Optimal. Leaf size=145 \[ -\frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b d}+\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 b d}+\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{4 b d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{8 b d} \]
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Rubi [A] time = 0.314128, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4805, 12, 4635, 4406, 3303, 3299, 3302} \[ -\frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{4 b d}+\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{8 b d}+\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{4 b d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{8 b d} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4635
Rule 4406
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^3}{a+b \sin ^{-1}(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \left (\frac{\sin (2 x)}{4 (a+b x)}-\frac{\sin (4 x)}{8 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=-\frac{e^3 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d}\\ &=\frac{\left (e^3 \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d}-\frac{\left (e^3 \cos \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac{\left (e^3 \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 d}+\frac{\left (e^3 \sin \left (\frac{4 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac{e^3 \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{2 a}{b}\right )}{4 b d}+\frac{e^3 \text{Ci}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{4 a}{b}\right )}{8 b d}+\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{4 b d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{8 b d}\\ \end{align*}
Mathematica [A] time = 0.236173, size = 109, normalized size = 0.75 \[ \frac{e^3 \left (-2 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )}{8 b d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.034, size = 112, normalized size = 0.8 \begin{align*} -{\frac{{e}^{3}}{8\,bd} \left ({\it Si} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) -{\it Ci} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) -2\,{\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) +2\,{\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) \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*} \int \frac{{\left (d e x + c e\right )}^{3}}{b \arcsin \left (d x + c\right ) + a}\,{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{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b \arcsin \left (d x + c\right ) + a}, 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*} e^{3} \left (\int \frac{c^{3}}{a + b \operatorname{asin}{\left (c + d x \right )}}\, dx + \int \frac{d^{3} x^{3}}{a + b \operatorname{asin}{\left (c + d x \right )}}\, dx + \int \frac{3 c d^{2} x^{2}}{a + b \operatorname{asin}{\left (c + d x \right )}}\, dx + \int \frac{3 c^{2} d x}{a + b \operatorname{asin}{\left (c + d x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.21758, size = 363, normalized size = 2.5 \begin{align*} \frac{\cos \left (\frac{a}{b}\right )^{3} \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right ) e^{3} \sin \left (\frac{a}{b}\right )}{b d} - \frac{\cos \left (\frac{a}{b}\right )^{4} e^{3} \operatorname{Si}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b d} - \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right ) e^{3} \sin \left (\frac{a}{b}\right )}{2 \, b d} - \frac{\cos \left (\frac{a}{b}\right ) \operatorname{Ci}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right ) e^{3} \sin \left (\frac{a}{b}\right )}{2 \, b d} + \frac{\cos \left (\frac{a}{b}\right )^{2} e^{3} \operatorname{Si}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{b d} + \frac{\cos \left (\frac{a}{b}\right )^{2} e^{3} \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{2 \, b d} - \frac{e^{3} \operatorname{Si}\left (\frac{4 \, a}{b} + 4 \, \arcsin \left (d x + c\right )\right )}{8 \, b d} - \frac{e^{3} \operatorname{Si}\left (\frac{2 \, a}{b} + 2 \, \arcsin \left (d x + c\right )\right )}{4 \, b d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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