Optimal. Leaf size=249 \[ \frac {e^3 \sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}-\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (\frac {4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}-\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{2 b^3 d}+\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{b^3 d}+\frac {2 e^3 (c+d x)^4}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {3 e^3 (c+d x)^2}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^3 \sqrt {1-(c+d x)^2} (c+d x)^3}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]
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Rubi [A] time = 0.66, antiderivative size = 249, normalized size of antiderivative = 1.00, 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 12
Rule 3299
Rule 3302
Rule 3303
Rule 4406
Rule 4633
Rule 4635
Rule 4719
Rule 4805
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*}
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Mathematica [A] time = 0.75, size = 181, normalized size = 0.73 \[ \frac {e^3 \left (-\frac {b^2 \sqrt {1-(c+d x)^2} (c+d x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+\sin \left (\frac {2 a}{b}\right ) \text {Ci}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-2 \sin \left (\frac {4 a}{b}\right ) \text {Ci}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\frac {b \left (4 (c+d x)^4-3 (c+d x)^2\right )}{a+b \sin ^{-1}(c+d x)}\right )}{2 b^3 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.83, size = 2169, normalized size = 8.71 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 506, normalized size = 2.03 \[ \frac {e^{3} \left (16 \arcsin \left (d x +c \right )^{2} \Si \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) b^{2}-16 \arcsin \left (d x +c \right )^{2} \Ci \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) b^{2}-8 \arcsin \left (d x +c \right )^{2} \Si \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) b^{2}+8 \arcsin \left (d x +c \right )^{2} \Ci \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) b^{2}+32 \arcsin \left (d x +c \right ) \Si \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a b -32 \arcsin \left (d x +c \right ) \Ci \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a b -16 \arcsin \left (d x +c \right ) \Si \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a b +16 \arcsin \left (d x +c \right ) \Ci \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a b +4 \arcsin \left (d x +c \right ) \cos \left (4 \arcsin \left (d x +c \right )\right ) b^{2}-4 \arcsin \left (d x +c \right ) \cos \left (2 \arcsin \left (d x +c \right )\right ) b^{2}+16 \Si \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \cos \left (\frac {4 a}{b}\right ) a^{2}-16 \Ci \left (4 \arcsin \left (d x +c \right )+\frac {4 a}{b}\right ) \sin \left (\frac {4 a}{b}\right ) a^{2}-8 \Si \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) a^{2}+8 \Ci \left (2 \arcsin \left (d x +c \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right ) a^{2}+\sin \left (4 \arcsin \left (d x +c \right )\right ) b^{2}+4 \cos \left (4 \arcsin \left (d x +c \right )\right ) a b -2 \sin \left (2 \arcsin \left (d x +c \right )\right ) b^{2}-4 \cos \left (2 \arcsin \left (d x +c \right )\right ) a b \right )}{16 d \left (a +b \arcsin \left (d x +c \right )\right )^{2} b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ e^{3} \left (\int \frac {c^{3}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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