Optimal. Leaf size=110 \[ \frac {\text {Li}_3\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{8 b^2}-\frac {x \text {Li}_2\left (-\left ((1-d) e^{2 a+2 b x}\right )\right )}{4 b}-\frac {1}{4} x^2 \log \left ((1-d) e^{2 a+2 b x}+1\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^3}{6} \]
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Rubi [A] time = 0.24, antiderivative size = 110, 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 = {6240, 2184, 2190, 2531, 2282, 6589} \[ \frac {\text {PolyLog}\left (3,-(1-d) e^{2 a+2 b x}\right )}{8 b^2}-\frac {x \text {PolyLog}\left (2,-(1-d) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{4} x^2 \log \left ((1-d) e^{2 a+2 b x}+1\right )+\frac {1}{2} x^2 \coth ^{-1}(d (-\tanh (a+b x))-d+1)+\frac {b x^3}{6} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 6240
Rule 6589
Rubi steps
\begin {align*} \int x \coth ^{-1}(1-d-d \tanh (a+b x)) \, dx &=\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))+\frac {1}{2} b \int \frac {x^2}{1+(1-d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{2} (b (1-d)) \int \frac {e^{2 a+2 b x} x^2}{1+(1-d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1-d) e^{2 a+2 b x}\right )+\frac {1}{2} \int x \log \left (1+(1-d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x \text {Li}_2\left (-(1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {\int \text {Li}_2\left (-(1-d) e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x \text {Li}_2\left (-(1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2((-1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac {b x^3}{6}+\frac {1}{2} x^2 \coth ^{-1}(1-d-d \tanh (a+b x))-\frac {1}{4} x^2 \log \left (1+(1-d) e^{2 a+2 b x}\right )-\frac {x \text {Li}_2\left (-(1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {\text {Li}_3\left (-(1-d) e^{2 a+2 b x}\right )}{8 b^2}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 93, normalized size = 0.85 \[ \frac {2 b^2 x^2 \left (2 \coth ^{-1}(d (-\tanh (a+b x))-d+1)-\log \left (1-\frac {e^{-2 (a+b x)}}{d-1}\right )\right )+2 b x \text {Li}_2\left (\frac {e^{-2 (a+b x)}}{d-1}\right )+\text {Li}_3\left (\frac {e^{-2 (a+b x)}}{d-1}\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 1.01, size = 305, normalized size = 2.77 \[ \frac {2 \, b^{3} x^{3} - 3 \, b^{2} x^{2} \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{{\left (d - 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d - 1}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d - 1}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\sqrt {d - 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \]
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 \[ \int x \operatorname {arcoth}\left (-d \tanh \left (b x + a\right ) - d + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 4.90, size = 1664, normalized size = 15.13 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.10, size = 100, normalized size = 0.91 \[ \frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d - \frac {1}{2} \, x^{2} \operatorname {arcoth}\left (d \tanh \left (b x + a\right ) + d - 1\right ) \]
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 \[ \int -x\,\mathrm {acoth}\left (d+d\,\mathrm {tanh}\left (a+b\,x\right )-1\right ) \,d x \]
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 \[ \int x \operatorname {acoth}{\left (- d \tanh {\left (a + b x \right )} - d + 1 \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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