Optimal. Leaf size=288 \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^3 \cos \left (\frac {4 a}{b}\right ) C\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {\sqrt {\pi } \sqrt {b} e^3 \cos \left (\frac {2 a}{b}\right ) C\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{16 d}+\frac {\sqrt {\pi } \sqrt {b} e^3 \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{16 d}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^3 \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d} \]
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Rubi [A] time = 0.72, antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4805, 12, 4629, 4723, 3312, 3306, 3305, 3351, 3304, 3352} \[ -\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^3 \cos \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {\sqrt {\pi } \sqrt {b} e^3 \cos \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{16 d}+\frac {\sqrt {\pi } \sqrt {b} e^3 \sin \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{16 d}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {b} e^3 \sin \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{64 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 3304
Rule 3305
Rule 3306
Rule 3312
Rule 3351
Rule 3352
Rule 4629
Rule 4723
Rule 4805
Rubi steps
\begin {align*} \int (c e+d e x)^3 \sqrt {a+b \sin ^{-1}(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \sqrt {a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \sqrt {a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1-x^2} \sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{8 d}\\ &=\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {\sin ^4(x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}\\ &=\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \left (\frac {3}{8 \sqrt {a+b x}}-\frac {\cos (2 x)}{2 \sqrt {a+b x}}+\frac {\cos (4 x)}{8 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}\\ &=-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {\cos (4 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{64 d}+\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}\\ &=-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}+\frac {\left (b e^3 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {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 )}{16 d}-\frac {\left (b e^3 \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {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 )}{64 d}+\frac {\left (b e^3 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {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 )}{16 d}-\frac {\left (b e^3 \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {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 )}{64 d}\\ &=-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}+\frac {\left (e^3 \cos \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{8 d}-\frac {\left (e^3 \cos \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{32 d}+\frac {\left (e^3 \sin \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {2 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{8 d}-\frac {\left (e^3 \sin \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {4 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{32 d}\\ &=-\frac {3 e^3 \sqrt {a+b \sin ^{-1}(c+d x)}}{32 d}+\frac {e^3 (c+d x)^4 \sqrt {a+b \sin ^{-1}(c+d x)}}{4 d}-\frac {\sqrt {b} 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 )}{64 d}+\frac {\sqrt {b} 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 )}{16 d}+\frac {\sqrt {b} 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 )}{16 d}-\frac {\sqrt {b} 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 )}{64 d}\\ \end {align*}
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Mathematica [C] time = 0.20, size = 269, normalized size = 0.93 \[ \frac {e^3 e^{-\frac {4 i a}{b}} \sqrt {a+b \sin ^{-1}(c+d x)} \left (-4 \sqrt {2} e^{\frac {2 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {3}{2},-\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-4 \sqrt {2} e^{\frac {6 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {3}{2},\frac {2 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {3}{2},-\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+e^{\frac {8 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {3}{2},\frac {4 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{128 d \sqrt {\frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{b^2}}} \]
Warning: Unable to verify antiderivative.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 2.19, size = 1111, normalized size = 3.86 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.39, size = 372, normalized size = 1.29 \[ -\frac {e^{3} \left (\sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b +\sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) b -8 \cos \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, b -8 \sin \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, b +16 \arcsin \left (d x +c \right ) \cos \left (\frac {2 a +2 b \arcsin \left (d x +c \right )}{b}-\frac {2 a}{b}\right ) b +16 \cos \left (\frac {2 a +2 b \arcsin \left (d x +c \right )}{b}-\frac {2 a}{b}\right ) a -4 \arcsin \left (d x +c \right ) \cos \left (\frac {4 a +4 b \arcsin \left (d x +c \right )}{b}-\frac {4 a}{b}\right ) b -4 \cos \left (\frac {4 a +4 b \arcsin \left (d x +c \right )}{b}-\frac {4 a}{b}\right ) a \right )}{128 d \sqrt {a +b \arcsin \left (d x +c \right )}} \]
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 \[ \int {\left (d e x + c e\right )}^{3} \sqrt {b \arcsin \left (d x + c\right ) + a}\,{d x} \]
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 \[ \int {\left (c\,e+d\,e\,x\right )}^3\,\sqrt {a+b\,\mathrm {asin}\left (c+d\,x\right )} \,d x \]
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 \[ e^{3} \left (\int c^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int d^{3} x^{3} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int 3 c d^{2} x^{2} \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}\, dx + \int 3 c^{2} d x \sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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