Optimal. Leaf size=210 \[ -\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {5 a^3}{256 d (a \sin (c+d x)+a)^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}-\frac {5 a^2}{128 d (a \sin (c+d x)+a)^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}-\frac {35 a}{512 d (a \sin (c+d x)+a)^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {35}{256 d (a \sin (c+d x)+a)}+\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d} \]
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Rubi [A] time = 0.17, antiderivative size = 210, 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} \[ -\frac {a^4}{160 d (a \sin (c+d x)+a)^5}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}-\frac {5 a^3}{256 d (a \sin (c+d x)+a)^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}-\frac {5 a^2}{128 d (a \sin (c+d x)+a)^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}-\frac {35 a}{512 d (a \sin (c+d x)+a)^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {35}{256 d (a \sin (c+d x)+a)}+\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d} \]
Antiderivative was successfully verified.
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Rule 44
Rule 206
Rule 2667
Rubi steps
\begin {align*} \int \frac {\sec ^9(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {a^9 \operatorname {Subst}\left (\int \frac {1}{(a-x)^5 (a+x)^6} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^9 \operatorname {Subst}\left (\int \left (\frac {1}{64 a^6 (a-x)^5}+\frac {3}{64 a^7 (a-x)^4}+\frac {21}{256 a^8 (a-x)^3}+\frac {7}{64 a^9 (a-x)^2}+\frac {1}{32 a^5 (a+x)^6}+\frac {5}{64 a^6 (a+x)^5}+\frac {15}{128 a^7 (a+x)^4}+\frac {35}{256 a^8 (a+x)^3}+\frac {35}{256 a^9 (a+x)^2}+\frac {63}{256 a^9 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35 a}{512 d (a+a \sin (c+d x))^2}-\frac {35}{256 d (a+a \sin (c+d x))}+\frac {63 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{256 d}\\ &=\frac {63 \tanh ^{-1}(\sin (c+d x))}{256 a d}+\frac {a^3}{256 d (a-a \sin (c+d x))^4}+\frac {a^2}{64 d (a-a \sin (c+d x))^3}+\frac {21 a}{512 d (a-a \sin (c+d x))^2}+\frac {7}{64 d (a-a \sin (c+d x))}-\frac {a^4}{160 d (a+a \sin (c+d x))^5}-\frac {5 a^3}{256 d (a+a \sin (c+d x))^4}-\frac {5 a^2}{128 d (a+a \sin (c+d x))^3}-\frac {35 a}{512 d (a+a \sin (c+d x))^2}-\frac {35}{256 d (a+a \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 1.40, size = 165, normalized size = 0.79 \[ \frac {\sec ^8(c+d x) \left (-315 \sin ^8(c+d x)-315 \sin ^7(c+d x)+1155 \sin ^6(c+d x)+1155 \sin ^5(c+d x)-1533 \sin ^4(c+d x)-1533 \sin ^3(c+d x)+837 \sin ^2(c+d x)+837 \sin (c+d x)+315 \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^8 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^{10}-128\right )}{1280 a d (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.52, size = 187, normalized size = 0.89 \[ -\frac {630 \, \cos \left (d x + c\right )^{8} - 210 \, \cos \left (d x + c\right )^{6} - 84 \, \cos \left (d x + c\right )^{4} - 48 \, \cos \left (d x + c\right )^{2} - 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 315 \, {\left (\cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{8}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (105 \, \cos \left (d x + c\right )^{6} + 70 \, \cos \left (d x + c\right )^{4} + 56 \, \cos \left (d x + c\right )^{2} + 48\right )} \sin \left (d x + c\right ) - 32}{2560 \, {\left (a d \cos \left (d x + c\right )^{8} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{8}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.24, size = 156, normalized size = 0.74 \[ \frac {\frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {1260 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {5 \, {\left (525 \, \sin \left (d x + c\right )^{4} - 2324 \, \sin \left (d x + c\right )^{3} + 3906 \, \sin \left (d x + c\right )^{2} - 2972 \, \sin \left (d x + c\right ) + 873\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{4}} - \frac {2877 \, \sin \left (d x + c\right )^{5} + 15785 \, \sin \left (d x + c\right )^{4} + 35070 \, \sin \left (d x + c\right )^{3} + 39670 \, \sin \left (d x + c\right )^{2} + 23085 \, \sin \left (d x + c\right ) + 5641}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{5}}}{10240 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.41, size = 198, normalized size = 0.94 \[ \frac {1}{256 a d \left (\sin \left (d x +c \right )-1\right )^{4}}-\frac {1}{64 a d \left (\sin \left (d x +c \right )-1\right )^{3}}+\frac {21}{512 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {7}{64 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {63 \ln \left (\sin \left (d x +c \right )-1\right )}{512 a d}-\frac {1}{160 a d \left (1+\sin \left (d x +c \right )\right )^{5}}-\frac {5}{256 a d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{128 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {35}{512 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {35}{256 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {63 \ln \left (1+\sin \left (d x +c \right )\right )}{512 a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 214, normalized size = 1.02 \[ -\frac {\frac {2 \, {\left (315 \, \sin \left (d x + c\right )^{8} + 315 \, \sin \left (d x + c\right )^{7} - 1155 \, \sin \left (d x + c\right )^{6} - 1155 \, \sin \left (d x + c\right )^{5} + 1533 \, \sin \left (d x + c\right )^{4} + 1533 \, \sin \left (d x + c\right )^{3} - 837 \, \sin \left (d x + c\right )^{2} - 837 \, \sin \left (d x + c\right ) + 128\right )}}{a \sin \left (d x + c\right )^{9} + a \sin \left (d x + c\right )^{8} - 4 \, a \sin \left (d x + c\right )^{7} - 4 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} + 6 \, a \sin \left (d x + c\right )^{4} - 4 \, a \sin \left (d x + c\right )^{3} - 4 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {315 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {315 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{2560 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 9.41, size = 199, normalized size = 0.95 \[ \frac {63\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{256\,a\,d}-\frac {\frac {63\,{\sin \left (c+d\,x\right )}^8}{256}+\frac {63\,{\sin \left (c+d\,x\right )}^7}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^6}{256}-\frac {231\,{\sin \left (c+d\,x\right )}^5}{256}+\frac {1533\,{\sin \left (c+d\,x\right )}^4}{1280}+\frac {1533\,{\sin \left (c+d\,x\right )}^3}{1280}-\frac {837\,{\sin \left (c+d\,x\right )}^2}{1280}-\frac {837\,\sin \left (c+d\,x\right )}{1280}+\frac {1}{10}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^9+a\,{\sin \left (c+d\,x\right )}^8-4\,a\,{\sin \left (c+d\,x\right )}^7-4\,a\,{\sin \left (c+d\,x\right )}^6+6\,a\,{\sin \left (c+d\,x\right )}^5+6\,a\,{\sin \left (c+d\,x\right )}^4-4\,a\,{\sin \left (c+d\,x\right )}^3-4\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]
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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