Integrand size = 22, antiderivative size = 54 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {2401}{1936 (1-2 x)^2}-\frac {10633}{5324 (1-2 x)}-\frac {1}{33275 (3+5 x)}-\frac {47481 \log (1-2 x)}{117128}+\frac {138 \log (3+5 x)}{366025} \] Output:
2401/1936/(1-2*x)^2-10633/(5324-10648*x)-1/(99825+166375*x)-47481/117128*l n(1-2*x)+138/366025*ln(3+5*x)
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {\frac {7263025}{(1-2 x)^2}+\frac {11696300}{-1+2 x}-\frac {176}{3+5 x}-2374050 \log (1-2 x)+2208 \log (6+10 x)}{5856400} \] Input:
Integrate[(2 + 3*x)^4/((1 - 2*x)^3*(3 + 5*x)^2),x]
Output:
(7263025/(1 - 2*x)^2 + 11696300/(-1 + 2*x) - 176/(3 + 5*x) - 2374050*Log[1 - 2*x] + 2208*Log[6 + 10*x])/5856400
Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(3 x+2)^4}{(1-2 x)^3 (5 x+3)^2} \, dx\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \int \left (\frac {138}{73205 (5 x+3)}+\frac {1}{6655 (5 x+3)^2}-\frac {47481}{58564 (2 x-1)}-\frac {10633}{2662 (2 x-1)^2}-\frac {2401}{484 (2 x-1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {10633}{5324 (1-2 x)}-\frac {1}{33275 (5 x+3)}+\frac {2401}{1936 (1-2 x)^2}-\frac {47481 \log (1-2 x)}{117128}+\frac {138 \log (5 x+3)}{366025}\) |
Input:
Int[(2 + 3*x)^4/((1 - 2*x)^3*(3 + 5*x)^2),x]
Output:
2401/(1936*(1 - 2*x)^2) - 10633/(5324*(1 - 2*x)) - 1/(33275*(3 + 5*x)) - ( 47481*Log[1 - 2*x])/117128 + (138*Log[3 + 5*x])/366025
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\frac {1329117}{66550} x^{2}+\frac {4364739}{532400} x -\frac {1209091}{532400}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {47481 \ln \left (-1+2 x \right )}{117128}+\frac {138 \ln \left (3+5 x \right )}{366025}\) | \(44\) |
default | \(-\frac {1}{33275 \left (3+5 x \right )}+\frac {138 \ln \left (3+5 x \right )}{366025}+\frac {2401}{1936 \left (-1+2 x \right )^{2}}+\frac {10633}{5324 \left (-1+2 x \right )}-\frac {47481 \ln \left (-1+2 x \right )}{117128}\) | \(45\) |
norman | \(\frac {\frac {277826}{19965} x^{2}+\frac {231529}{79860} x +\frac {1209091}{79860} x^{3}}{\left (-1+2 x \right )^{2} \left (3+5 x \right )}-\frac {47481 \ln \left (-1+2 x \right )}{117128}+\frac {138 \ln \left (3+5 x \right )}{366025}\) | \(47\) |
parallelrisch | \(-\frac {71221500 \ln \left (x -\frac {1}{2}\right ) x^{3}-66240 \ln \left (x +\frac {3}{5}\right ) x^{3}-28488600 \ln \left (x -\frac {1}{2}\right ) x^{2}+26496 \ln \left (x +\frac {3}{5}\right ) x^{2}-133000010 x^{3}-24927525 \ln \left (x -\frac {1}{2}\right ) x +23184 \ln \left (x +\frac {3}{5}\right ) x -122243440 x^{2}+10683225 \ln \left (x -\frac {1}{2}\right )-9936 \ln \left (x +\frac {3}{5}\right )-25468190 x}{8784600 \left (-1+2 x \right )^{2} \left (3+5 x \right )}\) | \(93\) |
Input:
int((2+3*x)^4/(1-2*x)^3/(3+5*x)^2,x,method=_RETURNVERBOSE)
Output:
20*(1329117/1331000*x^2+4364739/10648000*x-1209091/10648000)/(-1+2*x)^2/(3 +5*x)-47481/117128*ln(-1+2*x)+138/366025*ln(3+5*x)
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.39 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {116962296 \, x^{2} + 2208 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 2374050 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (2 \, x - 1\right ) + 48012129 \, x - 13300001}{5856400 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \] Input:
