Optimal. Leaf size=312 \[ -\frac {3 b \sqrt {2 i d x^2+d^2 x^4} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}+\frac {3 \sqrt {i b} b \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )}-\frac {3 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {i b} \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )} \]
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Rubi [A]
time = 0.08, antiderivative size = 312, 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, 4903}
\begin {gather*} -\frac {3 \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 {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right )}{\sqrt {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 {3 b \sqrt {d^2 x^4+2 i d x^2} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{d x}+\frac {3 \sqrt {\pi } \sqrt {i b} b x \left (-\sinh \left (\frac {a}{2 b}\right )+i \cosh \left (\frac {a}{2 b}\right )\right ) \text {FresnelC}\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi } \sqrt {i b}}\right )}{\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 )}+x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2} \end {gather*}
Antiderivative was successfully verified.
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Rule 4898
Rule 4903
Rubi steps
\begin {align*} \int \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2} \, dx &=-\frac {3 b \sqrt {2 i d x^2+d^2 x^4} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}+\left (3 b^2\right ) \int \frac {1}{\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}} \, dx\\ &=-\frac {3 b \sqrt {2 i d x^2+d^2 x^4} \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}+\frac {3 \sqrt {i b} b \sqrt {\pi } x C\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (i \cosh \left (\frac {a}{2 b}\right )-\sinh \left (\frac {a}{2 b}\right )\right )}{\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 )}-\frac {3 b^2 \sqrt {\pi } x S\left (\frac {\sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{\sqrt {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 = 0.15, size = 258, normalized size = 0.83 \begin {gather*} -\frac {3 b \sqrt {d x^2 \left (2 i+d x^2\right )} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{d x}+x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}+\frac {3 b^2 \sqrt {\pi } x \left (-S\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )-\text {FresnelC}\left (\frac {\sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {i b} \sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )\right )}{\sqrt {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 = 0.92, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsinh \left (d \,x^{2}+i\right )\right )^{\frac {3}{2}}\, dx\]
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+1{}\mathrm {i}\right )\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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