Optimal. Leaf size=395 \[ \frac {i \text {Li}_4\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{8 b^3}-\frac {i \text {Li}_4\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{8 b^3}+\frac {x \text {Li}_3\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b^2}-\frac {i x^2 \text {Li}_2\left (-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \tan (a+b x)+c) \]
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Rubi [A] time = 0.50, antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6267, 2190, 2531, 6609, 2282, 6589} \[ \frac {x \text {PolyLog}\left (3,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b^2}-\frac {x \text {PolyLog}\left (3,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b^2}+\frac {i \text {PolyLog}\left (4,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{8 b^3}-\frac {i \text {PolyLog}\left (4,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{8 b^3}-\frac {i x^2 \text {PolyLog}\left (2,-\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )}{4 b}+\frac {i x^2 \text {PolyLog}\left (2,-\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1+\frac {(-c+i d+1) e^{2 i a+2 i b x}}{-c-i d+1}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(c-i d+1) e^{2 i a+2 i b x}}{c+i d+1}\right )+\frac {1}{3} x^3 \tanh ^{-1}(d \tan (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 6267
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \tanh ^{-1}(c+d \tan (a+b x)) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{3} (b (i (1-c)-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-c-i d+(1-c+i d) e^{2 i a+2 i b x}} \, dx-\frac {1}{3} (b (i+i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+c+i d+(1+c-i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {1}{2} \int x^2 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right ) \, dx+\frac {1}{2} \int x^2 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {i \int x \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right ) \, dx}{2 b}-\frac {i \int x \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}-\frac {\int \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right ) \, dx}{4 b^2}+\frac {\int \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(1-c+i d) x}{-1+c+i d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {(1+c-i d) x}{1+c+i d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \tanh ^{-1}(c+d \tan (a+b x))+\frac {1}{6} x^3 \log \left (1+\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )-\frac {1}{6} x^3 \log \left (1+\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )-\frac {i x^2 \text {Li}_2\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b}+\frac {x \text {Li}_3\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{4 b^2}+\frac {i \text {Li}_4\left (-\frac {(1-c+i d) e^{2 i a+2 i b x}}{1-c-i d}\right )}{8 b^3}-\frac {i \text {Li}_4\left (-\frac {(1+c-i d) e^{2 i a+2 i b x}}{1+c+i d}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A] time = 0.93, size = 346, normalized size = 0.88 \[ \frac {1}{3} x^3 \tanh ^{-1}(d \tan (a+b x)+c)+\frac {4 b^3 x^3 \log \left (1+\frac {(c-i d-1) e^{2 i (a+b x)}}{c+i d-1}\right )-4 b^3 x^3 \log \left (1+\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )-6 i b^2 x^2 \text {Li}_2\left (\frac {(-c+i d+1) e^{2 i (a+b x)}}{c+i d-1}\right )+6 i b^2 x^2 \text {Li}_2\left (-\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )+6 b x \text {Li}_3\left (\frac {(-c+i d+1) e^{2 i (a+b x)}}{c+i d-1}\right )-6 b x \text {Li}_3\left (-\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )+3 i \text {Li}_4\left (\frac {(-c+i d+1) e^{2 i (a+b x)}}{c+i d-1}\right )-3 i \text {Li}_4\left (-\frac {(c-i d+1) e^{2 i (a+b x)}}{c+i d+1}\right )}{24 b^3} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.75, size = 2161, normalized size = 5.47 \[ \text {result too large to display} \]
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 \[ \int x^{2} \operatorname {artanh}\left (d \tan \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 44.58, size = 6967, normalized size = 17.64 \[ \text {output too large to display} \]
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}{12} \, x^{3} \log \left ({\left (c^{2} + d^{2} + 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + 4 \, {\left (c + 1\right )} d \sin \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} + d^{2} + 2 \, c + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + c^{2} + d^{2} + 2 \, {\left (c^{2} - d^{2} + 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, c + 1\right ) - \frac {1}{12} \, x^{3} \log \left ({\left (c^{2} + d^{2} - 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + 4 \, {\left (c - 1\right )} d \sin \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} + d^{2} - 2 \, c + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + c^{2} + d^{2} + 2 \, {\left (c^{2} - d^{2} - 2 \, c + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 2 \, c + 1\right ) - 4 \, b d \int -\frac {2 \, {\left (c^{2} + d^{2} - 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c d x^{3} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (c^{2} + d^{2} - 1\right )} x^{3} \sin \left (2 \, b x + 2 \, a\right )^{2} + {\left (c^{2} - d^{2} - 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right ) - {\left (2 \, c d x^{3} \sin \left (2 \, b x + 2 \, a\right ) - {\left (c^{2} - d^{2} - 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (2 \, c d x^{3} \cos \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} - d^{2} - 1\right )} x^{3} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right )}{3 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, c^{2} + 2 \, {\left (c^{4} + d^{4} - 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} - 2 \, c^{2} + 2 \, {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \, {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (2 \, c d^{3} - 2 \, {\left (c^{3} - c\right )} d - 2 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (c^{4} - d^{4} - 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) + 8 \, {\left (c d^{3} + {\left (c^{3} - c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1\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 \[ \int x^2\,\mathrm {atanh}\left (c+d\,\mathrm {tan}\left (a+b\,x\right )\right ) \,d x \]
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 \[ \int x^{2} \operatorname {atanh}{\left (c + d \tan {\left (a + b x \right )} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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