Optimal. Leaf size=213 \[ -\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d^2 \sqrt {1+c^2 x^2}}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b^2 \text {ArcTan}(c x)}{c^3 d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}+\frac {i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d^2} \]
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Rubi [A]
time = 0.20, 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 = {5810, 5789,
4265, 2611, 2320, 6724, 5798, 209} \begin {gather*} \frac {\text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{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 {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 {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 {b^2 \text {ArcTan}(c x)}{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 {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 2320
Rule 2611
Rule 4265
Rule 5789
Rule 5798
Rule 5810
Rule 6724
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 {\text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 1.10, size = 385, normalized size = 1.81 \begin {gather*} -\frac {\frac {a^2 c x}{1+c^2 x^2}+\frac {2 b^2 \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}+\frac {b^2 c x \sinh ^{-1}(c x)^2}{1+c^2 x^2}+\frac {a b \left (-i \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{-i+c x}+\frac {a b \left (i \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{i+c x}-a^2 \text {ArcTan}(c x)-\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 {PolyLog}\left (2,-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 {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )+i b^2 \left (4 i \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\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 )+2 \sinh ^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+2 \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-2 \text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )\right )}{2 c^3 d^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.11, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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}} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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