Optimal. Leaf size=113 \[ \frac {i \text {Li}_3\left (-i c e^{2 a+2 b x}\right )}{8 b^2}-\frac {i x \text {Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))+\frac {1}{6} i b x^3 \]
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Rubi [A] time = 0.20, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {5196, 2184, 2190, 2531, 2282, 6589} \[ \frac {i \text {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2}-\frac {i x \text {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))+\frac {1}{6} i b x^3 \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 5196
Rule 6589
Rubi steps
\begin {align*} \int x \cot ^{-1}(c+(i+c) \tanh (a+b x)) \, dx &=\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))+\frac {1}{2} b \int \frac {x^2}{-i+c e^{2 a+2 b x}} \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{2} (i b c) \int \frac {e^{2 a+2 b x} x^2}{-i+c e^{2 a+2 b x}} \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {1}{2} i \int x \log \left (1+i c e^{2 a+2 b x}\right ) \, dx\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac {i x \text {Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac {i \int \text {Li}_2\left (-i c e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac {i x \text {Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac {i \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac {1}{6} i b x^3+\frac {1}{2} x^2 \cot ^{-1}(c+(i+c) \tanh (a+b x))-\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )-\frac {i x \text {Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}+\frac {i \text {Li}_3\left (-i c e^{2 a+2 b x}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 102, normalized size = 0.90 \[ \frac {1}{2} x^2 \cot ^{-1}(c+(c+i) \tanh (a+b x))-\frac {i \left (2 b^2 x^2 \log \left (1-\frac {i e^{-2 (a+b x)}}{c}\right )-2 b x \text {Li}_2\left (\frac {i e^{-2 (a+b x)}}{c}\right )-\text {Li}_3\left (\frac {i e^{-2 (a+b x)}}{c}\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.43, size = 244, normalized size = 2.16 \[ \frac {2 i \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (\frac {{\left (c e^{\left (2 \, b x + 2 \, a\right )} - i\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{c + i}\right ) + 2 i \, a^{3} - 6 i \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 6 i \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) + {\left (-3 i \, b^{2} x^{2} + 3 i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{2}} \]
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 \[ \int x \operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 5.46, size = 1513, normalized size = 13.39 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.02, size = 107, normalized size = 0.95 \[ -{\left (\frac {2 \, x^{3}}{3 i \, c - 3} - \frac {2 \, b^{2} x^{2} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{3} {\left (-i \, c + 1\right )}}\right )} b {\left (c + i\right )} + \frac {1}{2} \, x^{2} \operatorname {arccot}\left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {acot}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c+1{}\mathrm {i}\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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