Optimal. Leaf size=339 \[ \frac{\sqrt{2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac{x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac{\sqrt{2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{7/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{7/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )} \]
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Rubi [A] time = 0.0984169, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {4828, 4822} \[ \frac{\sqrt{2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac{x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac{\sqrt{2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{7/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{7/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )} \]
Antiderivative was successfully verified.
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Rule 4828
Rule 4822
Rubi steps
\begin{align*} \int \frac{1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{7/2}} \, dx &=-\frac{\sqrt{2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac{x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}-\frac{\int \frac{1}{\left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}} \, dx}{15 b^2}\\ &=-\frac{\sqrt{2 d x^2-d^2 x^4}}{5 b d x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}+\frac{x}{15 b^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{3/2}}+\frac{\sqrt{2 d x^2-d^2 x^4}}{15 b^3 d x \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}+\frac{\left (-\frac{1}{b}\right )^{7/2} \sqrt{\pi } x C\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}-\frac{\left (-\frac{1}{b}\right )^{7/2} \sqrt{\pi } x S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right ) \left (\cos \left (\frac{a}{2 b}\right )+\sin \left (\frac{a}{2 b}\right )\right )}{15 \left (\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )\right )}\\ \end{align*}
Mathematica [A] time = 0.904952, size = 319, normalized size = 0.94 \[ \frac{\frac{\sqrt{\pi } \left (-\frac{1}{b}\right )^{3/2} x \left (\cos \left (\frac{a}{2 b}\right )-\sin \left (\frac{a}{2 b}\right )\right ) \text{FresnelC}\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{\sqrt{\pi } b \left (-\frac{1}{b}\right )^{5/2} x \left (\sin \left (\frac{a}{2 b}\right )+\cos \left (\frac{a}{2 b}\right )\right ) S\left (\frac{\sqrt{-\frac{1}{b}} \sqrt{a-b \sin ^{-1}\left (1-d x^2\right )}}{\sqrt{\pi }}\right )}{\cos \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )-\sin \left (\frac{1}{2} \sin ^{-1}\left (1-d x^2\right )\right )}+\frac{x^2 \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )+\frac{\sqrt{d x^2 \left (2-d x^2\right )} \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^2}{b d}-\frac{3 b \sqrt{d x^2 \left (2-d x^2\right )}}{d}}{x \left (a-b \sin ^{-1}\left (1-d x^2\right )\right )^{5/2}}}{15 b^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arcsin \left ( d{x}^{2}-1 \right ) \right ) ^{-{\frac{7}{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 \frac{1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac{7}{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 \frac{1}{{\left (b \arcsin \left (d x^{2} - 1\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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