Optimal. Leaf size=460 \[ -\frac{i b (e g-2 d h) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{i b (e g-2 d h) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^3}-\frac{\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac{b c \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 \sqrt{c^2 d^2-e^2}}+\frac{b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}+\frac{b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^3}+\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b \sin ^{-1}(c x)^2 (e g-2 d h)}{2 e^3}-\frac{b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]
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Rubi [A] time = 0.847327, antiderivative size = 460, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 14, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {698, 4753, 12, 6742, 261, 725, 204, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac{i b (e g-2 d h) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{i b (e g-2 d h) \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^3}-\frac{\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac{b c \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 \sqrt{c^2 d^2-e^2}}+\frac{b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}+\frac{b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^3}+\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b \sin ^{-1}(c x)^2 (e g-2 d h)}{2 e^3}-\frac{b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]
Antiderivative was successfully verified.
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Rule 698
Rule 4753
Rule 12
Rule 6742
Rule 261
Rule 725
Rule 204
Rule 216
Rule 2404
Rule 4741
Rule 4519
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int \frac{\left (f+g x+h x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{(d+e x)^2} \, dx &=\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-(b c) \int \frac{e h x-\frac{e^2 f-d e g+d^2 h}{d+e x}+(e g-2 d h) \log (d+e x)}{e^3 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac{(b c) \int \frac{e h x-\frac{e^2 f-d e g+d^2 h}{d+e x}+(e g-2 d h) \log (d+e x)}{\sqrt{1-c^2 x^2}} \, dx}{e^3}\\ &=\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac{(b c) \int \left (\frac{e h x}{\sqrt{1-c^2 x^2}}+\frac{-e^2 f+d e g-d^2 h}{(d+e x) \sqrt{1-c^2 x^2}}+\frac{(e g-2 d h) \log (d+e x)}{\sqrt{1-c^2 x^2}}\right ) \, dx}{e^3}\\ &=\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac{(b c h) \int \frac{x}{\sqrt{1-c^2 x^2}} \, dx}{e^2}-\frac{(b c (e g-2 d h)) \int \frac{\log (d+e x)}{\sqrt{1-c^2 x^2}} \, dx}{e^3}+\frac{\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \int \frac{1}{(d+e x) \sqrt{1-c^2 x^2}} \, dx}{e^3}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac{(b c (e g-2 d h)) \int \frac{\sin ^{-1}(c x)}{c d+c e x} \, dx}{e^2}-\frac{\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-c^2 d^2+e^2-x^2} \, dx,x,\frac{e+c^2 d x}{\sqrt{1-c^2 x^2}}\right )}{e^3}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \left (e^2 f-d e g+d^2 h\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{e^3 \sqrt{c^2 d^2-e^2}}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac{(b c (e g-2 d h)) \operatorname{Subst}\left (\int \frac{x \cos (x)}{c^2 d+c e \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{e^2}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b (e g-2 d h) \sin ^{-1}(c x)^2}{2 e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \left (e^2 f-d e g+d^2 h\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{e^3 \sqrt{c^2 d^2-e^2}}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac{(b c (e g-2 d h)) \operatorname{Subst}\left (\int \frac{e^{i x} x}{c^2 d-c \sqrt{c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e^2}+\frac{(b c (e g-2 d h)) \operatorname{Subst}\left (\int \frac{e^{i x} x}{c^2 d+c \sqrt{c^2 d^2-e^2}-i c e e^{i x}} \, dx,x,\sin ^{-1}(c x)\right )}{e^2}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b (e g-2 d h) \sin ^{-1}(c x)^2}{2 e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \left (e^2 f-d e g+d^2 h\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{e^3 \sqrt{c^2 d^2-e^2}}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac{(b (e g-2 d h)) \operatorname{Subst}\left (\int \log \left (1-\frac{i c e e^{i x}}{c^2 d-c \sqrt{c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^3}-\frac{(b (e g-2 d h)) \operatorname{Subst}\left (\int \log \left (1-\frac{i c e e^{i x}}{c^2 d+c \sqrt{c^2 d^2-e^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{e^3}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b (e g-2 d h) \sin ^{-1}(c x)^2}{2 e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \left (e^2 f-d e g+d^2 h\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{e^3 \sqrt{c^2 d^2-e^2}}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac{(i b (e g-2 d h)) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i c e x}{c^2 d-c \sqrt{c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^3}+\frac{(i b (e g-2 d h)) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{i c e x}{c^2 d+c \sqrt{c^2 d^2-e^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{e^3}\\ &=\frac{b h \sqrt{1-c^2 x^2}}{c e^2}-\frac{i b (e g-2 d h) \sin ^{-1}(c x)^2}{2 e^3}+\frac{h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}-\frac{\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac{b c \left (e^2 f-d e g+d^2 h\right ) \tan ^{-1}\left (\frac{e+c^2 d x}{\sqrt{c^2 d^2-e^2} \sqrt{1-c^2 x^2}}\right )}{e^3 \sqrt{c^2 d^2-e^2}}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}+\frac{b (e g-2 d h) \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{b (e g-2 d h) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac{(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac{i b (e g-2 d h) \text{Li}_2\left (\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^3}-\frac{i b (e g-2 d h) \text{Li}_2\left (\frac{i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt{c^2 d^2-e^2}}\right )}{e^3}\\ \end{align*}
Mathematica [A] time = 0.968613, size = 392, normalized size = 0.85 \[ \frac{-\frac{1}{2} i b (e g-2 d h) \left (2 \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )+2 \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )\right )\right )\right )-\frac{\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{d+e x}+(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )+e h x \left (a+b \sin ^{-1}(c x)\right )+\frac{b c \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{\sqrt{c^2 d^2-e^2}}+\frac{b e h \sqrt{1-c^2 x^2}}{c}-b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.245, size = 1922, normalized size = 4.2 \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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{a h x^{2} + a g x + a f +{\left (b h x^{2} + b g x + b f\right )} \arcsin \left (c x\right )}{e^{2} x^{2} + 2 \, d e x + 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*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (f + g x + h x^{2}\right )}{\left (d + e x\right )^{2}}\, 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 (h x^{2} + g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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