Optimal. Leaf size=198 \[ \frac {2 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 x}{c^2 d} \]
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Rubi [A] time = 0.29, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5767, 5693, 4180, 2531, 2282, 6589, 5717, 8} \[ \frac {2 i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d}+\frac {2 b^2 x}{c^2 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 2282
Rule 2531
Rule 4180
Rule 5693
Rule 5717
Rule 5767
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {\int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^2}-\frac {(2 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{c d}\\ &=-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {\operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 d}+\frac {\left (2 b^2\right ) \int 1 \, dx}{c^2 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {(2 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 d}-\frac {(2 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^3 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^3 d}\\ &=\frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}\\ \end {align*}
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Mathematica [A] time = 0.59, size = 293, normalized size = 1.48 \[ -\frac {a^2 \tan ^{-1}(c x)}{c^3 d}+\frac {a^2 x}{c^2 d}+\frac {2 a b \left (-\sqrt {c^2 x^2+1}+i \left (\text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )\right )+c x \sinh ^{-1}(c x)+i \sinh ^{-1}(c x) \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )\right )}{c^3 d}+\frac {b^2 \left (-2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)-i \left (-2 \sinh ^{-1}(c x) \left (\text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )\right )-2 \left (\text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )\right )-\left (\sinh ^{-1}(c x)^2 \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )+c x \left (\sinh ^{-1}(c x)^2+2\right )\right )}{c^3 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{2}}{c^{2} d x^{2} + d}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{2}}{c^{2} d x^{2} + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{c^{2} d \,x^{2}+d}\, dx \]
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 \[ a^{2} {\left (\frac {x}{c^{2} d} - \frac {\arctan \left (c x\right )}{c^{3} d}\right )} + \int \frac {b^{2} x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{2} d x^{2} + d} + \frac {2 \, a b x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d x^{2} + d}\,{d x} \]
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 \[ \int \frac {x^2\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{d\,c^2\,x^2+d} \,d x \]
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 \[ \frac {\int \frac {a^{2} x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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