integrate((2+3*x)^4/(1-2*x)^3/(3+5*x)^2,x, algorithm="fricas")
Output:
1/5856400*(116962296*x^2 + 2208*(20*x^3 - 8*x^2 - 7*x + 3)*log(5*x + 3) - 2374050*(20*x^3 - 8*x^2 - 7*x + 3)*log(2*x - 1) + 48012129*x - 13300001)/( 20*x^3 - 8*x^2 - 7*x + 3)
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=- \frac {- 10632936 x^{2} - 4364739 x + 1209091}{10648000 x^{3} - 4259200 x^{2} - 3726800 x + 1597200} - \frac {47481 \log {\left (x - \frac {1}{2} \right )}}{117128} + \frac {138 \log {\left (x + \frac {3}{5} \right )}}{366025} \] Input:
integrate((2+3*x)**4/(1-2*x)**3/(3+5*x)**2,x)
Output:
-(-10632936*x**2 - 4364739*x + 1209091)/(10648000*x**3 - 4259200*x**2 - 37 26800*x + 1597200) - 47481*log(x - 1/2)/117128 + 138*log(x + 3/5)/366025
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {10632936 \, x^{2} + 4364739 \, x - 1209091}{532400 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} + \frac {138}{366025} \, \log \left (5 \, x + 3\right ) - \frac {47481}{117128} \, \log \left (2 \, x - 1\right ) \] Input:
integrate((2+3*x)^4/(1-2*x)^3/(3+5*x)^2,x, algorithm="maxima")
Output:
1/532400*(10632936*x^2 + 4364739*x - 1209091)/(20*x^3 - 8*x^2 - 7*x + 3) + 138/366025*log(5*x + 3) - 47481/117128*log(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.28 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=-\frac {1}{33275 \, {\left (5 \, x + 3\right )}} - \frac {1715 \, {\left (\frac {297}{5 \, x + 3} - 89\right )}}{58564 \, {\left (\frac {11}{5 \, x + 3} - 2\right )}^{2}} + \frac {81}{200} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) - \frac {47481}{117128} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \] Input:
integrate((2+3*x)^4/(1-2*x)^3/(3+5*x)^2,x, algorithm="giac")
Output:
-1/33275/(5*x + 3) - 1715/58564*(297/(5*x + 3) - 89)/(11/(5*x + 3) - 2)^2 + 81/200*log(1/5*abs(5*x + 3)/(5*x + 3)^2) - 47481/117128*log(abs(-11/(5*x + 3) + 2))
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {138\,\ln \left (x+\frac {3}{5}\right )}{366025}-\frac {47481\,\ln \left (x-\frac {1}{2}\right )}{117128}-\frac {\frac {1329117\,x^2}{1331000}+\frac {4364739\,x}{10648000}-\frac {1209091}{10648000}}{-x^3+\frac {2\,x^2}{5}+\frac {7\,x}{20}-\frac {3}{20}} \] Input:
int(-(3*x + 2)^4/((2*x - 1)^3*(5*x + 3)^2),x)
Output:
(138*log(x + 3/5))/366025 - (47481*log(x - 1/2))/117128 - ((4364739*x)/106 48000 + (1329117*x^2)/1331000 - 1209091/10648000)/((7*x)/20 + (2*x^2)/5 - x^3 - 3/20)
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.96 \[ \int \frac {(2+3 x)^4}{(1-2 x)^3 (3+5 x)^2} \, dx=\frac {22080 \,\mathrm {log}\left (5 x +3\right ) x^{3}-8832 \,\mathrm {log}\left (5 x +3\right ) x^{2}-7728 \,\mathrm {log}\left (5 x +3\right ) x +3312 \,\mathrm {log}\left (5 x +3\right )-23740500 \,\mathrm {log}\left (2 x -1\right ) x^{3}+9496200 \,\mathrm {log}\left (2 x -1\right ) x^{2}+8309175 \,\mathrm {log}\left (2 x -1\right ) x -3561075 \,\mathrm {log}\left (2 x -1\right )+146202870 x^{3}-27164940 x +15280430}{58564000 x^{3}-23425600 x^{2}-20497400 x +8784600} \] Input:
int((2+3*x)^4/(1-2*x)^3/(3+5*x)^2,x)
Output:
(22080*log(5*x + 3)*x**3 - 8832*log(5*x + 3)*x**2 - 7728*log(5*x + 3)*x + 3312*log(5*x + 3) - 23740500*log(2*x - 1)*x**3 + 9496200*log(2*x - 1)*x**2 + 8309175*log(2*x - 1)*x - 3561075*log(2*x - 1) + 146202870*x**3 - 271649 40*x + 15280430)/(2928200*(20*x**3 - 8*x**2 - 7*x + 3))