Optimal. Leaf size=300 \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac{3 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d^2}-\frac{2 b^2 x}{c^4 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2} \]
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Rubi [A] time = 0.525494, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 14, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.518, Rules used = {4703, 4715, 4657, 4181, 2531, 2282, 6589, 4677, 8, 266, 43, 4689, 388, 208} \[ -\frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac{3 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^5 d^2}-\frac{2 b^2 x}{c^4 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2} \]
Antiderivative was successfully verified.
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Rule 4703
Rule 4715
Rule 4657
Rule 4181
Rule 2531
Rule 2282
Rule 6589
Rule 4677
Rule 8
Rule 266
Rule 43
Rule 4689
Rule 388
Rule 208
Rubi steps
\begin{align*} \int \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx &=\frac{x^3 \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^3 \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{c d^2}-\frac{3 \int \frac{x^2 \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^5 d^2 \sqrt{1-c^2 x^2}}-\frac{b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{b^2 \int \frac{2-c^2 x^2}{c^4-c^6 x^2} \, dx}{d^2}-\frac{(3 b) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{c^3 d^2}-\frac{3 \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{2 c^4 d}\\ &=\frac{b^2 x}{c^4 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{b^2 \int \frac{1}{c^4-c^6 x^2} \, dx}{d^2}-\frac{3 \operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^5 d^2}-\frac{\left (3 b^2\right ) \int 1 \, dx}{c^4 d^2}\\ &=-\frac{2 b^2 x}{c^4 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2}+\frac{(3 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d^2}-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^5 d^2}\\ &=-\frac{2 b^2 x}{c^4 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{\left (3 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^5 d^2}-\frac{\left (3 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^5 d^2}\\ &=-\frac{2 b^2 x}{c^4 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}\\ &=-\frac{2 b^2 x}{c^4 d^2}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2 \sqrt{1-c^2 x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c^5 d^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^4 d^2}+\frac{x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{b^2 \tanh ^{-1}(c x)}{c^5 d^2}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}+\frac{3 b^2 \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}-\frac{3 b^2 \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{c^5 d^2}\\ \end{align*}
Mathematica [B] time = 3.10413, size = 614, normalized size = 2.05 \[ \frac{-12 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+12 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+12 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )-12 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )-\frac{2 a^2 c x}{c^2 x^2-1}+4 a^2 c x+3 a^2 \log (1-c x)-3 a^2 \log (c x+1)+8 a b \sqrt{1-c^2 x^2}+\frac{2 a b \sqrt{1-c^2 x^2}}{c x-1}-\frac{2 a b \sqrt{1-c^2 x^2}}{c x+1}+8 a b c x \sin ^{-1}(c x)-\frac{2 a b \sin ^{-1}(c x)}{c x-1}-\frac{2 a b \sin ^{-1}(c x)}{c x+1}+6 i \pi a b \sin ^{-1}(c x)-12 a b \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+12 a b \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )-6 \pi a b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+6 \pi a b \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+6 \pi a b \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\frac{8 b^2 c^3 x^3}{1-c^2 x^2}+\frac{8 b^2 c x}{c^2 x^2-1}+\frac{4 b^2 c^3 x^3 \sin ^{-1}(c x)^2}{c^2 x^2-1}-\frac{6 b^2 c^2 x^2 \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}+\frac{6 b^2 c x \sin ^{-1}(c x)^2}{1-c^2 x^2}+2 b^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)+\frac{2 b^2 \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}+4 b^2 \tanh ^{-1}(c x)+12 i b^2 \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^5 d^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.409, size = 705, normalized size = 2.4 \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^{6} d^{2} x^{2} - c^{4} d^{2}} - \frac{4 \, x}{c^{4} d^{2}} + \frac{3 \, \log \left (c x + 1\right )}{c^{5} d^{2}} - \frac{3 \, \log \left (c x - 1\right )}{c^{5} d^{2}}\right )} - \frac{3 \,{\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 ) - 3 \,{\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 (2 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} - 2 \,{\left (c^{7} d^{2} x^{2} - c^{5} d^{2}\right )} \int \frac{4 \, a b c^{4} x^{4} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (3 \,{\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 ) - 3 \,{\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 ) - 2 \,{\left (2 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{8} d^{2} x^{4} - 2 \, c^{6} d^{2} x^{2} + c^{4} d^{2}}\,{d x}}{4 \,{\left (c^{7} d^{2} x^{2} - c^{5} 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^{4} \arcsin \left (c x\right )^{2} + 2 \, a b x^{4} \arcsin \left (c x\right ) + a^{2} x^{4}}{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^{4}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{b^{2} x^{4} \operatorname{asin}^{2}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{2 a b x^{4} \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^{4}}{{\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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