Optimal. Leaf size=207 \[ \frac {3}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b^2 \text {Li}_3\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b \text {Li}_2\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {3}{4} b \text {Li}_2\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {3}{8} b^3 \text {Li}_4\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {Li}_4\left (\frac {2}{1-c x^2}-1\right ) \]
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Rubi [A] time = 0.56, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6095, 5914, 6052, 5948, 6058, 6062, 6610} \[ \frac {3}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b^2 \text {PolyLog}\left (3,\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {3}{4} b \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {3}{4} b \text {PolyLog}\left (2,\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3}{8} b^3 \text {PolyLog}\left (4,1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {PolyLog}\left (4,\frac {2}{1-c x^2}-1\right )+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \]
Antiderivative was successfully verified.
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Rule 5914
Rule 5948
Rule 6052
Rule 6058
Rule 6062
Rule 6095
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-(3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} (3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} (3 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{2} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )-\frac {1}{4} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )+\frac {1}{4} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{4} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {3}{4} b^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )-\frac {3}{8} b^3 \text {Li}_4\left (1-\frac {2}{1-c x^2}\right )+\frac {3}{8} b^3 \text {Li}_4\left (-1+\frac {2}{1-c x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.21, size = 211, normalized size = 1.02 \[ \frac {3}{8} b \left (2 \text {Li}_2\left (\frac {c x^2+1}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-2 \text {Li}_2\left (\frac {c x^2+1}{c x^2-1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+b \left (-2 \text {Li}_3\left (\frac {c x^2+1}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+2 \text {Li}_3\left (\frac {c x^2+1}{c x^2-1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+b \left (\text {Li}_4\left (\frac {c x^2+1}{1-c x^2}\right )-\text {Li}_4\left (\frac {c x^2+1}{c x^2-1}\right )\right )\right )\right )+\tanh ^{-1}\left (\frac {2}{c x^2-1}+1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3 \]
Antiderivative was successfully verified.
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fricas [F] time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (c x^{2}\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (c x^{2}\right ) + a^{3}}{x}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.21, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x}\, dx \]
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 \[ a^{3} \log \relax (x) + \int \frac {b^{3} {\left (\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )\right )}^{3}}{8 \, x} + \frac {3 \, a b^{2} {\left (\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )\right )}^{2}}{4 \, x} + \frac {3 \, a^{2} b {\left (\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )\right )}}{2 \, x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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