Optimal. Leaf size=348 \[ -\frac{15 \sqrt{\pi } \sqrt{-\frac{i}{b}} b^3 x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac{15 \sqrt{\pi } b^2 x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )}{\sqrt{-\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+15 b^2 x \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}-\frac{5 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2} \]
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Rubi [A] time = 0.111911, antiderivative size = 348, normalized size of antiderivative = 1., 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 = {4814, 4811} \[ -\frac{15 \sqrt{\pi } \sqrt{-\frac{i}{b}} b^3 x \left (\sinh \left (\frac{a}{2 b}\right )+i \cosh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )}+\frac{15 \sqrt{\pi } b^2 x \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )}{\sqrt{-\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+15 b^2 x \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}-\frac{5 b \sqrt{d^2 x^4+2 i d x^2} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}{d x}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2} \]
Antiderivative was successfully verified.
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Rule 4814
Rule 4811
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 \sqrt{-\frac{i}{b}} b^3 \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 (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 )}\\ \end{align*}
Mathematica [A] time = 0.270713, size = 337, normalized size = 0.97 \[ \frac{15 b^2 x \left (-\sqrt{\pi } \left (\cosh \left (\frac{a}{2 b}\right )-i \sinh \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )+\sqrt{\pi } \left (\cosh \left (\frac{a}{2 b}\right )+i \sinh \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{i}{b}} \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}}{\sqrt{\pi }}\right )+\sqrt{-\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right ) \sqrt{a+i b \sin ^{-1}\left (1-i d x^2\right )}\right )}{\sqrt{-\frac{i}{b}} \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-i d x^2\right )\right )\right )}+x \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{5/2}-\frac{5 b \sqrt{d x^2 \left (d x^2+2 i\right )} \left (a+i b \sin ^{-1}\left (1-i d x^2\right )\right )^{3/2}}{d x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.102, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( i+d{x}^{2} \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arsinh}\left (d x^{2} + i\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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