Optimal. Leaf size=249 \[ \frac{30 b^2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x}-\frac{5 b \sqrt{-d^2 x^4-2 d x^2} \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac{30 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}-\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2} \]
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Rubi [A] time = 0.0949448, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4815, 4812} \[ \frac{30 b^2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x}-\frac{5 b \sqrt{-d^2 x^4-2 d x^2} \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{3/2}}{d x}+\frac{30 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}-\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+x \left (a+b \cos ^{-1}\left (d x^2+1\right )\right )^{5/2} \]
Antiderivative was successfully verified.
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Rule 4815
Rule 4812
Rubi steps
\begin{align*} \int \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{5/2} \, dx &=-\frac{5 b \sqrt{-2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{5/2}-\left (15 b^2\right ) \int \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )} \, dx\\ &=-\frac{5 b \sqrt{-2 d x^2-d^2 x^4} \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{3/2}}{d x}+x \left (a+b \cos ^{-1}\left (1+d x^2\right )\right )^{5/2}-\frac{30 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}{\sqrt{\pi }}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+\frac{30 \sqrt{\pi } C\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )}}{\sqrt{\pi }}\right ) \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{\left (\frac{1}{b}\right )^{5/2} d x}+\frac{30 b^2 \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )} \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (1+d x^2\right )\right )}{d x}\\ \end{align*}
Mathematica [A] time = 2.61709, size = 256, normalized size = 1.03 \[ -\frac{2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt{a+b \cos ^{-1}\left (d x^2+1\right )} \left (\left (a^2-15 b^2\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+5 a b \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+b \cos ^{-1}\left (d x^2+1\right ) \left (2 a \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )+5 b \cos \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )+b^2 \cos ^{-1}\left (d x^2+1\right )^2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right )\right )-\frac{15 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2}}+\frac{15 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\left (\frac{1}{b}\right )^{5/2}}\right )}{d x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.076, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\arccos \left ( d{x}^{2}+1 \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 \arccos \left (d x^{2} + 1\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 \arccos \left (d x^{2} + 1\right ) + a\right )}^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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