Optimal. Leaf size=190 \[ \frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}+\frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^2 d}-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
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Rubi [A] time = 0.294796, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4631
Rule 3303
Rule 3299
Rule 3302
Rubi steps
\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \left (\frac{\cos (2 x)}{2 (a+b x)}-\frac{\cos (4 x)}{2 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}-\frac{e^3 \operatorname{Subst}\left (\int \frac{\cos (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (e^3 \cos \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 )}{2 b d}-\frac{\left (e^3 \cos \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 )}{2 b d}+\frac{\left (e^3 \sin \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 )}{2 b d}-\frac{\left (e^3 \sin \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 )}{2 b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Ci}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ \end{align*}
Mathematica [A] time = 0.875986, size = 220, normalized size = 1.16 \[ -\frac{e^3 \left (3 \left (\cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\log \left (a+b \sin ^{-1}(c+d x)\right )\right )-4 \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\cos \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-4 \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac{2 b \sqrt{1-(c+d x)^2} (c+d x)^3}{a+b \sin ^{-1}(c+d x)}+3 \log \left (a+b \sin ^{-1}(c+d x)\right )\right )}{2 b^2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 281, normalized size = 1.5 \begin{align*} -{\frac{{e}^{3}}{8\,d \left ( a+b\arcsin \left ( dx+c \right ) \right ){b}^{2}} \left ( 4\,\arcsin \left ( dx+c \right ){\it Si} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) b+4\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) b-4\,\arcsin \left ( dx+c \right ){\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) b-4\,\arcsin \left ( dx+c \right ){\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) b+4\,{\it Si} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \sin \left ( 4\,{\frac{a}{b}} \right ) a+4\,{\it Ci} \left ( 4\,\arcsin \left ( dx+c \right ) +4\,{\frac{a}{b}} \right ) \cos \left ( 4\,{\frac{a}{b}} \right ) a-4\,{\it Si} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \sin \left ( 2\,{\frac{a}{b}} \right ) a-4\,{\it Ci} \left ( 2\,\arcsin \left ( dx+c \right ) +2\,{\frac{a}{b}} \right ) \cos \left ( 2\,{\frac{a}{b}} \right ) a-\sin \left ( 4\,\arcsin \left ( dx+c \right ) \right ) b+2\,\sin \left ( 2\,\arcsin \left ( dx+c \right ) \right ) b \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\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*} e^{3} \left (\int \frac{c^{3}}{a^{2} + 2 a b \operatorname{asin}{\left (c + d x \right )} + b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac{d^{3} x^{3}}{a^{2} + 2 a b \operatorname{asin}{\left (c + d x \right )} + b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac{3 c d^{2} x^{2}}{a^{2} + 2 a b \operatorname{asin}{\left (c + d x \right )} + b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}\, dx + \int \frac{3 c^{2} d x}{a^{2} + 2 a b \operatorname{asin}{\left (c + d x \right )} + b^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.49088, size = 1229, normalized size = 6.47 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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