Optimal. Leaf size=100 \[ \frac {\text {Li}_3\left ((d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac {x \text {Li}_2\left ((d+1) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{4} x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac {1}{2} x^2 \tanh ^{-1}(d \coth (a+b x)+d+1)+\frac {b x^3}{6} \]
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Rubi [A] time = 0.23, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6241, 2184, 2190, 2531, 2282, 6589} \[ \frac {\text {PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac {x \text {PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac {1}{4} x^2 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac {1}{2} x^2 \tanh ^{-1}(d \coth (a+b x)+d+1)+\frac {b x^3}{6} \]
Antiderivative was successfully verified.
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Rule 2184
Rule 2190
Rule 2282
Rule 2531
Rule 6241
Rule 6589
Rubi steps
\begin {align*} \int x \tanh ^{-1}(1+d+d \coth (a+b x)) \, dx &=\frac {1}{2} x^2 \tanh ^{-1}(1+d+d \coth (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 \tanh ^{-1}(1+d+d \coth (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 \tanh ^{-1}(1+d+d \coth (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 \tanh ^{-1}(1+d+d \coth (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 \tanh ^{-1}(1+d+d \coth (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 \tanh ^{-1}(1+d+d \coth (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 = 5.22, size = 90, normalized size = 0.90 \[ \frac {2 b^2 x^2 \left (2 \tanh ^{-1}(d \coth (a+b x)+d+1)-\log \left (1-\frac {e^{-2 (a+b x)}}{d+1}\right )\right )+2 b x \text {Li}_2\left (\frac {e^{-2 (a+b x)}}{d+1}\right )+\text {Li}_3\left (\frac {e^{-2 (a+b x)}}{d+1}\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [C] time = 0.58, size = 306, normalized size = 3.06 \[ \frac {2 \, b^{3} x^{3} + 3 \, b^{2} x^{2} \log \left (-\frac {d \cosh \left (b x + a\right ) + {\left (d + 2\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_2\left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b x {\rm Li}_2\left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt {d + 1}\right ) - 3 \, a^{2} \log \left (2 \, {\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \, {\left (d + 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt {d + 1}\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 3 \, {\left (b^{2} x^{2} - a^{2}\right )} \log \left (-\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 6 \, {\rm polylog}\left (3, \sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 6 \, {\rm polylog}\left (3, -\sqrt {d + 1} {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{2}} \]
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 \[ \int x \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 4.35, size = 1603, normalized size = 16.03 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.09, size = 100, normalized size = 1.00 \[ \frac {1}{24} \, {\left (\frac {4 \, x^{3}}{d} - \frac {3 \, {\left (2 \, b^{2} x^{2} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{3} d}\right )} b d + \frac {1}{2} \, x^{2} \operatorname {artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) \]
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 \[ \int x\,\mathrm {atanh}\left (d+d\,\mathrm {coth}\left (a+b\,x\right )+1\right ) \,d x \]
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 \[ \int x \operatorname {atanh}{\left (d \coth {\left (a + b x \right )} + d + 1 \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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