Optimal. Leaf size=218 \[ -\frac {8 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {8 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
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Rubi [A] time = 0.43, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {4803, 4621, 4719, 4723, 3306, 3305, 3351, 3304, 3352} \[ -\frac {8 \sqrt {2 \pi } \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {8 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3304
Rule 3305
Rule 3306
Rule 3351
Rule 3352
Rule 4621
Rule 4719
Rule 4723
Rule 4803
Rubi steps
\begin {align*} \int \frac {1}{\left (a+b \sin ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b \sin ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}-\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {x}{\sqrt {1-x^2} \sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{15 b^3 d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (8 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}-\frac {\left (8 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (16 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}-\frac {\left (16 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{15 b^4 d}\\ &=-\frac {2 \sqrt {1-(c+d x)^2}}{5 b d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}}+\frac {4 (c+d x)}{15 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )^{3/2}}+\frac {8 \sqrt {1-(c+d x)^2}}{15 b^3 d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {8 \sqrt {2 \pi } \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {8 \sqrt {2 \pi } C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{15 b^{7/2} d}\\ \end {align*}
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Mathematica [C] time = 0.74, size = 287, normalized size = 1.32 \[ \frac {e^{-i \sin ^{-1}(c+d x)} \left (8 a^2+4 a b \left (4 \sin ^{-1}(c+d x)+i\right )-8 e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (a+b \sin ^{-1}(c+d x)\right )^2 \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 b^2 \left (4 \sin ^{-1}(c+d x)^2+2 i \sin ^{-1}(c+d x)-3\right )\right )+4 e^{-\frac {i a}{b}} \left (a+b \sin ^{-1}(c+d x)\right ) \left (e^{\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (2 a+b \left (2 \sin ^{-1}(c+d x)-i\right )\right )-2 i b \left (-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )-6 b^2 e^{i \sin ^{-1}(c+d x)}}{30 b^3 d \left (a+b \sin ^{-1}(c+d x)\right )^{5/2}} \]
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 [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.20, size = 600, normalized size = 2.75 \[ \frac {\frac {8 \arcsin \left (d x +c \right )^{2} \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b^{2}}{15}-\frac {8 \arcsin \left (d x +c \right )^{2} \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b^{2}}{15}+\frac {16 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, a b}{15}-\frac {16 \arcsin \left (d x +c \right ) \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, a b}{15}+\frac {8 \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, a^{2}}{15}-\frac {8 \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {2}\, \sqrt {\pi }\, \sqrt {\frac {1}{b}}\, a^{2}}{15}+\frac {8 \arcsin \left (d x +c \right )^{2} \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) b^{2}}{15}+\frac {16 \arcsin \left (d x +c \right ) \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) a b}{15}+\frac {4 \arcsin \left (d x +c \right ) \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) b^{2}}{15}+\frac {8 \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) a^{2}}{15}-\frac {2 \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) b^{2}}{5}+\frac {4 \sin \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right ) a b}{15}}{d \,b^{3} \left (a +b \arcsin \left (d x +c \right )\right )^{\frac {5}{2}}} \]
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 \[ \int \frac {1}{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}\,{d x} \]
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 \[ \int \frac {1}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{7/2}} \,d x \]
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 \[ \int \frac {1}{\left (a + b \operatorname {asin}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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