Optimal. Leaf size=200 \[ \frac {181 \tan (c+d x)}{63 a^5 d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {5 \tan (c+d x)}{d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {67 \tan (c+d x) \sec ^2(c+d x)}{63 a^3 d (a \sec (c+d x)+a)^2}-\frac {29 \tan (c+d x) \sec ^3(c+d x)}{63 a^2 d (a \sec (c+d x)+a)^3}-\frac {\tan (c+d x) \sec ^5(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {5 \tan (c+d x) \sec ^4(c+d x)}{21 a d (a \sec (c+d x)+a)^4} \]
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Rubi [A] time = 0.48, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {181 \tan (c+d x)}{63 a^5 d}-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {29 \tan (c+d x) \sec ^3(c+d x)}{63 a^2 d (a \sec (c+d x)+a)^3}-\frac {67 \tan (c+d x) \sec ^2(c+d x)}{63 a^3 d (a \sec (c+d x)+a)^2}+\frac {5 \tan (c+d x)}{d \left (a^5 \sec (c+d x)+a^5\right )}-\frac {\tan (c+d x) \sec ^5(c+d x)}{9 d (a \sec (c+d x)+a)^5}-\frac {5 \tan (c+d x) \sec ^4(c+d x)}{21 a d (a \sec (c+d x)+a)^4} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3787
Rule 3816
Rule 4008
Rule 4019
Rubi steps
\begin {align*} \int \frac {\sec ^7(c+d x)}{(a+a \sec (c+d x))^5} \, dx &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {\int \frac {\sec ^5(c+d x) (5 a-10 a \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {\int \frac {\sec ^4(c+d x) \left (60 a^2-85 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx}{63 a^4}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\sec ^3(c+d x) \left (435 a^3-570 a^3 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{315 a^6}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\sec ^2(c+d x) \left (2010 a^4-2715 a^4 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{945 a^8}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}+\frac {\int \sec (c+d x) \left (-4725 a^5+2715 a^5 \sec (c+d x)\right ) \, dx}{945 a^{10}}\\ &=-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}+\frac {181 \int \sec ^2(c+d x) \, dx}{63 a^5}-\frac {5 \int \sec (c+d x) \, dx}{a^5}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}-\frac {181 \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{63 a^5 d}\\ &=-\frac {5 \tanh ^{-1}(\sin (c+d x))}{a^5 d}+\frac {181 \tan (c+d x)}{63 a^5 d}-\frac {\sec ^5(c+d x) \tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {5 \sec ^4(c+d x) \tan (c+d x)}{21 a d (a+a \sec (c+d x))^4}-\frac {29 \sec ^3(c+d x) \tan (c+d x)}{63 a^2 d (a+a \sec (c+d x))^3}-\frac {67 \sec ^2(c+d x) \tan (c+d x)}{63 a^3 d (a+a \sec (c+d x))^2}+\frac {5 \tan (c+d x)}{d \left (a^5+a^5 \sec (c+d x)\right )}\\ \end {align*}
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Mathematica [B] time = 1.97, size = 401, normalized size = 2.00 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (\sec \left (\frac {c}{2}\right ) \sec (c) \left (-56952 \sin \left (c-\frac {d x}{2}\right )+43722 \sin \left (c+\frac {d x}{2}\right )-47208 \sin \left (2 c+\frac {d x}{2}\right )-18144 \sin \left (c+\frac {3 d x}{2}\right )+41796 \sin \left (2 c+\frac {3 d x}{2}\right )-28350 \sin \left (3 c+\frac {3 d x}{2}\right )+34578 \sin \left (c+\frac {5 d x}{2}\right )-5691 \sin \left (2 c+\frac {5 d x}{2}\right )+28719 \sin \left (3 c+\frac {5 d x}{2}\right )-11550 \sin \left (4 c+\frac {5 d x}{2}\right )+15517 \sin \left (2 c+\frac {7 d x}{2}\right )-504 \sin \left (3 c+\frac {7 d x}{2}\right )+13186 \sin \left (4 c+\frac {7 d x}{2}\right )-2835 \sin \left (5 c+\frac {7 d x}{2}\right )+4149 \sin \left (3 c+\frac {9 d x}{2}\right )+252 \sin \left (4 c+\frac {9 d x}{2}\right )+3582 \sin \left (5 c+\frac {9 d x}{2}\right )-315 \sin \left (6 c+\frac {9 d x}{2}\right )+496 \sin \left (4 c+\frac {11 d x}{2}\right )+63 \sin \left (5 c+\frac {11 d x}{2}\right )+433 \sin \left (6 c+\frac {11 d x}{2}\right )-33978 \sin \left (\frac {d x}{2}\right )+52002 \sin \left (\frac {3 d x}{2}\right )\right ) \sec (c+d x)+322560 \cos ^9\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2016 a^5 d (\sec (c+d x)+1)^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 1.31, size = 278, normalized size = 1.39 \[ -\frac {315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{6} + 5 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 5 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (496 \, \cos \left (d x + c\right )^{5} + 2165 \, \cos \left (d x + c\right )^{4} + 3633 \, \cos \left (d x + c\right )^{3} + 2840 \, \cos \left (d x + c\right )^{2} + 946 \, \cos \left (d x + c\right ) + 63\right )} \sin \left (d x + c\right )}{126 \, {\left (a^{5} d \cos \left (d x + c\right )^{6} + 5 \, a^{5} d \cos \left (d x + c\right )^{5} + 10 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 5 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d \cos \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.98, size = 155, normalized size = 0.78 \[ -\frac {\frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac {5040 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} + \frac {2016 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{5}} - \frac {7 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 72 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 378 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1512 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8127 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{1008 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 177, normalized size = 0.88 \[ \frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{144 d \,a^{5}}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{14 d \,a^{5}}+\frac {3 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{5}}+\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{5}}+\frac {129 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \,a^{5}}-\frac {1}{d \,a^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{5}}-\frac {1}{d \,a^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.76, size = 206, normalized size = 1.03 \[ \frac {\frac {2016 \, \sin \left (d x + c\right )}{{\left (a^{5} - \frac {a^{5} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {8127 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1512 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {378 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {72 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac {5040 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{1008 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.72, size = 149, normalized size = 0.74 \[ \frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2\,a^5\,d}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{14\,a^5\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{144\,a^5\,d}-\frac {10\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^5\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^5\right )}+\frac {129\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^5\,d} \]
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 \[ \frac {\int \frac {\sec ^{7}{\left (c + d x \right )}}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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