Optimal. Leaf size=233 \[ -\frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac{b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2} \]
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Rubi [A] time = 0.29931, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {4703, 4657, 4181, 2531, 2282, 6589, 4677, 206} \[ -\frac{i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac{i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2}+\frac{b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2} \]
Antiderivative was successfully verified.
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Rule 4703
Rule 4657
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rule 4677
Rule 206
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx &=\frac{x \left (a+b \sin ^{-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 \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{c d^2}-\frac{\int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{2 c^2 d}\\ &=-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-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 \sec (x) \, dx,x,\sin ^{-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 \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2}+\frac{b \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d^2}-\frac{b \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d^2}\\ &=-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d^2}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d^2}\\ &=-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}\\ &=-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^3 d^2 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^3 d^2}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac{b^2 \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac{b^2 \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}\\ \end{align*}
Mathematica [A] time = 2.63084, size = 383, normalized size = 1.64 \[ -\frac{4 i a b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )-4 i a b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-4 b^2 \left (-i \sin ^{-1}(c x) \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+i \sin ^{-1}(c x) \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+\text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )-\text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )+\tanh ^{-1}(c x)+i \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )\right )+\frac{2 a^2 c x}{c^2 x^2-1}+a^2 (-\log (1-c x))+a^2 \log (c x+1)+\frac{2 a b \left (-2 \sqrt{1-c^2 x^2}+\cos \left (2 \sin ^{-1}(c x)\right )+\sin ^{-1}(c x) \left (2 c x-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )+\log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\left (\log \left (1+i e^{i \sin ^{-1}(c x)}\right )-\log \left (1-i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (2 \sin ^{-1}(c x)\right )\right )+1\right )}{c^2 x^2-1}+\frac{2 b^2 \sin ^{-1}(c x) \left (c x \sin ^{-1}(c x)-2 \sqrt{1-c^2 x^2}\right )}{c^2 x^2-1}}{4 c^3 d^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.267, size = 599, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{4} \, a^{2}{\left (\frac{2 \, x}{c^{4} d^{2} x^{2} - c^{2} d^{2}} + \frac{\log \left (c x + 1\right )}{c^{3} d^{2}} - \frac{\log \left (c x - 1\right )}{c^{3} d^{2}}\right )} - \frac{2 \, b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (c x + 1\right ) -{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \,{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )} \int \frac{4 \, a b c^{2} x^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (2 \, b^{2} c x \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) -{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{6} d^{2} x^{4} - 2 \, c^{4} d^{2} x^{2} + c^{2} d^{2}}\,{d x}}{4 \,{\left (c^{5} d^{2} x^{2} - c^{3} d^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} x^{2} \arcsin \left (c x\right )^{2} + 2 \, a b x^{2} \arcsin \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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{asin}^{2}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{2 a b x^{2} \operatorname{asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c^{2} d x^{2} - d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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