Optimal. Leaf size=128 \[ \frac{9 \tan (c+d x)}{5 a^3 d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{\tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{3 \tan (c+d x) \sec ^2(c+d x)}{5 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.264708, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3816, 4019, 4008, 3787, 3770, 3767, 8} \[ \frac{9 \tan (c+d x)}{5 a^3 d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{3 \tan (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{\tan (c+d x) \sec ^3(c+d x)}{5 d (a \sec (c+d x)+a)^3}-\frac{3 \tan (c+d x) \sec ^2(c+d x)}{5 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3816
Rule 4019
Rule 4008
Rule 3787
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int \frac{\sec ^5(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{\sec ^3(c+d x) (3 a-6 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\sec ^2(c+d x) \left (18 a^2-27 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \sec (c+d x) \left (-45 a^3+27 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{9 \int \sec ^2(c+d x) \, dx}{5 a^3}-\frac{3 \int \sec (c+d x) \, dx}{a^3}\\ &=-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{9 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{5 a^3 d}\\ &=-\frac{3 \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{9 \tan (c+d x)}{5 a^3 d}-\frac{\sec ^3(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sec ^2(c+d x) \tan (c+d x)}{5 a d (a+a \sec (c+d x))^2}+\frac{3 \tan (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [B] time = 1.25505, size = 294, normalized size = 2.3 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (8 \tan \left (\frac{c}{2}\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+20 \cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (\frac{\sin (d x)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+76 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right )+8 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{5 a^3 d (\sec (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.039, size = 139, normalized size = 1.1 \begin{align*}{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}}-{\frac{1}{d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+3\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.19071, size = 223, normalized size = 1.74 \begin{align*} \frac{\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac{a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} + \frac{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{20 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.7734, size = 506, normalized size = 3.95 \begin{align*} -\frac{15 \,{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (\cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (24 \, \cos \left (d x + c\right )^{3} + 57 \, \cos \left (d x + c\right )^{2} + 39 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{10 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\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*} \frac{\int \frac{\sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.4012, size = 165, normalized size = 1.29 \begin{align*} -\frac{\frac{60 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac{a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{20 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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