Optimal. Leaf size=270 \[ -\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \text {ArcSin}(c+d x)}}-\frac {e^3 \sqrt {\frac {\pi }{2}} \cos \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{b^{3/2} d}-\frac {e^3 \sqrt {\frac {\pi }{2}} S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{b^{3/2} d} \]
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Rubi [A]
time = 0.32, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps
used = 14, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4889, 12,
4727, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {\sqrt {\frac {\pi }{2}} e^3 \cos \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {\sqrt {\pi } e^3 \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{b^{3/2} d}+\frac {\sqrt {\pi } e^3 \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} d}-\frac {\sqrt {\frac {\pi }{2}} e^3 \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3385
Rule 3386
Rule 3387
Rule 3432
Rule 3433
Rule 4727
Rule 4889
Rubi steps
\begin {align*} \int \frac {(c e+d e x)^3}{\left (a+b \sin ^{-1}(c+d x)\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {e^3 x^3}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \text {Subst}\left (\int \frac {x^3}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (2 e^3\right ) \text {Subst}\left (\int \left (\frac {\cos (2 x)}{2 \sqrt {a+b x}}-\frac {\cos (4 x)}{2 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {e^3 \text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}-\frac {e^3 \text {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (e^3 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}-\frac {\left (e^3 \cos \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {4 a}{b}+4 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}+\frac {\left (e^3 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {2 a}{b}+2 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}-\frac {\left (e^3 \sin \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {4 a}{b}+4 x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (2 e^3 \cos \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}-\frac {\left (2 e^3 \cos \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}+\frac {\left (2 e^3 \sin \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}-\frac {\left (2 e^3 \sin \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{b^2 d}\\ &=-\frac {2 e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {e^3 \sqrt {\frac {\pi }{2}} \cos \left (\frac {4 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{b^{3/2} d}+\frac {e^3 \sqrt {\pi } S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{b^{3/2} d}-\frac {e^3 \sqrt {\frac {\pi }{2}} S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{b^{3/2} d}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 0.21, size = 300, normalized size = 1.11 \begin {gather*} -\frac {i e^3 e^{-\frac {4 i a}{b}} \left (\sqrt {2} e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )-\sqrt {2} e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )-\sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {4 i (a+b \text {ArcSin}(c+d x))}{b}\right )+e^{\frac {8 i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {4 i (a+b \text {ArcSin}(c+d x))}{b}\right )-2 i e^{\frac {4 i a}{b}} \sin (2 \text {ArcSin}(c+d x))+i e^{\frac {4 i a}{b}} \sin (4 \text {ArcSin}(c+d x))\right )}{4 b d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.35, size = 335, normalized size = 1.24
method | result | size |
default | \(\frac {e^{3} \left (-2 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \cos \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}+2 \sqrt {-\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sin \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}+2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}-2 \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {2}{b}}+2 \sin \left (-\frac {2 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {2 a}{b}\right )-\sin \left (-\frac {4 \left (a +b \arcsin \left (d x +c \right )\right )}{b}+\frac {4 a}{b}\right )\right )}{4 d b \sqrt {a +b \arcsin \left (d x +c \right )}}\) | \(335\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
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 \begin {gather*} e^{3} \left (\int \frac {c^{3}}{a \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {d^{3} x^{3}}{a \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {3 c d^{2} x^{2}}{a \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )}}\, dx + \int \frac {3 c^{2} d x}{a \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} + b \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}} \operatorname {asin}{\left (c + d x \right )}}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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