Optimal. Leaf size=125 \[ -\frac {3}{2} b^2 c \text {Li}_2\left (\frac {2}{c x^2+1}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{2 x^2}+\frac {3}{2} b c \log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {3}{4} b^3 c \text {Li}_3\left (\frac {2}{c x^2+1}-1\right ) \]
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Rubi [F] time = 0.78, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^3}{x^3} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{8 x^3}+\frac {3 b \left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{8 x^3}-\frac {3 b^2 \left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{8 x^3}+\frac {b^3 \log ^3\left (1+c x^2\right )}{8 x^3}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^3}{x^3} \, dx+\frac {1}{8} (3 b) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right )^2 \log \left (1+c x^2\right )}{x^3} \, dx-\frac {1}{8} \left (3 b^2\right ) \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log ^2\left (1+c x^2\right )}{x^3} \, dx+\frac {1}{8} b^3 \int \frac {\log ^3\left (1+c x^2\right )}{x^3} \, dx\\ &=\frac {1}{16} \operatorname {Subst}\left (\int \frac {(2 a-b \log (1-c x))^3}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} b^3 \operatorname {Subst}\left (\int \frac {\log ^3(1+c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{16} (3 b c) \operatorname {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x} \, dx,x,x^2\right )+\frac {1}{16} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log ^2(1+c x)}{x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\log (c x) (2 a-b \log (1-c x))}{1-c x} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c^2\right ) \operatorname {Subst}\left (\int \frac {\log (-c x) \log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{8} \left (3 b^2 c\right ) \operatorname {Subst}\left (\int \frac {(2 a-b \log (x)) \log \left (c \left (\frac {1}{c}-\frac {x}{c}\right )\right )}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\log (x) \log \left (-c \left (-\frac {1}{c}+\frac {x}{c}\right )\right )}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1-c x^2\right )-\frac {1}{8} \left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,1+c x^2\right )\\ &=\frac {3}{16} b c \log \left (c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^3}{16 x^2}+\frac {3}{16} b^3 c \log \left (-c x^2\right ) \log ^2\left (1+c x^2\right )-\frac {b^3 \left (1+c x^2\right ) \log ^3\left (1+c x^2\right )}{16 x^2}-\frac {3}{8} b^2 c \left (2 a-b \log \left (1-c x^2\right )\right ) \text {Li}_2\left (1-c x^2\right )+\frac {3}{8} b^3 c \log \left (1+c x^2\right ) \text {Li}_2\left (1+c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1-c x^2\right )-\frac {3}{8} b^3 c \text {Li}_3\left (1+c x^2\right )+\frac {1}{16} (3 b) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x))^2 \log (1+c x)}{x^2} \, dx,x,x^2\right )-\frac {1}{16} \left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log ^2(1+c x)}{x^2} \, dx,x,x^2\right )\\ \end {align*}
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Mathematica [C] time = 0.43, size = 222, normalized size = 1.78 \[ \frac {1}{4} \left (-\frac {2 a^3}{x^2}-3 a^2 b c \log \left (1-c^2 x^4\right )-\frac {6 a^2 b \tanh ^{-1}\left (c x^2\right )}{x^2}+12 a^2 b c \log (x)+6 a b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (\left (1-\frac {1}{c x^2}\right ) \tanh ^{-1}\left (c x^2\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text {Li}_2\left (e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )+2 b^3 c \left (3 \tanh ^{-1}\left (c x^2\right ) \text {Li}_2\left (e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac {3}{2} \text {Li}_3\left (e^{2 \tanh ^{-1}\left (c x^2\right )}\right )-\frac {\tanh ^{-1}\left (c x^2\right )^3}{c x^2}-\tanh ^{-1}\left (c x^2\right )^3+3 \tanh ^{-1}\left (c x^2\right )^2 \log \left (1-e^{2 \tanh ^{-1}\left (c x^2\right )}\right )+\frac {i \pi ^3}{8}\right )\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.29, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (c x^{2}\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (c x^{2}\right ) + a^{3}}{x^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{3}}{x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.29, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{3}}{x^{3}}\, dx \]
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 \[ -\frac {3}{4} \, {\left (c {\left (\log \left (c^{2} x^{4} - 1\right ) - \log \left (x^{4}\right )\right )} + \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{2}}\right )} a^{2} b - \frac {a^{3}}{2 \, x^{2}} - \frac {{\left (b^{3} c x^{2} - b^{3}\right )} \log \left (-c x^{2} + 1\right )^{3} + 3 \, {\left (2 \, a b^{2} + {\left (b^{3} c x^{2} + b^{3}\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )^{2}}{16 \, x^{2}} - \int -\frac {{\left (b^{3} c x^{2} - b^{3}\right )} \log \left (c x^{2} + 1\right )^{3} + 6 \, {\left (a b^{2} c x^{2} - a b^{2}\right )} \log \left (c x^{2} + 1\right )^{2} + 3 \, {\left (4 \, a b^{2} c x^{2} - {\left (b^{3} c x^{2} - b^{3}\right )} \log \left (c x^{2} + 1\right )^{2} + 2 \, {\left (b^{3} c^{2} x^{4} + 2 \, a b^{2} - {\left (2 \, a b^{2} c - b^{3} c\right )} x^{2}\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )}{8 \, {\left (c x^{5} - x^{3}\right )}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^3}{x^3} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{3}}{x^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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