Optimal. Leaf size=100 \[ -\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a^5} \]
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Rubi [A]
time = 0.43, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps
used = 21, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6153, 6175,
6179, 6181, 3393, 3382, 6115, 5556} \begin {gather*} -\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a^5}-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 3382
Rule 3393
Rule 5556
Rule 6115
Rule 6153
Rule 6175
Rule 6179
Rule 6181
Rubi steps
\begin {align*} \int \frac {x^4}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^3} \, dx &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2 \int \frac {x^3}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{a}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2 \int \frac {x}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)^2} \, dx}{a^3}-\frac {2 \int \frac {x}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2} \, dx}{a^3}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \int \frac {1}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a^4}-\frac {2 \int \frac {1}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx}{a^4}-\frac {2 \int \frac {x^2}{\left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \, dx}{a^2}+\frac {6 \int \frac {x^2}{\left (1-a^2 x^2\right )^3 \tanh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 \text {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {2 \text {Subst}\left (\int \frac {\cosh ^4(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}-\frac {2 \text {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {6 \text {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {2 \text {Subst}\left (\int \left (\frac {1}{2 x}-\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}-\frac {2 \text {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {2 \text {Subst}\left (\int \left (\frac {3}{8 x}+\frac {\cosh (2 x)}{2 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {6 \text {Subst}\left (\int \left (-\frac {1}{8 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}+\frac {3 \text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}-\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}\\ &=-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a^5}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 60, normalized size = 0.60 \begin {gather*} -\frac {\frac {a^3 x^3 \left (a x+4 \tanh ^{-1}(a x)\right )}{\left (-1+a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+2 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )-2 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.73, size = 90, normalized size = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {3}{16 \arctanh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )^{2}}+\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}-\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{a^{5}}\) | \(90\) |
default | \(\frac {-\frac {3}{16 \arctanh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )^{2}}+\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}-\hyperbolicCosineIntegral \left (2 \arctanh \left (a x \right )\right )-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\hyperbolicCosineIntegral \left (4 \arctanh \left (a x \right )\right )}{a^{5}}\) | \(90\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 256 vs.
\(2 (95) = 190\).
time = 0.35, size = 256, normalized size = 2.56 \begin {gather*} -\frac {4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{2 \, {\left (a^{9} x^{4} - 2 \, a^{7} x^{2} + a^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \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}}{a^{6} x^{6} \operatorname {atanh}^{3}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{3}{\left (a x \right )} - \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} -\int \frac {x^4}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (a^2\,x^2-1\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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