Optimal. Leaf size=98 \[ \frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5} \]
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Rubi [A]
time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2199, 2190,
2189, 2188, 29} \begin {gather*} \frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 x^3}{3 b^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 2188
Rule 2189
Rule 2190
Rule 2199
Rubi steps
\begin {align*} \int \frac {x^4}{\tanh ^{-1}(\tanh (a+b x))^2} \, dx &=-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 \int \frac {x^3}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b}\\ &=\frac {4 x^3}{3 b^2}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )\right ) \int \frac {x^2}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^2}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2\right ) \int \frac {x}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^3}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \int \frac {1}{\tanh ^{-1}(\tanh (a+b x))} \, dx}{b^4}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}-\frac {\left (4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ &=\frac {4 x^3}{3 b^2}+\frac {2 x^2 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {4 x \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {x^4}{b \tanh ^{-1}(\tanh (a+b x))}+\frac {4 \left (b x-\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5}\\ \end {align*}
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Mathematica [A]
time = 0.06, size = 106, normalized size = 1.08 \begin {gather*} \frac {x^3}{3 b^2}-\frac {x^2 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )}{b^3}+\frac {3 x \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^2}{b^4}-\frac {\left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^4}{b^5 \tanh ^{-1}(\tanh (a+b x))}-\frac {4 \left (-b x+\tanh ^{-1}(\tanh (a+b x))\right )^3 \log \left (\tanh ^{-1}(\tanh (a+b x))\right )}{b^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs.
\(2(96)=192\).
time = 2.87, size = 350, normalized size = 3.57
method | result | size |
default | \(\frac {x^{3}}{3 b^{2}}-\frac {a \,x^{2}}{b^{3}}-\frac {x^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{3}}+\frac {3 x \,a^{2}}{b^{4}}+\frac {6 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right ) x}{b^{4}}+\frac {3 \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2} x}{b^{4}}-\frac {a^{4}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {4 a^{3} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {6 a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {4 a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {\left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{4}}{b^{5} \arctanh \left (\tanh \left (b x +a \right )\right )}-\frac {4 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{3}}{b^{5}}-\frac {12 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a^{2} \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )}{b^{5}}-\frac {12 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) a \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{2}}{b^{5}}-\frac {4 \ln \left (\arctanh \left (\tanh \left (b x +a \right )\right )\right ) \left (\arctanh \left (\tanh \left (b x +a \right )\right )-b x -a \right )^{3}}{b^{5}}\) | \(350\) |
risch | \(\text {Expression too large to display}\) | \(51088\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.70, size = 70, normalized size = 0.71 \begin {gather*} \frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4}}{3 \, {\left (b^{6} x + a b^{5}\right )}} - \frac {4 \, a^{3} \log \left (b x + a\right )}{b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.32, size = 73, normalized size = 0.74 \begin {gather*} \frac {b^{4} x^{4} - 2 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 9 \, a^{3} b x - 3 \, a^{4} - 12 \, {\left (a^{3} b x + a^{4}\right )} \log \left (b x + a\right )}{3 \, {\left (b^{6} x + a b^{5}\right )}} \end {gather*}
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 \begin {gather*} \int \frac {x^{4}}{\operatorname {atanh}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.39, size = 62, normalized size = 0.63 \begin {gather*} -\frac {4 \, a^{3} \log \left ({\left | b x + a \right |}\right )}{b^{5}} - \frac {a^{4}}{{\left (b x + a\right )} b^{5}} + \frac {b^{4} x^{3} - 3 \, a b^{3} x^{2} + 9 \, a^{2} b^{2} x}{3 \, b^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.18, size = 669, normalized size = 6.83 \begin {gather*} \frac {x^3}{3\,b^2}-\frac {{\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 )}^4+24\,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 )}^2+16\,a^4-8\,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 )}^3-32\,a^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\,b\,\left (8\,a\,b^4+8\,b^5\,x-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 )}+\frac {x^2\,\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}{2\,b^3}+\frac {3\,x\,{\left (\ln \left (\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}+1}\right )+2\,b\,x\right )}^2}{4\,b^4}+\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 ({\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 )}{2\,b^5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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