Optimal. Leaf size=90 \[ -i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \]
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Rubi [A] time = 0.122593, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4625, 3717, 2190, 2531, 2282, 6589} \[ -i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \]
Antiderivative was successfully verified.
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Rule 4625
Rule 3717
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x} \, dx &=\operatorname{Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 i \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac{i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} b^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end{align*}
Mathematica [A] time = 0.155712, size = 143, normalized size = 1.59 \[ 2 a b \left (\sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac{1}{2} i \left (\sin ^{-1}(c x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )\right )\right )+b^2 \left (i \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c x)}\right )+\frac{1}{3} i \sin ^{-1}(c x)^3+\sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+a^2 \log (c x) \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.03, size = 319, normalized size = 3.5 \begin{align*}{a}^{2}\ln \left ( cx \right ) -{\frac{i}{3}}{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{3}+{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,i{b}^{2}\arcsin \left ( cx \right ){\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,{b}^{2}{\it polylog} \left ( 3,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,i{b}^{2}\arcsin \left ( cx \right ){\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,{b}^{2}{\it polylog} \left ( 3,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -iab \left ( \arcsin \left ( cx \right ) \right ) ^{2}-2\,iab{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,iab{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,ab\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,ab\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) \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*} a^{2} \log \left (x\right ) + \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 )}{x}\,{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}}{x}, 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}}{x}\, 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}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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