Optimal. Leaf size=210 \[ \frac{i b c^2 \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{i b c^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 c^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+\frac{b^2 c^2 \log (x)}{d} \]
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Rubi [A] time = 0.382664, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4701, 4679, 4419, 4183, 2531, 2282, 6589, 4681, 29} \[ \frac{i b c^2 \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{i b c^2 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 c^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{2 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+\frac{b^2 c^2 \log (x)}{d} \]
Antiderivative was successfully verified.
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Rule 4701
Rule 4679
Rule 4419
Rule 4183
Rule 2531
Rule 2282
Rule 6589
Rule 4681
Rule 29
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x^3 \left (d-c^2 d x^2\right )} \, dx &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+c^2 \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )} \, dx+\frac{(b c) \int \frac{a+b \sin ^{-1}(c x)}{x^2 \sqrt{1-c^2 x^2}} \, dx}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+\frac{c^2 \operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (b^2 c^2\right ) \int \frac{1}{x} \, dx}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}+\frac{b^2 c^2 \log (x)}{d}+\frac{\left (2 c^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{b^2 c^2 \log (x)}{d}-\frac{\left (2 b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (2 b c^2\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{b^2 c^2 \log (x)}{d}+\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (i b^2 c^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (i b^2 c^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{b^2 c^2 \log (x)}{d}+\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}\\ &=-\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{d x}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 d x^2}-\frac{2 c^2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{b^2 c^2 \log (x)}{d}+\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b c^2 \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 c^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 1.19558, size = 353, normalized size = 1.68 \[ -\frac{2 a b c^2 \left (-i \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )+i \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac{\sqrt{1-c^2 x^2}}{c x}+\frac{\sin ^{-1}(c x)}{c^2 x^2}-2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )+2 b^2 c^2 \left (-i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c x)}\right )-i \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )+\frac{\sin ^{-1}(c x)^2}{2 c^2 x^2}+\frac{\sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c x}-\log (c x)-\frac{2}{3} i \sin ^{-1}(c x)^3-\sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+\frac{i \pi ^3}{24}\right )+a^2 c^2 \log \left (1-c^2 x^2\right )-2 a^2 c^2 \log (x)+\frac{a^2}{x^2}}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.262, size = 793, normalized size = 3.8 \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}{2} \,{\left (\frac{c^{2} \log \left (c x + 1\right )}{d} + \frac{c^{2} \log \left (c x - 1\right )}{d} - \frac{2 \, c^{2} \log \left (x\right )}{d} + \frac{1}{d x^{2}}\right )} a^{2} - \int \frac{b^{2} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{c^{2} d x^{5} - d x^{3}}\,{d x} \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} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{2} d x^{5} - d x^{3}}, 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}}{c^{2} x^{5} - x^{3}}\, dx + \int \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx + \int \frac{2 a b \operatorname{asin}{\left (c x \right )}}{c^{2} x^{5} - x^{3}}\, dx}{d} \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}}{{\left (c^{2} d x^{2} - d\right )} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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