Optimal. Leaf size=120 \[ -\frac {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d} \]
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Rubi [A] time = 0.11, antiderivative size = 120, 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} \[ -\frac {a^2}{24 d (a \sin (c+d x)+a)^3}+\frac {a}{32 d (a-a \sin (c+d x))^2}-\frac {3 a}{32 d (a \sin (c+d x)+a)^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {3}{16 d (a \sin (c+d x)+a)}+\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d} \]
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)} \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^4} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{16 a^4 (a-x)^3}+\frac {1}{8 a^5 (a-x)^2}+\frac {1}{8 a^3 (a+x)^4}+\frac {3}{16 a^4 (a+x)^3}+\frac {3}{16 a^5 (a+x)^2}+\frac {5}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=\frac {5 \tanh ^{-1}(\sin (c+d x))}{16 a d}+\frac {a}{32 d (a-a \sin (c+d x))^2}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a^2}{24 d (a+a \sin (c+d x))^3}-\frac {3 a}{32 d (a+a \sin (c+d x))^2}-\frac {3}{16 d (a+a \sin (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.15, size = 97, normalized size = 0.81 \[ \frac {\sec ^4(c+d x) \left (-15 \sin ^4(c+d x)-15 \sin ^3(c+d x)+25 \sin ^2(c+d x)+25 \sin (c+d x)+15 (\sin (c+d x)-1)^2 (\sin (c+d x)+1)^3 \tanh ^{-1}(\sin (c+d x))-8\right )}{48 a d (\sin (c+d x)+1)} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.64, size = 147, normalized size = 1.22 \[ -\frac {30 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 10 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 4}{96 \, {\left (a d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{4}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.75, size = 116, normalized size = 0.97 \[ \frac {\frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {30 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {3 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 38 \, \sin \left (d x + c\right ) + 25\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {55 \, \sin \left (d x + c\right )^{3} + 201 \, \sin \left (d x + c\right )^{2} + 255 \, \sin \left (d x + c\right ) + 117}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{192 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.17, size = 126, normalized size = 1.05 \[ \frac {1}{32 a d \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {1}{8 a d \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{32 a d}-\frac {1}{24 a d \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 a d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 a d \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{32 a d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.30, size = 130, normalized size = 1.08 \[ -\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 15 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 25 \, \sin \left (d x + c\right ) + 8\right )}}{a \sin \left (d x + c\right )^{5} + a \sin \left (d x + c\right )^{4} - 2 \, a \sin \left (d x + c\right )^{3} - 2 \, a \sin \left (d x + c\right )^{2} + a \sin \left (d x + c\right ) + a} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{96 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.14, size = 115, normalized size = 0.96 \[ \frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{16\,a\,d}-\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{16}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{16}-\frac {25\,{\sin \left (c+d\,x\right )}^2}{48}-\frac {25\,\sin \left (c+d\,x\right )}{48}+\frac {1}{6}}{d\,\left (a\,{\sin \left (c+d\,x\right )}^5+a\,{\sin \left (c+d\,x\right )}^4-2\,a\,{\sin \left (c+d\,x\right )}^3-2\,a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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