Optimal. Leaf size=184 \[ -\frac{2 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}+\frac{2 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}-\frac{2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x} \]
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Rubi [A] time = 0.0215428, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {4812} \[ -\frac{2 \sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}+\frac{2 \sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )}{\sqrt{\frac{1}{b}} d x}-\frac{2 \sin ^2\left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{d x} \]
Antiderivative was successfully verified.
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Rule 4812
Rubi steps
\begin{align*} \int \sqrt{a+b \cos ^{-1}\left (1+d x^2\right )} \, dx &=\frac{2 \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 )}{\sqrt{\frac{1}{b}} d x}-\frac{2 \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 )}{\sqrt{\frac{1}{b}} d x}-\frac{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 = 0.0848811, size = 157, normalized size = 0.85 \[ -\frac{2 \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \left (\sqrt{\pi } \sin \left (\frac{a}{2 b}\right ) \text{FresnelC}\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )-\sqrt{\pi } \cos \left (\frac{a}{2 b}\right ) S\left (\frac{\sqrt{\frac{1}{b}} \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}}{\sqrt{\pi }}\right )+\sqrt{\frac{1}{b}} \sin \left (\frac{1}{2} \cos ^{-1}\left (d x^2+1\right )\right ) \sqrt{a+b \cos ^{-1}\left (d x^2+1\right )}\right )}{\sqrt{\frac{1}{b}} d x} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.063, size = 0, normalized size = 0. \begin{align*} \int \sqrt{a+b\arccos \left ( d{x}^{2}+1 \right ) }\, 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 \sqrt{b \arccos \left (d x^{2} + 1\right ) + a}\,{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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \operatorname{acos}{\left (d x^{2} + 1 \right )}}\, dx \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 \sqrt{b \arccos \left (d x^{2} + 1\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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