Optimal. Leaf size=214 \[ \frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2 c^4}+\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac{5 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}+\frac{\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2 c^4}+\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac{5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )} \]
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Rubi [A] time = 0.633116, antiderivative size = 210, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4721, 4635, 4406, 3303, 3299, 3302} \[ \frac{\cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac{5 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}+\frac{\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac{5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
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Rule 4721
Rule 4635
Rule 4406
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{x^3 \sqrt{1-c^2 x^2}}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \int \frac{x^2}{a+b \sin ^{-1}(c x)} \, dx}{b c}-\frac{(5 c) \int \frac{x^4}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{4 (a+b x)}-\frac{\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}-\frac{5 \operatorname{Subst}\left (\int \left (\frac{\cos (x)}{8 (a+b x)}-\frac{3 \cos (3 x)}{16 (a+b x)}+\frac{\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^4}\\ &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac{3 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{3 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}+\frac{15 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac{\left (5 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac{\left (3 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{\left (3 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}+\frac{\left (15 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}-\frac{\left (5 \cos \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}-\frac{\left (5 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b c^4}+\frac{\left (3 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}-\frac{\left (3 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^4}+\frac{\left (15 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}-\frac{\left (5 \sin \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac{x^3 \left (1-c^2 x^2\right )}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac{\cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac{3 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac{5 \cos \left (\frac{5 a}{b}\right ) \text{Ci}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}+\frac{\sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2 c^4}+\frac{3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2 c^4}-\frac{5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2 c^4}\\ \end{align*}
Mathematica [A] time = 0.543693, size = 175, normalized size = 0.82 \[ \frac{\frac{16 b c^5 x^5}{a+b \sin ^{-1}(c x)}-\frac{16 b c^3 x^3}{a+b \sin ^{-1}(c x)}+2 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )+3 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-5 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )+2 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )+3 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )-5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (5 \left (\frac{a}{b}+\sin ^{-1}(c x)\right )\right )}{16 b^2 c^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.056, size = 340, normalized size = 1.6 \begin{align*} -{\frac{1}{16\,{c}^{4} \left ( a+b\arcsin \left ( cx \right ) \right ){b}^{2}} \left ( 5\,\arcsin \left ( cx \right ) \sin \left ( 5\,{\frac{a}{b}} \right ){\it Si} \left ( 5\,\arcsin \left ( cx \right ) +5\,{\frac{a}{b}} \right ) b+5\,\arcsin \left ( cx \right ){\it Ci} \left ( 5\,\arcsin \left ( cx \right ) +5\,{\frac{a}{b}} \right ) \cos \left ( 5\,{\frac{a}{b}} \right ) b-2\,\arcsin \left ( cx \right ){\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) b-2\,\arcsin \left ( cx \right ){\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) b-3\,\arcsin \left ( cx \right ){\it Si} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) b-3\,\arcsin \left ( cx \right ){\it Ci} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) b+5\,\sin \left ( 5\,{\frac{a}{b}} \right ){\it Si} \left ( 5\,\arcsin \left ( cx \right ) +5\,{\frac{a}{b}} \right ) a+5\,{\it Ci} \left ( 5\,\arcsin \left ( cx \right ) +5\,{\frac{a}{b}} \right ) \cos \left ( 5\,{\frac{a}{b}} \right ) a-2\,{\it Si} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \sin \left ({\frac{a}{b}} \right ) a-2\,{\it Ci} \left ( \arcsin \left ( cx \right ) +{\frac{a}{b}} \right ) \cos \left ({\frac{a}{b}} \right ) a-3\,{\it Si} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) a-3\,{\it Ci} \left ( 3\,\arcsin \left ( cx \right ) +3\,{\frac{a}{b}} \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a+2\,xbc-\sin \left ( 5\,\arcsin \left ( cx \right ) \right ) b+\sin \left ( 3\,\arcsin \left ( cx \right ) \right ) b \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*} \frac{c^{2} x^{5} - x^{3} - \frac{{\left (b^{2} c \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c\right )}{\left (5 \, c^{2} \int \frac{x^{4}}{b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a}\,{d x} - 3 \, \int \frac{x^{2}}{b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a}\,{d x}\right )}}{b c}}{b^{2} c \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) + a b c} \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} x^{3}}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{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{x^{3} \sqrt{- \left (c x - 1\right ) \left (c x + 1\right )}}{\left (a + b \operatorname{asin}{\left (c x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.70335, size = 1686, normalized size = 7.88 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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