Optimal. Leaf size=71 \[ \frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]
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Rubi [A] time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2168, 2158, 2157, 29} \[ -\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 x}{b^3} \]
Antiderivative was successfully verified.
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Rule 29
Rule 2157
Rule 2158
Rule 2168
Rubi steps
\begin {align*} \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))^3} \, dx &=-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}+\frac {3 \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx}{2 b}\\ &=-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (3 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ &=\frac {3 x}{b^3}-\frac {x^3}{2 b \tanh ^{-1}(\tanh (a+b x))^2}-\frac {3 x^2}{2 b^2 \tanh ^{-1}(\tanh (a+b x))}+\frac {3 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right ) \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^4}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 86, normalized size = 1.21 \[ -\frac {3 b^2 x^2 \tanh ^{-1}(\tanh (a+b x))-b x \tanh ^{-1}(\tanh (a+b x))^2 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+11\right )+\tanh ^{-1}(\tanh (a+b x))^3 \left (6 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )+5\right )+b^3 x^3}{2 b^4 \tanh ^{-1}(\tanh (a+b x))^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.56, size = 83, normalized size = 1.17 \[ \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 44, normalized size = 0.62 \[ \frac {x}{b^{3}} - \frac {3 \, a \log \left ({\left | b x + a \right |}\right )}{b^{4}} - \frac {6 \, a^{2} b x + 5 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.15, size = 239, normalized size = 3.37 \[ \frac {x}{b^{3}}+\frac {a^{3}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {3 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}+\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{2 b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )^{2}}-\frac {3 a^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{4} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a}{b^{4}}-\frac {3 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.41, size = 69, normalized size = 0.97 \[ \frac {2 \, b^{3} x^{3} + 4 \, a b^{2} x^{2} - 4 \, a^{2} b x - 5 \, a^{3}}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} - \frac {3 \, a \log \left (b x + a\right )}{b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.47, size = 620, normalized size = 8.73 \[ \frac {x}{b^3}-\frac {x\,\left (3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2-12\,a\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )+12\,a^2\right )-\frac {5\,\left ({\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^3-8\,a^3-6\,a\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+12\,a^2\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )}{4\,b}}{b^3\,{\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2+x\,\left (8\,a\,b^4-4\,b^4\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )\right )+4\,a^2\,b^3+4\,b^5\,x^2-4\,a\,b^3\,\left (2\,a-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}+\frac {\ln \left (\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )\right )\,\left (3\,\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-3\,\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+6\,b\,x\right )}{2\,b^4} \]
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 \[ \int \frac {x^{3}}{\operatorname {atanh}^{3}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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