Optimal. Leaf size=197 \[ -\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a^5}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^5}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac {5 \sqrt {1-a^2 x^2}}{8 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4} \]
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Rubi [A] time = 0.22, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6016, 266, 43, 261, 5950} \[ -\frac {3 i \text {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a^5}+\frac {3 i \text {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{8 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac {5 \sqrt {1-a^2 x^2}}{8 a^5}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \tanh ^{-1}(a x)}{4 a^5} \]
Antiderivative was successfully verified.
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Rule 43
Rule 261
Rule 266
Rule 5950
Rule 6016
Rubi steps
\begin {align*} \int \frac {x^4 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx &=-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}+\frac {3 \int \frac {x^2 \tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{4 a^2}+\frac {\int \frac {x^3}{\sqrt {1-a^2 x^2}} \, dx}{4 a}\\ &=-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}+\frac {3 \int \frac {\tanh ^{-1}(a x)}{\sqrt {1-a^2 x^2}} \, dx}{8 a^4}+\frac {3 \int \frac {x}{\sqrt {1-a^2 x^2}} \, dx}{8 a^3}+\frac {\operatorname {Subst}\left (\int \frac {x}{\sqrt {1-a^2 x}} \, dx,x,x^2\right )}{8 a}\\ &=-\frac {3 \sqrt {1-a^2 x^2}}{8 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^5}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5}+\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a^2 \sqrt {1-a^2 x}}-\frac {\sqrt {1-a^2 x}}{a^2}\right ) \, dx,x,x^2\right )}{8 a}\\ &=-\frac {5 \sqrt {1-a^2 x^2}}{8 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{4 a^2}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \tanh ^{-1}(a x)}{4 a^5}-\frac {3 i \text {Li}_2\left (-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5}+\frac {3 i \text {Li}_2\left (\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5}\\ \end {align*}
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Mathematica [A] time = 0.51, size = 160, normalized size = 0.81 \[ \frac {\sqrt {1-a^2 x^2} \left (-\frac {9 i \left (\text {Li}_2\left (-i e^{-\tanh ^{-1}(a x)}\right )-\text {Li}_2\left (i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-2 a^2 x^2-6 a x \left (a^2 x^2-1\right ) \tanh ^{-1}(a x)-\frac {9 i \tanh ^{-1}(a x) \left (\log \left (1-i e^{-\tanh ^{-1}(a x)}\right )-\log \left (1+i e^{-\tanh ^{-1}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-15 a x \tanh ^{-1}(a x)-13\right )}{24 a^5} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} x^{4} \operatorname {artanh}\left (a x\right )}{a^{2} x^{2} - 1}, x\right ) \]
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 \[ \int \frac {x^{4} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.47, size = 175, normalized size = 0.89 \[ -\frac {\sqrt {-\left (a x -1\right ) \left (a x +1\right )}\, \left (6 a^{3} x^{3} \arctanh \left (a x \right )+2 a^{2} x^{2}+9 a x \arctanh \left (a x \right )+13\right )}{24 a^{5}}-\frac {3 i \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{8 a^{5}}+\frac {3 i \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right ) \arctanh \left (a x \right )}{8 a^{5}}-\frac {3 i \dilog \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}}+\frac {3 i \dilog \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}} \]
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 \[ \int \frac {x^{4} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}}\,{d x} \]
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 \[ \int \frac {x^4\,\mathrm {atanh}\left (a\,x\right )}{\sqrt {1-a^2\,x^2}} \,d x \]
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 \[ \int \frac {x^{4} \operatorname {atanh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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