Optimal. Leaf size=249 \[ \frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}+\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]
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Rubi [A] time = 0.659939, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302} \[ \frac{e^3 \sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^3 d}-\frac{e^3 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{b^3 d}-\frac{e^3 \cos \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right )}{2 b^3 d}+\frac{e^3 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (\frac{4 a}{b}+4 \sin ^{-1}(c+d x)\right )}{b^3 d}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac{e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]
Antiderivative was successfully verified.
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Rule 4805
Rule 12
Rule 4633
Rule 4719
Rule 4635
Rule 4406
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 )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\cos (x) \sin ^3(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 (a+b x)} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (8 e^3\right ) \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 )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{\sin (4 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{\sin (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{\left (3 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 )}{2 b^2 d}-\frac{\left (2 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 )}{b^2 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 )}{b^2 d}-\frac{\left (3 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 )}{2 b^2 d}+\frac{\left (2 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 )}{b^2 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 )}{b^2 d}\\ &=-\frac{e^3 (c+d x)^3 \sqrt{1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac{3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac{e^3 \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c+d x)\right ) \sin \left (\frac{2 a}{b}\right )}{2 b^3 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 )}{b^3 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 )}{2 b^3 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 )}{b^3 d}\\ \end{align*}
Mathematica [A] time = 0.705064, size = 181, normalized size = 0.73 \[ \frac{e^3 \left (-\frac{b^2 \sqrt{1-(c+d x)^2} (c+d x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+\sin \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-2 \sin \left (\frac{4 a}{b}\right ) \text{CosIntegral}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac{2 a}{b}\right ) \text{Si}\left (2 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \cos \left (\frac{4 a}{b}\right ) \text{Si}\left (4 \left (\frac{a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac{b \left (4 (c+d x)^4-3 (c+d x)^2\right )}{a+b \sin ^{-1}(c+d x)}\right )}{2 b^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.044, size = 506, normalized size = 2. \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(-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^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{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*} e^{3} \left (\int \frac{c^{3}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{d^{3} x^{3}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{3 c d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac{3 c^{2} d x}{a^{3} + 3 a^{2} b \operatorname{asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname{asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname{asin}^{3}{\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 = 2.01396, size = 2928, normalized size = 11.76 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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