Optimal. Leaf size=99 \[ \frac{2 (a \sec (c+d x)+a)^{5/2}}{5 a^2 d}-\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a d}-\frac{2 \sqrt{a \sec (c+d x)+a}}{d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d} \]
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Rubi [A] time = 0.0837909, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3880, 80, 50, 63, 207} \[ \frac{2 (a \sec (c+d x)+a)^{5/2}}{5 a^2 d}-\frac{2 (a \sec (c+d x)+a)^{3/2}}{3 a d}-\frac{2 \sqrt{a \sec (c+d x)+a}}{d}+\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 3880
Rule 80
Rule 50
Rule 63
Rule 207
Rubi steps
\begin{align*} \int \sqrt{a+a \sec (c+d x)} \tan ^3(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(-a+a x) (a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{(a+a x)^{3/2}}{x} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac{2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+a x}}{x} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{2 \sqrt{a+a \sec (c+d x)}}{d}-\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac{2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{2 \sqrt{a+a \sec (c+d x)}}{d}-\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac{2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{2 \sqrt{a+a \sec (c+d x)}}{d}-\frac{2 (a+a \sec (c+d x))^{3/2}}{3 a d}+\frac{2 (a+a \sec (c+d x))^{5/2}}{5 a^2 d}\\ \end{align*}
Mathematica [A] time = 0.156921, size = 80, normalized size = 0.81 \[ \frac{2 \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{\sec (c+d x)+1} \left (3 \sec ^2(c+d x)+\sec (c+d x)-17\right )+15 \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )\right )}{15 d \sqrt{\sec (c+d x)+1}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.235, size = 221, normalized size = 2.2 \begin{align*} -{\frac{1}{60\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}+30\,\cos \left ( dx+c \right ) \sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}+15\,\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}+136\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-8\,\cos \left ( dx+c \right ) -24 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.88456, size = 684, normalized size = 6.91 \begin{align*} \left [\frac{15 \, \sqrt{a} \cos \left (d x + c\right )^{2} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \,{\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \,{\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{30 \, d \cos \left (d x + c\right )^{2}}, -\frac{15 \, \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{2} + 2 \,{\left (17 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}}\right ] \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 \left (\sec{\left (c + d x \right )} + 1\right )} \tan ^{3}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 5.126, size = 205, normalized size = 2.07 \begin{align*} -\frac{\sqrt{2}{\left (\frac{15 \, \sqrt{2} a^{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a}} + \frac{2 \,{\left (15 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} a^{2} - 10 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} a^{3} - 12 \, a^{4}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{15 \, a d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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