Optimal. Leaf size=94 \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{2 c}-\frac {b \log \left (\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{c}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )}{2 c} \]
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Rubi [B] time = 0.51, antiderivative size = 207, normalized size of antiderivative = 2.20, number of steps used = 28, number of rules used = 12, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6099, 2454, 2389, 2296, 2295, 6715, 2430, 43, 2416, 2394, 2393, 2391} \[ -\frac {b^2 \text {PolyLog}\left (2,\frac {1}{2} \left (1-c x^2\right )\right )}{4 c}+\frac {b^2 \text {PolyLog}\left (2,\frac {1}{2} \left (c x^2+1\right )\right )}{4 c}+\frac {b \log \left (\frac {1}{2} \left (c x^2+1\right )\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{4 c}+\frac {1}{4} b x^2 \log \left (c x^2+1\right ) \left (2 a-b \log \left (1-c x^2\right )\right )-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}+\frac {b^2 \left (c x^2+1\right ) \log ^2\left (c x^2+1\right )}{8 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (c x^2+1\right )}{4 c} \]
Warning: Unable to verify antiderivative.
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Rule 43
Rule 2295
Rule 2296
Rule 2389
Rule 2391
Rule 2393
Rule 2394
Rule 2416
Rule 2430
Rule 2454
Rule 6099
Rule 6715
Rubi steps
\begin {align*} \int x \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac {1}{4} x \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac {1}{2} b x \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {1}{4} b^2 x \log ^2\left (1+c x^2\right )\right ) \, dx\\ &=\frac {1}{4} \int x \left (2 a-b \log \left (1-c x^2\right )\right )^2 \, dx-\frac {1}{2} b \int x \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right ) \, dx+\frac {1}{4} b^2 \int x \log ^2\left (1+c x^2\right ) \, dx\\ &=\frac {1}{8} \operatorname {Subst}\left (\int (2 a-b \log (1-c x))^2 \, dx,x,x^2\right )-\frac {1}{4} b \operatorname {Subst}\left (\int (-2 a+b \log (1-c x)) \log (1+c x) \, dx,x,x^2\right )+\frac {1}{8} b^2 \operatorname {Subst}\left (\int \log ^2(1+c x) \, dx,x,x^2\right )\\ &=\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {\operatorname {Subst}\left (\int (2 a-b \log (x))^2 \, dx,x,1-c x^2\right )}{8 c}+\frac {b^2 \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,1+c x^2\right )}{8 c}+\frac {1}{4} (b c) \operatorname {Subst}\left (\int \frac {x (-2 a+b \log (1-c x))}{1+c x} \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {x \log (1+c x)}{1-c x} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}+\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c}-\frac {b \operatorname {Subst}\left (\int (2 a-b \log (x)) \, dx,x,1-c x^2\right )}{4 c}-\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{4 c}+\frac {1}{4} (b c) \operatorname {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{c}-\frac {-2 a+b \log (1-c x)}{c (1+c x)}\right ) \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \operatorname {Subst}\left (\int \left (-\frac {\log (1+c x)}{c}-\frac {\log (1+c x)}{c (-1+c x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{2} a b x^2+\frac {b^2 x^2}{4}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}-\frac {b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{4 c}+\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c}+\frac {1}{4} b \operatorname {Subst}\left (\int (-2 a+b \log (1-c x)) \, dx,x,x^2\right )-\frac {1}{4} b \operatorname {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )+\frac {1}{4} b^2 \operatorname {Subst}\left (\int \log (1+c x) \, dx,x,x^2\right )+\frac {1}{4} b^2 \operatorname {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{4 c}\\ &=\frac {b^2 x^2}{2}+\frac {b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{4 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}+\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}-\frac {b^2 \left (1+c x^2\right ) \log \left (1+c x^2\right )}{4 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 c}+\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c}-\frac {1}{4} b^2 \operatorname {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )+\frac {1}{4} b^2 \operatorname {Subst}\left (\int \log (1-c x) \, dx,x,x^2\right )-\frac {1}{4} b^2 \operatorname {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )+\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1+c x^2\right )}{4 c}\\ &=\frac {b^2 x^2}{4}+\frac {b^2 \left (1-c x^2\right ) \log \left (1-c x^2\right )}{4 c}-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}+\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 c}+\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )}{4 c}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )}{4 c}-\frac {b^2 \operatorname {Subst}\left (\int \log (x) \, dx,x,1-c x^2\right )}{4 c}\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 c}+\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}+\frac {b^2 \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 c}+\frac {1}{4} b x^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )+\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 c}-\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )}{4 c}+\frac {b^2 \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )}{4 c}\\ \end {align*}
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Mathematica [A] time = 0.16, size = 99, normalized size = 1.05 \[ \frac {a \left (a c x^2+b \log \left (1-c^2 x^4\right )\right )+2 b \tanh ^{-1}\left (c x^2\right ) \left (a c x^2-b \log \left (e^{-2 \tanh ^{-1}\left (c x^2\right )}+1\right )\right )+b^2 \text {Li}_2\left (-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )+b^2 \left (c x^2-1\right ) \tanh ^{-1}\left (c x^2\right )^2}{2 c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x \operatorname {artanh}\left (c x^{2}\right )^{2} + 2 \, a b x \operatorname {artanh}\left (c x^{2}\right ) + a^{2} x, x\right ) \]
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 \[ \int {\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 144, normalized size = 1.53 \[ \frac {x^{2} b^{2} \arctanh \left (c \,x^{2}\right )^{2}}{2}+x^{2} a b \arctanh \left (c \,x^{2}\right )+\frac {a^{2} x^{2}}{2}+\frac {b^{2} \arctanh \left (c \,x^{2}\right )^{2}}{2 c}-\frac {\arctanh \left (c \,x^{2}\right ) \ln \left (1+\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) b^{2}}{c}+\frac {a b \ln \left (-c^{2} x^{4}+1\right )}{2 c}-\frac {\polylog \left (2, -\frac {\left (c \,x^{2}+1\right )^{2}}{-c^{2} x^{4}+1}\right ) b^{2}}{2 c} \]
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 \[ \frac {1}{2} \, a^{2} x^{2} + \frac {1}{8} \, {\left (x^{2} \log \left (-c x^{2} + 1\right )^{2} - c^{2} {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )} - 2 \, c {\left (\frac {x^{2}}{c} + \frac {\log \left (c x^{2} - 1\right )}{c^{2}}\right )} \log \left (-c x^{2} + 1\right ) + 12 \, c \int \frac {x^{3} \log \left (c x^{2} + 1\right )}{c^{2} x^{4} - 1}\,{d x} + \frac {c x^{2} \log \left (c x^{2} + 1\right )^{2} + 2 \, {\left (c x^{2} - {\left (c x^{2} + 1\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )}{c} + \frac {2 \, c x^{2} + \log \left (c x^{2} - 1\right )^{2} + 2 \, \log \left (c x^{2} - 1\right )}{c} - \frac {\log \left (c^{2} x^{4} - 1\right )}{c} + 4 \, \int \frac {x \log \left (c x^{2} + 1\right )}{c^{2} x^{4} - 1}\,{d x}\right )} b^{2} + \frac {{\left (2 \, c x^{2} \operatorname {artanh}\left (c x^{2}\right ) + \log \left (-c^{2} x^{4} + 1\right )\right )} a b}{2 \, c} \]
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 \[ \int x\,{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^2 \,d x \]
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 \[ \int x \left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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