Optimal. Leaf size=96 \[ -\frac{e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}+\sqrt{x} \]
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Rubi [A] time = 0.0681866, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3402, 3358, 3356, 2208} \[ -\frac{e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}+\sqrt{x} \]
Antiderivative was successfully verified.
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Rule 3402
Rule 3358
Rule 3356
Rule 2208
Rubi steps
\begin{align*} \int \frac{\cos ^2\left (a+b x^2\right )}{\sqrt{x}} \, dx &=2 \operatorname{Subst}\left (\int \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{1}{2} \cos \left (2 a+2 b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}+\operatorname{Subst}\left (\int \cos \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}+\frac{1}{2} \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} \, dx,x,\sqrt{x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}-\frac{e^{2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}\\ \end{align*}
Mathematica [A] time = 0.232448, size = 120, normalized size = 1.25 \[ -\frac{\sqrt{x} \left (2^{3/4} \sqrt [4]{-i b x^2} (\cos (2 a)-i \sin (2 a)) \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )+2^{3/4} \sqrt [4]{i b x^2} (\cos (2 a)+i \sin (2 a)) \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )-16 \sqrt [4]{b^2 x^4}\right )}{16 \sqrt [4]{b^2 x^4}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}{\frac{1}{\sqrt{x}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.45108, size = 393, normalized size = 4.09 \begin{align*} -\frac{2^{\frac{3}{4}}{\left ({\left ({\left ({\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (-i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left ({\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \sqrt{x} - 16 \cdot 2^{\frac{1}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}} \sqrt{x}\right )}}{32 \, \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.76987, size = 170, normalized size = 1.77 \begin{align*} \frac{i \, \left (2 i \, b\right )^{\frac{3}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \left (-2 i \, b\right )^{\frac{3}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right ) + 16 \, b \sqrt{x}}{16 \, b} \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 \frac{\cos ^{2}{\left (a + b x^{2} \right )}}{\sqrt{x}}\, 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 \frac{\cos \left (b x^{2} + a\right )^{2}}{\sqrt{x}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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