Optimal. Leaf size=204 \[ \frac {2 i b c \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {2 i b c \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}-\frac {4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b^2 c \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b^2 c \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d} \]
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Rubi [A] time = 0.34, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5747, 5693, 4180, 2531, 2282, 6589, 5760, 4182, 2279, 2391} \[ \frac {2 i b c \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {2 i b c \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {2 b^2 c \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b^2 c \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b^2 c \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}-\frac {4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2282
Rule 2391
Rule 2531
Rule 4180
Rule 4182
Rule 5693
Rule 5747
Rule 5760
Rule 6589
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-c^2 \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{d}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {c \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac {(2 b c) \operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {(2 i b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac {(2 i b c) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {\left (2 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {\left (2 i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac {2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 b^2 c \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 b^2 c \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {2 i b^2 c \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {2 i b^2 c \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d}\\ \end {align*}
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Mathematica [A] time = 1.06, size = 363, normalized size = 1.78 \[ -\frac {a^2 c \tan ^{-1}(c x)+\frac {a^2}{x}+2 a b c \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )+\frac {1}{2} i a b c \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )\right )-\frac {1}{2} i a b c \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )\right )+\frac {2 a b \sinh ^{-1}(c x)}{x}-b^2 c \left (2 i \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 i \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+2 \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-2 \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+2 i \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 i \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )-\frac {\sinh ^{-1}(c x)^2}{c x}+i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+2 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )}{d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{c^{2} d x^{4} + d x^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} + d\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.31, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{x^{2} \left (c^{2} d \,x^{2}+d \right )}\, dx \]
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 \[ -a^{2} {\left (\frac {c \arctan \left (c x\right )}{d} + \frac {1}{d x}\right )} + \int \frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{2} d x^{4} + d x^{2}} + \frac {2 \, a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d x^{4} + d x^{2}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{x^2\,\left (d\,c^2\,x^2+d\right )} \,d x \]
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 \[ \frac {\int \frac {a^{2}}{c^{2} x^{4} + x^{2}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{4} + x^{2}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{4} + x^{2}}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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