Optimal. Leaf size=213 \[ -\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2}+\frac {\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {c^2 x^2+1}}+\frac {i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2} \]
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Rubi [A] time = 0.30, antiderivative size = 213, 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 = {5751, 5693, 4180, 2531, 2282, 6589, 5717, 203} \[ -\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2}+\frac {i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {c^2 x^2+1}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2} \]
Antiderivative was successfully verified.
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Rule 203
Rule 2282
Rule 2531
Rule 4180
Rule 5693
Rule 5717
Rule 5751
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx &=-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {b \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{c d^2}+\frac {\int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{2 c^2 d}\\ &=-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c^3 d^2}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{c^2 d^2}\\ &=-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2}-\frac {(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^2}+\frac {(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^2}\\ &=-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {\left (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^2}-\frac {\left (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^2}\\ &=-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {\left (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^2}-\frac {\left (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^2}\\ &=-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \tan ^{-1}(c x)}{c^3 d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}\\ \end {align*}
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Mathematica [A] time = 1.86, size = 385, normalized size = 1.81 \[ -\frac {\frac {a^2 c x}{c^2 x^2+1}+a^2 \left (-\tan ^{-1}(c x)\right )+\frac {a b \left (\sinh ^{-1}(c x)-i \sqrt {c^2 x^2+1}\right )}{c x-i}+\frac {a b \left (\sinh ^{-1}(c x)+i \sqrt {c^2 x^2+1}\right )}{c x+i}-\frac {1}{2} i a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )\right )+\frac {1}{2} i a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )\right )+\frac {b^2 c x \sinh ^{-1}(c x)^2}{c^2 x^2+1}+\frac {2 b^2 \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+i b^2 \left (2 \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+2 \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{2 c^3 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{2} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{2}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (c^{2} d \,x^{2}+d \right )^{2}}\, dx \]
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 \[ -\frac {1}{2} \, a^{2} {\left (\frac {x}{c^{4} d^{2} x^{2} + c^{2} d^{2}} - \frac {\arctan \left (c x\right )}{c^{3} d^{2}}\right )} + \int \frac {b^{2} x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}} + \frac {2 \, a b x^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}\,{d x} \]
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 \[ \int \frac {x^2\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a^{2} x^{2}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{2} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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