Optimal. Leaf size=137 \[ -\frac {\text {Li}_4\left ((1-d) e^{2 a+2 b x}\right )}{8 b^3}+\frac {x \text {Li}_3\left ((1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )+\frac {1}{3} x^3 \coth ^{-1}(d (-\coth (a+b x))-d+1)+\frac {b x^4}{12} \]
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Rubi [A] time = 0.27, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {6242, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac {x \text {PolyLog}\left (3,(1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\text {PolyLog}\left (4,(1-d) e^{2 a+2 b x}\right )}{8 b^3}-\frac {x^2 \text {PolyLog}\left (2,(1-d) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )+\frac {1}{3} x^3 \coth ^{-1}(d (-\coth (a+b x))-d+1)+\frac {b x^4}{12} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 6242
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int x^2 \coth ^{-1}(1-d-d \coth (a+b x)) \, dx &=\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))+\frac {1}{3} b \int \frac {x^3}{1+(-1+d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))+\frac {1}{3} (b (1-d)) \int \frac {e^{2 a+2 b x} x^3}{1+(-1+d) e^{2 a+2 b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )+\frac {1}{2} \int x^2 \log \left (1+(-1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {\int x \text {Li}_2\left (-(-1+d) e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left ((1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\int \text {Li}_3\left ((1-d) e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left ((1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3((1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \coth ^{-1}(1-d-d \coth (a+b x))-\frac {1}{6} x^3 \log \left (1-(1-d) e^{2 a+2 b x}\right )-\frac {x^2 \text {Li}_2\left ((1-d) e^{2 a+2 b x}\right )}{4 b}+\frac {x \text {Li}_3\left ((1-d) e^{2 a+2 b x}\right )}{4 b^2}-\frac {\text {Li}_4\left ((1-d) e^{2 a+2 b x}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 121, normalized size = 0.88 \[ \frac {1}{24} \left (\frac {3 \text {Li}_4\left (-\frac {e^{-2 (a+b x)}}{d-1}\right )}{b^3}+\frac {6 x \text {Li}_3\left (-\frac {e^{-2 (a+b x)}}{d-1}\right )}{b^2}+\frac {6 x^2 \text {Li}_2\left (-\frac {e^{-2 (a+b x)}}{d-1}\right )}{b}-4 x^3 \log \left (\frac {e^{-2 (a+b x)}}{d-1}+1\right )+8 x^3 \coth ^{-1}(d (-\coth (a+b x))-d+1)\right ) \]
Antiderivative was successfully verified.
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fricas [C] time = 0.53, size = 381, normalized size = 2.78 \[ \frac {b^{4} x^{4} - 2 \, b^{3} x^{3} \log \left (\frac {d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + {\left (d - 2\right )} \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) + \sqrt {-4 \, d + 4}\right ) + 2 \, a^{3} \log \left (2 \, {\left (d - 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d - 1\right )} \sinh \left (b x + a\right ) - \sqrt {-4 \, d + 4}\right ) + 12 \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 \, d + 4} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}} \]
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 \[ \int x^{2} \operatorname {arcoth}\left (-d \coth \left (b x + a\right ) - d + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 5.46, size = 1771, normalized size = 12.93 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.10, size = 125, normalized size = 0.91 \[ -\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (d \coth \left (b x + a\right ) + d - 1\right ) + \frac {1}{36} \, {\left (\frac {3 \, x^{4}}{d} - \frac {2 \, {\left (4 \, b^{3} x^{3} \log \left ({\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-{\left (d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d \]
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 \[ \int -x^2\,\mathrm {acoth}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )-1\right ) \,d x \]
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 \[ \int x^{2} \operatorname {acoth}{\left (- d \coth {\left (a + b x \right )} - d + 1 \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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