Optimal. Leaf size=265 \[ \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (c x+1)^2}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (c x+1)}-\frac {b^2}{16 c^3 d^3 (c x+1)^2}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3} \]
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Rubi [A] time = 0.54, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5940, 5928, 5926, 627, 44, 207, 5948, 5918, 6056, 6610} \[ \frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^3 d^3}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (c x+1)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (c x+1)^2}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {13 b^2}{16 c^3 d^3 (c x+1)}-\frac {b^2}{16 c^3 d^3 (c x+1)^2}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3} \]
Antiderivative was successfully verified.
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Rule 44
Rule 207
Rule 627
Rule 5918
Rule 5926
Rule 5928
Rule 5940
Rule 5948
Rule 6056
Rule 6610
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\int \left (\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)^3}-\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{c^2 d^3}+\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{c^2 d^3}-\frac {2 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{c^2 d^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}+\frac {(2 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^3}-\frac {(4 b) \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c^2 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c^2 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c^2 d^3}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^2 d^3}+\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{c^2 d^3}-\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 c^2 d^3}+\frac {b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 c^2 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^2 d^3}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}-\frac {\left (2 b^2\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=-\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 c^2 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^2 d^3}+\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {b^2}{16 c^3 d^3 (1+c x)^2}+\frac {13 b^2}{16 c^3 d^3 (1+c x)}-\frac {13 b^2 \tanh ^{-1}(c x)}{16 c^3 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)^2}+\frac {7 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^3 d^3 (1+c x)}-\frac {7 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3 (1+c x)^2}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^3 d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^3 d^3}\\ \end {align*}
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Mathematica [A] time = 1.40, size = 310, normalized size = 1.17 \[ \frac {\frac {32 a^2}{c x+1}-\frac {8 a^2}{(c x+1)^2}+16 a^2 \log (c x+1)+a b \left (16 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )-12 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+12 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (-8 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+6 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )+16 b^2 \left (\tanh ^{-1}(c x) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )+\frac {1}{64} \left (\sinh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (2 \tanh ^{-1}(c x)\right )\right ) \left (-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )-12\right )+8 \tanh ^{-1}(c x)^2 \left (\left (8 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-1\right ) \sinh \left (2 \tanh ^{-1}(c x)\right )+\left (8 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+1\right ) \cosh \left (2 \tanh ^{-1}(c x)\right )-6\right )-24\right )\right )}{16 c^3 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {artanh}\left (c x\right ) + a^{2} x^{2}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.77, size = 1241, normalized size = 4.68 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {4 \, c x + 3}{c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}} + \frac {2 \, \log \left (c x + 1\right )}{c^{3} d^{3}}\right )} + \frac {{\left (4 \, b^{2} c x + 3 \, b^{2} + 2 \, {\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, {\left (c^{5} d^{3} x^{2} + 2 \, c^{4} d^{3} x + c^{3} d^{3}\right )}} - \int -\frac {{\left (b^{2} c^{3} x^{3} - b^{2} c^{2} x^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c^{3} x^{3} - a b c^{2} x^{2}\right )} \log \left (c x + 1\right ) - {\left (4 \, a b c^{3} x^{3} + 7 \, b^{2} c x - 4 \, {\left (a b c^{2} - b^{2} c^{2}\right )} x^{2} + 3 \, b^{2} + 2 \, {\left (2 \, b^{2} c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{6} d^{3} x^{4} + 2 \, c^{5} d^{3} x^{3} - 2 \, c^{3} d^{3} x - c^{2} d^{3}\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^2\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{2}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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