Optimal. Leaf size=389 \[ -\frac {\sqrt {2 i d x^2+d^2 x^4}}{5 b d x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{5/2}}-\frac {x}{15 b^2 \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}}-\frac {\sqrt {2 i d x^2+d^2 x^4}}{15 b^3 d x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \sqrt {\pi } x \text {FresnelC}\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )-i \sinh \left (\frac {a}{2 b}\right )\right )}{15 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )}+\frac {\left (-\frac {i}{b}\right )^{3/2} \sqrt {\pi } x S\left (\frac {\sqrt {-\frac {i}{b}} \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}{\sqrt {\pi }}\right ) \left (\cosh \left (\frac {a}{2 b}\right )+i \sinh \left (\frac {a}{2 b}\right )\right )}{15 b^2 \left (\cos \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )-\sin \left (\frac {1}{2} \text {ArcSin}\left (1-i d x^2\right )\right )\right )} \]
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Rubi [A]
time = 0.06, antiderivative size = 389, 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 = {4912, 4906}
\begin {gather*} -\frac {\sqrt {d^2 x^4+2 i d x^2}}{15 b^3 d x \sqrt {a+i b \text {ArcSin}\left (1-i d x^2\right )}}-\frac {\sqrt {\pi } \left (-\frac {i}{b}\right )^{3/2} 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 )}{15 b^2 \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 {\sqrt {\pi } \left (-\frac {i}{b}\right )^{3/2} 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 (1-i d x^2\right )}}{\sqrt {\pi }}\right )}{15 b^2 \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 {x}{15 b^2 \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{3/2}}-\frac {\sqrt {d^2 x^4+2 i d x^2}}{5 b d 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 4906
Rule 4912
Rubi steps
\begin {align*} \int \frac {1}{\left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{7/2}} \, dx &=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{5 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2}}-\frac {x}{15 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}+\frac {\int \frac {1}{\left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac {\sqrt {2 i d x^2+d^2 x^4}}{5 b d x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2}}-\frac {x}{15 b^2 \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}-\frac {\sqrt {2 i d x^2+d^2 x^4}}{15 b^3 d x \sqrt {a+i b \sin ^{-1}\left (1-i d x^2\right )}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \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 )}{15 b^2 \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 {\left (-\frac {i}{b}\right )^{3/2} \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 )}{15 b^2 \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 = 0.65, size = 365, normalized size = 0.94 \begin {gather*} \frac {\frac {-\frac {3 b \sqrt {d x^2 \left (2 i+d x^2\right )}}{d}-x^2 \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )+\frac {\sqrt {d x^2 \left (2 i+d x^2\right )} \left (-i a+b \text {ArcSin}\left (1-i d x^2\right )\right )^2}{b d}}{x \left (a+i b \text {ArcSin}\left (1-i d x^2\right )\right )^{5/2}}-\frac {\left (-\frac {i}{b}\right )^{3/2} \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 )}{\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 {\left (-\frac {i}{b}\right )^{3/2} \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 )}{\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 )}}{15 b^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.93, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d \,x^{2}+i\right )\right )^{\frac {7}{2}}}\, dx\]
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 \frac {1}{{\left (a+b\,\mathrm {asinh}\left (d\,x^2+1{}\mathrm {i}\right )\right )}^{7/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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