Optimal. Leaf size=100 \[ -\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a^5}-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]
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Rubi [A] time = 0.59, 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 = {6006, 6028, 6032, 6034, 3312, 3301, 5968, 5448} \[ -\frac {\text {Chi}\left (2 \tanh ^{-1}(a x)\right )}{a^5}+\frac {\text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{a^5}-\frac {x^4}{2 a \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)^2}+\frac {2 x}{a^4 \left (1-a^2 x^2\right ) \tanh ^{-1}(a x)}-\frac {2 x}{a^4 \left (1-a^2 x^2\right )^2 \tanh ^{-1}(a x)} \]
Antiderivative was successfully verified.
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Rule 3301
Rule 3312
Rule 5448
Rule 5968
Rule 6006
Rule 6028
Rule 6032
Rule 6034
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 \operatorname {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {2 \operatorname {Subst}\left (\int \frac {\cosh ^4(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}-\frac {2 \operatorname {Subst}\left (\int \frac {\sinh ^2(x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{a^5}+\frac {6 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 {\operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}+\frac {3 \operatorname {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\tanh ^{-1}(a x)\right )}{4 a^5}-\frac {\operatorname {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 = 0.22, size = 60, normalized size = 0.60 \[ -\frac {\frac {a^3 x^3 \left (a x+4 \tanh ^{-1}(a x)\right )}{\left (a^2 x^2-1\right )^2 \tanh ^{-1}(a x)^2}+2 \text {Chi}\left (2 \tanh ^{-1}(a x)\right )-2 \text {Chi}\left (4 \tanh ^{-1}(a x)\right )}{2 a^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.67, size = 256, normalized size = 2.56 \[ -\frac {4 \, a^{4} x^{4} + 8 \, a^{3} x^{3} \log \left (-\frac {a x + 1}{a x - 1}\right ) - {\left ({\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} + 2 \, a x + 1}{a^{2} x^{2} - 2 \, a x + 1}\right ) + {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (\frac {a^{2} x^{2} - 2 \, a x + 1}{a^{2} x^{2} + 2 \, a x + 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x + 1}{a x - 1}\right ) - {\left (a^{4} x^{4} - 2 \, a^{2} x^{2} + 1\right )} \operatorname {log\_integral}\left (-\frac {a x - 1}{a x + 1}\right )\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}}{2 \, {\left (a^{9} x^{4} - 2 \, a^{7} x^{2} + a^{5}\right )} \log \left (-\frac {a x + 1}{a x - 1}\right )^{2}} \]
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 \[ \int -\frac {x^{4}}{{\left (a^{2} x^{2} - 1\right )}^{3} \operatorname {artanh}\left (a x\right )^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 90, normalized size = 0.90 \[ \frac {-\frac {3}{16 \arctanh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )^{2}}+\frac {\sinh \left (2 \arctanh \left (a x \right )\right )}{2 \arctanh \left (a x \right )}-\Chi \left (2 \arctanh \left (a x \right )\right )-\frac {\cosh \left (4 \arctanh \left (a x \right )\right )}{16 \arctanh \left (a x \right )^{2}}-\frac {\sinh \left (4 \arctanh \left (a x \right )\right )}{4 \arctanh \left (a x \right )}+\Chi \left (4 \arctanh \left (a x \right )\right )}{a^{5}} \]
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 \[ -\frac {2 \, {\left (a x^{4} + 2 \, x^{3} \log \left (a x + 1\right ) - 2 \, x^{3} \log \left (-a x + 1\right )\right )}}{{\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right )^{2} - 2 \, {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right ) + {\left (a^{6} x^{4} - 2 \, a^{4} x^{2} + a^{2}\right )} \log \left (-a x + 1\right )^{2}} + \int -\frac {4 \, {\left (a^{2} x^{4} + 3 \, x^{2}\right )}}{{\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (a x + 1\right ) - {\left (a^{8} x^{6} - 3 \, a^{6} x^{4} + 3 \, a^{4} x^{2} - a^{2}\right )} \log \left (-a x + 1\right )}\,{d x} \]
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 \[ -\int \frac {x^4}{{\mathrm {atanh}\left (a\,x\right )}^3\,{\left (a^2\,x^2-1\right )}^3} \,d x \]
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 \[ - \int \frac {x^{4}}{a^{6} x^{6} \operatorname {atanh}^{3}{\left (a x \right )} - 3 a^{4} x^{4} \operatorname {atanh}^{3}{\left (a x \right )} + 3 a^{2} x^{2} \operatorname {atanh}^{3}{\left (a x \right )} - \operatorname {atanh}^{3}{\left (a x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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