Optimal. Leaf size=238 \[ \frac {1}{1024 d \left (a^8-a^8 \sin (c+d x)\right )}-\frac {9}{1024 d \left (a^8 \sin (c+d x)+a^8\right )}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{512 a^8 d}-\frac {7}{768 a^5 d (a \sin (c+d x)+a)^3}-\frac {1}{128 d \left (a^4 \sin (c+d x)+a^4\right )^2}-\frac {1}{64 a^3 d (a \sin (c+d x)+a)^5}-\frac {3}{256 d \left (a^2 \sin (c+d x)+a^2\right )^4}-\frac {1}{48 a^2 d (a \sin (c+d x)+a)^6}-\frac {a}{36 d (a \sin (c+d x)+a)^9}-\frac {1}{32 d (a \sin (c+d x)+a)^8}-\frac {3}{112 a d (a \sin (c+d x)+a)^7} \]
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Rubi [A] time = 0.17, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ \frac {1}{1024 d \left (a^8-a^8 \sin (c+d x)\right )}-\frac {9}{1024 d \left (a^8 \sin (c+d x)+a^8\right )}-\frac {1}{128 d \left (a^4 \sin (c+d x)+a^4\right )^2}-\frac {3}{256 d \left (a^2 \sin (c+d x)+a^2\right )^4}-\frac {1}{48 a^2 d (a \sin (c+d x)+a)^6}-\frac {1}{64 a^3 d (a \sin (c+d x)+a)^5}-\frac {7}{768 a^5 d (a \sin (c+d x)+a)^3}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{512 a^8 d}-\frac {a}{36 d (a \sin (c+d x)+a)^9}-\frac {1}{32 d (a \sin (c+d x)+a)^8}-\frac {3}{112 a d (a \sin (c+d x)+a)^7} \]
Antiderivative was successfully verified.
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Rule 44
Rule 206
Rule 2667
Rubi steps
\begin {align*} \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^8} \, dx &=\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{10}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {1}{1024 a^{10} (a-x)^2}+\frac {1}{4 a^2 (a+x)^{10}}+\frac {1}{4 a^3 (a+x)^9}+\frac {3}{16 a^4 (a+x)^8}+\frac {1}{8 a^5 (a+x)^7}+\frac {5}{64 a^6 (a+x)^6}+\frac {3}{64 a^7 (a+x)^5}+\frac {7}{256 a^8 (a+x)^4}+\frac {1}{64 a^9 (a+x)^3}+\frac {9}{1024 a^{10} (a+x)^2}+\frac {5}{512 a^{10} \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {a}{36 d (a+a \sin (c+d x))^9}-\frac {1}{32 d (a+a \sin (c+d x))^8}-\frac {3}{112 a d (a+a \sin (c+d x))^7}-\frac {1}{48 a^2 d (a+a \sin (c+d x))^6}-\frac {1}{64 a^3 d (a+a \sin (c+d x))^5}-\frac {7}{768 a^5 d (a+a \sin (c+d x))^3}-\frac {3}{256 d \left (a^2+a^2 \sin (c+d x)\right )^4}-\frac {1}{128 d \left (a^4+a^4 \sin (c+d x)\right )^2}+\frac {1}{1024 d \left (a^8-a^8 \sin (c+d x)\right )}-\frac {9}{1024 d \left (a^8+a^8 \sin (c+d x)\right )}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{512 a^7 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{512 a^8 d}-\frac {a}{36 d (a+a \sin (c+d x))^9}-\frac {1}{32 d (a+a \sin (c+d x))^8}-\frac {3}{112 a d (a+a \sin (c+d x))^7}-\frac {1}{48 a^2 d (a+a \sin (c+d x))^6}-\frac {1}{64 a^3 d (a+a \sin (c+d x))^5}-\frac {7}{768 a^5 d (a+a \sin (c+d x))^3}-\frac {3}{256 d \left (a^2+a^2 \sin (c+d x)\right )^4}-\frac {1}{128 d \left (a^4+a^4 \sin (c+d x)\right )^2}+\frac {1}{1024 d \left (a^8-a^8 \sin (c+d x)\right )}-\frac {9}{1024 d \left (a^8+a^8 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 1.83, size = 175, normalized size = 0.74 \[ -\frac {\sec ^2(c+d x) \left (-315 \sin ^9(c+d x)-2520 \sin ^8(c+d x)-8610 \sin ^7(c+d x)-15960 \sin ^6(c+d x)-16128 \sin ^5(c+d x)-5544 \sin ^4(c+d x)+7074 \sin ^3(c+d x)+11736 \sin ^2(c+d x)+9019 \sin (c+d x)-315 \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^{18}+5120\right )}{32256 a^8 d (\sin (c+d x)+1)^8} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.84, size = 446, normalized size = 1.87 \[ \frac {5040 \, \cos \left (d x + c\right )^{8} - 52080 \, \cos \left (d x + c\right )^{6} + 137088 \, \cos \left (d x + c\right )^{4} - 114624 \, \cos \left (d x + c\right )^{2} + 315 \, {\left (\cos \left (d x + c\right )^{10} - 32 \, \cos \left (d x + c\right )^{8} + 160 \, \cos \left (d x + c\right )^{6} - 256 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (\cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} + 24 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \, {\left (\cos \left (d x + c\right )^{10} - 32 \, \cos \left (d x + c\right )^{8} + 160 \, \cos \left (d x + c\right )^{6} - 256 \, \cos \left (d x + c\right )^{4} + 128 \, \cos \left (d x + c\right )^{2} - 8 \, {\left (\cos \left (d x + c\right )^{8} - 10 \, \cos \left (d x + c\right )^{6} + 24 \, \cos \left (d x + c\right )^{4} - 16 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (315 \, \cos \left (d x + c\right )^{8} - 9870 \, \cos \left (d x + c\right )^{6} + 43848 \, \cos \left (d x + c\right )^{4} - 52272 \, \cos \left (d x + c\right )^{2} + 8960\right )} \sin \left (d x + c\right ) + 14336}{64512 \, {\left (a^{8} d \cos \left (d x + c\right )^{10} - 32 \, a^{8} d \cos \left (d x + c\right )^{8} + 160 \, a^{8} d \cos \left (d x + c\right )^{6} - 256 \, a^{8} d \cos \left (d x + c\right )^{4} + 128 \, a^{8} d \cos \left (d x + c\right )^{2} - 8 \, {\left (a^{8} d \cos \left (d x + c\right )^{8} - 10 \, a^{8} d \cos \left (d x + c\right )^{6} + 24 \, a^{8} d \cos \left (d x + c\right )^{4} - 16 \, a^{8} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.57, size = 166, normalized size = 0.70 \[ \frac {\frac {2520 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{8}} - \frac {2520 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{8}} + \frac {504 \, {\left (5 \, \sin \left (d x + c\right ) - 6\right )}}{a^{8} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {7129 \, \sin \left (d x + c\right )^{9} + 68697 \, \sin \left (d x + c\right )^{8} + 296964 \, \sin \left (d x + c\right )^{7} + 758772 \, \sin \left (d x + c\right )^{6} + 1271214 \, \sin \left (d x + c\right )^{5} + 1465758 \, \sin \left (d x + c\right )^{4} + 1191540 \, \sin \left (d x + c\right )^{3} + 693828 \, \sin \left (d x + c\right )^{2} + 295425 \, \sin \left (d x + c\right ) + 89553}{a^{8} {\left (\sin \left (d x + c\right ) + 1\right )}^{9}}}{516096 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.34, size = 216, normalized size = 0.91 \[ -\frac {1}{1024 a^{8} d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{1024 a^{8} d}-\frac {1}{36 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{9}}-\frac {1}{32 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{8}}-\frac {3}{112 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{7}}-\frac {1}{48 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {1}{64 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {3}{256 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{768 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{128 a^{8} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {9}{1024 a^{8} d \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{1024 a^{8} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 248, normalized size = 1.04 \[ -\frac {\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{9} + 2520 \, \sin \left (d x + c\right )^{8} + 8610 \, \sin \left (d x + c\right )^{7} + 15960 \, \sin \left (d x + c\right )^{6} + 16128 \, \sin \left (d x + c\right )^{5} + 5544 \, \sin \left (d x + c\right )^{4} - 7074 \, \sin \left (d x + c\right )^{3} - 11736 \, \sin \left (d x + c\right )^{2} - 9019 \, \sin \left (d x + c\right ) - 5120\right )}}{a^{8} \sin \left (d x + c\right )^{10} + 8 \, a^{8} \sin \left (d x + c\right )^{9} + 27 \, a^{8} \sin \left (d x + c\right )^{8} + 48 \, a^{8} \sin \left (d x + c\right )^{7} + 42 \, a^{8} \sin \left (d x + c\right )^{6} - 42 \, a^{8} \sin \left (d x + c\right )^{4} - 48 \, a^{8} \sin \left (d x + c\right )^{3} - 27 \, a^{8} \sin \left (d x + c\right )^{2} - 8 \, a^{8} \sin \left (d x + c\right ) - a^{8}} - \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{8}} + \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{8}}}{64512 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.49, size = 231, normalized size = 0.97 \[ \frac {\frac {5\,{\sin \left (c+d\,x\right )}^9}{512}+\frac {5\,{\sin \left (c+d\,x\right )}^8}{64}+\frac {205\,{\sin \left (c+d\,x\right )}^7}{768}+\frac {95\,{\sin \left (c+d\,x\right )}^6}{192}+\frac {{\sin \left (c+d\,x\right )}^5}{2}+\frac {11\,{\sin \left (c+d\,x\right )}^4}{64}-\frac {393\,{\sin \left (c+d\,x\right )}^3}{1792}-\frac {163\,{\sin \left (c+d\,x\right )}^2}{448}-\frac {9019\,\sin \left (c+d\,x\right )}{32256}-\frac {10}{63}}{d\,\left (-a^8\,{\sin \left (c+d\,x\right )}^{10}-8\,a^8\,{\sin \left (c+d\,x\right )}^9-27\,a^8\,{\sin \left (c+d\,x\right )}^8-48\,a^8\,{\sin \left (c+d\,x\right )}^7-42\,a^8\,{\sin \left (c+d\,x\right )}^6+42\,a^8\,{\sin \left (c+d\,x\right )}^4+48\,a^8\,{\sin \left (c+d\,x\right )}^3+27\,a^8\,{\sin \left (c+d\,x\right )}^2+8\,a^8\,\sin \left (c+d\,x\right )+a^8\right )}+\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{512\,a^8\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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