Optimal. Leaf size=348 \[ 15 b^2 x \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}-\frac {15 b^2 \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )} \]
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Rubi [A]
time = 0.07, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4898, 4895}
\begin {gather*} -\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}+\frac {15 \sqrt {\pi } b^2 x \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right ) S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (i d x^2+1\right )}}{\sqrt {\pi }}\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )}+15 b^2 x \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}-\frac {5 b \sqrt {d^2 x^4-2 i d x^2} \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{5/2} \end {gather*}
Antiderivative was successfully verified.
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Rule 4895
Rule 4898
Rubi steps
\begin {align*} \int \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2} \, dx &=-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}+\left (15 b^2\right ) \int \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )} \, dx\\ &=15 b^2 x \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}-\frac {5 b \sqrt {-2 i d x^2+d^2 x^4} \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \sin ^{-1}\left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}-\frac {15 b^2 \sqrt {\pi } x C\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \sin ^{-1}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \sin ^{-1}\left (1+i d x^2\right )\right )\right )}\\ \end {align*}
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Mathematica [A]
time = 0.19, size = 337, normalized size = 0.97 \begin {gather*} -\frac {5 b \sqrt {d x^2 \left (-2 i+d x^2\right )} \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{3/2}}{d x}+x \left (a-i b \text {ArcSin}\left (1+i d x^2\right )\right )^{5/2}+\frac {15 b^2 x \left (\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )+\sqrt {\pi } S\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )-\sqrt {\pi } \text {FresnelC}\left (\frac {\sqrt {\frac {i}{b}} \sqrt {a-i b \text {ArcSin}\left (1+i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )\right )}{\sqrt {\frac {i}{b}} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1+i d x^2\right )\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 1.13, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsinh \left (d \,x^{2}-i\right )\right )^{\frac {5}{2}}\, dx\]
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(-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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Giac [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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {asinh}\left (d\,x^2-\mathrm {i}\right )\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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