Optimal. Leaf size=30 \[ \frac {x}{\frac {2 x}{3}-x \left (\frac {1}{4} (-1+x)+x\right )-\frac {4}{\log (1+x)}} \]
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Rubi [F] time = 3.57, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {36 \left (-16 x-16 (1+x) \log (1+x)+5 x^2 (1+x) \log ^2(1+x)\right )}{(1+x) (48+x (-11+15 x) \log (1+x))^2} \, dx\\ &=36 \int \frac {-16 x-16 (1+x) \log (1+x)+5 x^2 (1+x) \log ^2(1+x)}{(1+x) (48+x (-11+15 x) \log (1+x))^2} \, dx\\ &=36 \int \left (\frac {5}{(-11+15 x)^2}-\frac {16 \left (528-912 x-1319 x^2-330 x^3+225 x^4\right )}{x (1+x) (-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}-\frac {16 (-11+45 x)}{x (-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )}\right ) \, dx\\ &=\frac {12}{11-15 x}-576 \int \frac {528-912 x-1319 x^2-330 x^3+225 x^4}{x (1+x) (-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx-576 \int \frac {-11+45 x}{x (-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )} \, dx\\ &=\frac {12}{11-15 x}-576 \int \left (\frac {1}{\left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}+\frac {48}{11 x \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}-\frac {1}{(1+x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}-\frac {720}{(-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}-\frac {720}{11 (-11+15 x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2}\right ) \, dx-576 \int \left (-\frac {1}{11 x \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )}+\frac {30}{(-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )}+\frac {15}{11 (-11+15 x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )}\right ) \, dx\\ &=\frac {12}{11-15 x}+\frac {576}{11} \int \frac {1}{x \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )} \, dx-576 \int \frac {1}{\left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx+576 \int \frac {1}{(1+x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx-\frac {8640}{11} \int \frac {1}{(-11+15 x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )} \, dx-\frac {27648}{11} \int \frac {1}{x \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx-17280 \int \frac {1}{(-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )} \, dx+\frac {414720}{11} \int \frac {1}{(-11+15 x) \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx+414720 \int \frac {1}{(-11+15 x)^2 \left (48-11 x \log (1+x)+15 x^2 \log (1+x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 1.08, size = 23, normalized size = 0.77 \begin {gather*} -\frac {36 x \log (1+x)}{144+3 x (-11+15 x) \log (1+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 25, normalized size = 0.83 \begin {gather*} -\frac {12 \, x \log \left (x + 1\right )}{{\left (15 \, x^{2} - 11 \, x\right )} \log \left (x + 1\right ) + 48} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.21, size = 44, normalized size = 1.47 \begin {gather*} \frac {576}{225 \, x^{3} \log \left (x + 1\right ) - 330 \, x^{2} \log \left (x + 1\right ) + 121 \, x \log \left (x + 1\right ) + 720 \, x - 528} - \frac {12}{15 \, x - 11} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 33, normalized size = 1.10
method | result | size |
norman | \(\frac {-\frac {180 \ln \left (x +1\right ) x^{2}}{11}-\frac {576}{11}}{15 \ln \left (x +1\right ) x^{2}-11 \ln \left (x +1\right ) x +48}\) | \(33\) |
risch | \(-\frac {12}{15 x -11}+\frac {576}{\left (15 x -11\right ) \left (15 \ln \left (x +1\right ) x^{2}-11 \ln \left (x +1\right ) x +48\right )}\) | \(40\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 25, normalized size = 0.83 \begin {gather*} -\frac {12 \, x \log \left (x + 1\right )}{{\left (15 \, x^{2} - 11 \, x\right )} \log \left (x + 1\right ) + 48} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {\left (-180\,x^3-180\,x^2\right )\,{\ln \left (x+1\right )}^2+\left (576\,x+576\right )\,\ln \left (x+1\right )+576\,x}{\left (225\,x^5-105\,x^4-209\,x^3+121\,x^2\right )\,{\ln \left (x+1\right )}^2+\left (1440\,x^3+384\,x^2-1056\,x\right )\,\ln \left (x+1\right )+2304\,x+2304} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.30, size = 31, normalized size = 1.03 \begin {gather*} \frac {576}{720 x + \left (225 x^{3} - 330 x^{2} + 121 x\right ) \log {\left (x + 1 \right )} - 528} - \frac {180}{225 x - 165} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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