }
        }
    }
}

Any help is truly appreciated.