Optimal. Leaf size=514 \[ \frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{2 i b^2 m \text{PolyLog}\left (4,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 i b^2 m \text{PolyLog}\left (4,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c} \]
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Rubi [A] time = 0.752853, antiderivative size = 514, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {4641, 4779, 4741, 4519, 2190, 2531, 6609, 2282, 6589} \[ \frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{2 i b^2 m \text{PolyLog}\left (4,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 i b^2 m \text{PolyLog}\left (4,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c} \]
Antiderivative was successfully verified.
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Rule 4641
Rule 4779
Rule 4741
Rule 4519
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt{1-c^2 x^2}} \, dx &=\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac{(g m) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^3}{f+g x} \, dx}{3 b c}\\ &=\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{(a+b x)^3 \cos (x)}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^3}{c f-i e^{i x} g-\sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^3}{c f-i e^{i x} g+\sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac{m \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-\frac{i e^{i x} g}{c f-\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}+\frac{m \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-\frac{i e^{i x} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{(2 i b m) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (\frac{i e^{i x} g}{c f-\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}-\frac{(2 i b m) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (\frac{i e^{i x} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{\left (2 b^2 m\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (\frac{i e^{i x} g}{c f-\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}+\frac{\left (2 b^2 m\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (\frac{i e^{i x} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{\left (2 i b^2 m\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i g x}{c f-\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{\left (2 i b^2 m\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i g x}{c f+\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}\\ &=\frac{i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{3 b c}-\frac{m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{3 b c}+\frac{\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{i m \left (a+b \sin ^{-1}(c x)\right )^2 \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 b m \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 i b^2 m \text{Li}_4\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{2 i b^2 m \text{Li}_4\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}\\ \end{align*}
Mathematica [F] time = 76.4079, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt{1-c^2 x^2}} \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 6.27, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( h \left ( gx+f \right ) ^{m} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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{\sqrt{-c^{2} x^{2} + 1}{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, 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*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2} \log{\left (h \left (f + g x\right )^{m} \right )}}{\sqrt{- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \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} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{-c^{2} x^{2} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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