Optimal. Leaf size=105 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {29 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {7 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
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Rubi [A] time = 0.22, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3816, 4008, 3998, 3770, 3794} \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {29 \tan (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {7 \tan (c+d x)}{15 a d (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 3770
Rule 3794
Rule 3816
Rule 3998
Rule 4008
Rubi steps
\begin {align*} \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^2(c+d x) (2 a-5 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\sec (c+d x) \left (-14 a^2+15 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \sec (c+d x) \, dx}{a^3}-\frac {29 \int \frac {\sec (c+d x)}{a+a \sec (c+d x)} \, dx}{15 a^2}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {29 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.51, size = 209, normalized size = 1.99 \[ -\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (14 \tan \left (\frac {c}{2}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+3 \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+60 \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+88 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+14 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d (\sec (c+d x)+1)^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 158, normalized size = 1.50 \[ \frac {15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (22 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.93, size = 94, normalized size = 0.90 \[ \frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.41, size = 96, normalized size = 0.91 \[ -\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d \,a^{3}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \,a^{3}}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.75, size = 119, normalized size = 1.13 \[ -\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.69, size = 58, normalized size = 0.55 \[ -\frac {105\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{60\,a^3\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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