Optimal. Leaf size=146 \[ -\frac {a^2}{32 d (a \sin (c+d x)+a)^4}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {3}{32 d (a \sin (c+d x)+a)^2} \]
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Rubi [A] time = 0.11, antiderivative size = 146, 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} \[ -\frac {a^2}{32 d (a \sin (c+d x)+a)^4}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}-\frac {a}{16 d (a \sin (c+d x)+a)^3}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {3}{32 d (a \sin (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 44
Rule 206
Rule 2667
Rubi steps
\begin {align*} \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^5} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{32 a^5 (a-x)^3}+\frac {5}{64 a^6 (a-x)^2}+\frac {1}{8 a^3 (a+x)^5}+\frac {3}{16 a^4 (a+x)^4}+\frac {3}{16 a^5 (a+x)^3}+\frac {5}{32 a^6 (a+x)^2}+\frac {15}{64 a^6 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{64 a d}\\ &=\frac {15 \tanh ^{-1}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.33, size = 137, normalized size = 0.94 \[ \frac {(1-\sin (c+d x))^2 (\sin (c+d x)+1)^2 \sec ^4(c+d x) \left (\frac {5}{64 (1-\sin (c+d x))}-\frac {5}{32 (\sin (c+d x)+1)}+\frac {1}{64 (1-\sin (c+d x))^2}-\frac {3}{32 (\sin (c+d x)+1)^2}-\frac {1}{16 (\sin (c+d x)+1)^3}-\frac {1}{32 (\sin (c+d x)+1)^4}+\frac {15}{64} \tanh ^{-1}(\sin (c+d x))\right )}{a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.85, size = 198, normalized size = 1.36 \[ \frac {60 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) - 8}{128 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.54, size = 126, normalized size = 0.86 \[ \frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{2} - 110 \, \sin \left (d x + c\right ) + 69\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 580 \, \sin \left (d x + c\right )^{3} + 1038 \, \sin \left (d x + c\right )^{2} + 868 \, \sin \left (d x + c\right ) + 301}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{512 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 144, normalized size = 0.99 \[ \frac {1}{64 a^{2} d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 a^{2} d \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128 a^{2} d}-\frac {1}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 a^{2} d \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128 a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.37, size = 167, normalized size = 1.14 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} + 30 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{3} - 50 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{128 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.19, size = 151, normalized size = 1.03 \[ \frac {15\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{64\,a^2\,d}+\frac {-\frac {15\,{\sin \left (c+d\,x\right )}^5}{64}-\frac {15\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {25\,{\sin \left (c+d\,x\right )}^2}{32}+\frac {17\,\sin \left (c+d\,x\right )}{64}-\frac {1}{4}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]
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 \[ \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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