Optimal. Leaf size=148 \[ -\frac {b \text {Li}_2\left (1-\frac {2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac {b \text {Li}_2\left (\frac {2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-c x^n}\right )}{2 n}-\frac {b^2 \text {Li}_3\left (\frac {2}{1-c x^n}-1\right )}{2 n} \]
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Rubi [A] time = 0.31, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6095, 5914, 6052, 5948, 6058, 6610} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac {b \text {PolyLog}\left (2,\frac {2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x^n}\right )}{2 n}-\frac {b^2 \text {PolyLog}\left (3,\frac {2}{1-c x^n}-1\right )}{2 n}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]
Antiderivative was successfully verified.
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Rule 5914
Rule 5948
Rule 6052
Rule 6058
Rule 6095
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,x^n\right )}{n}\\ &=\frac {2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right )}{n}-\frac {(4 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^n\right )}{n}\\ &=\frac {2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right )}{n}+\frac {(2 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^n\right )}{n}-\frac {(2 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^n\right )}{n}\\ &=\frac {2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right )}{n}-\frac {b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^n}\right )}{n}+\frac {b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^n}\right )}{n}+\frac {\left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^n\right )}{n}-\frac {\left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^n\right )}{n}\\ &=\frac {2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right )}{n}-\frac {b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^n}\right )}{n}+\frac {b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^n}\right )}{n}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-c x^n}\right )}{2 n}-\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1-c x^n}\right )}{2 n}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 183, normalized size = 1.24 \[ \frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2-4 b c \left (\frac {1}{2} \left (\frac {\text {Li}_2\left (\frac {-c x^n-1}{c x^n-1}\right ) \left (-a-b \tanh ^{-1}\left (c x^n\right )\right )}{2 c}+\frac {b \text {Li}_3\left (\frac {-c x^n-1}{c x^n-1}\right )}{4 c}\right )+\frac {1}{2} \left (-\frac {\text {Li}_2\left (\frac {c x^n+1}{c x^n-1}\right ) \left (-a-b \tanh ^{-1}\left (c x^n\right )\right )}{2 c}-\frac {b \text {Li}_3\left (\frac {c x^n+1}{c x^n-1}\right )}{4 c}\right )\right )}{n} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x^{n}\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x^{n}\right ) + a^{2}}{x}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x^{n}\right ) + a\right )}^{2}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.24, size = 880, normalized size = 5.95 \[ \frac {a^{2} \ln \left (c \,x^{n}\right )}{n}+\frac {b^{2} \ln \left (c \,x^{n}\right ) \arctanh \left (c \,x^{n}\right )^{2}}{n}-\frac {b^{2} \arctanh \left (c \,x^{n}\right ) \polylog \left (2, -\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}\right )}{n}+\frac {b^{2} \polylog \left (3, -\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}\right )}{2 n}-\frac {b^{2} \arctanh \left (c \,x^{n}\right )^{2} \ln \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )}{n}+\frac {b^{2} \arctanh \left (c \,x^{n}\right )^{2} \ln \left (1-\frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}+\frac {2 b^{2} \arctanh \left (c \,x^{n}\right ) \polylog \left (2, \frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}-\frac {2 b^{2} \polylog \left (3, \frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}+\frac {b^{2} \arctanh \left (c \,x^{n}\right )^{2} \ln \left (1+\frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}+\frac {2 b^{2} \arctanh \left (c \,x^{n}\right ) \polylog \left (2, -\frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}-\frac {2 b^{2} \polylog \left (3, -\frac {c \,x^{n}+1}{\sqrt {-c^{2} x^{2 n}+1}}\right )}{n}+\frac {i b^{2} \pi \mathrm {csgn}\left (\frac {i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right )^{3} \arctanh \left (c \,x^{n}\right )^{2}}{2 n}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right )^{2} \arctanh \left (c \,x^{n}\right )^{2}}{2 n}-\frac {i b^{2} \pi \,\mathrm {csgn}\left (\frac {i}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right )^{2} \arctanh \left (c \,x^{n}\right )^{2}}{2 n}+\frac {i b^{2} \pi \,\mathrm {csgn}\left (i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}-1\right )}{1+\frac {\left (c \,x^{n}+1\right )^{2}}{-c^{2} x^{2 n}+1}}\right ) \arctanh \left (c \,x^{n}\right )^{2}}{2 n}+\frac {2 a b \ln \left (c \,x^{n}\right ) \arctanh \left (c \,x^{n}\right )}{n}-\frac {a b \ln \left (c \,x^{n}\right ) \ln \left (c \,x^{n}+1\right )}{n}-\frac {a b \dilog \left (c \,x^{n}\right )}{n}-\frac {a b \dilog \left (c \,x^{n}+1\right )}{n} \]
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 \[ \frac {1}{4} \, b^{2} \log \left (-c x^{n} + 1\right )^{2} \log \relax (x) + a^{2} \log \relax (x) - \int -\frac {{\left (b^{2} c x^{n} - b^{2}\right )} \log \left (c x^{n} + 1\right )^{2} + 4 \, {\left (a b c x^{n} - a b\right )} \log \left (c x^{n} + 1\right ) + 2 \, {\left (2 \, a b - {\left (b^{2} c n \log \relax (x) + 2 \, a b c\right )} x^{n} - {\left (b^{2} c x^{n} - b^{2}\right )} \log \left (c x^{n} + 1\right )\right )} \log \left (-c x^{n} + 1\right )}{4 \, {\left (c x x^{n} - x\right )}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^n\right )\right )}^2}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{n} \right )}\right )^{2}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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