Optimal. Leaf size=60 \[ -\frac {1}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {1}{4 d (a \sin (c+d x)+a)^2} \]
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Rubi [A] time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2667, 44, 206} \[ -\frac {1}{4 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac {\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {1}{4 d (a \sin (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 44
Rule 206
Rule 2667
Rubi steps
\begin {align*} \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac {a \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \left (\frac {1}{2 a (a+x)^3}+\frac {1}{4 a^2 (a+x)^2}+\frac {1}{4 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {1}{4 d (a+a \sin (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{4 a d}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{4 a^2 d}-\frac {1}{4 d (a+a \sin (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 38, normalized size = 0.63 \[ \frac {\tanh ^{-1}(\sin (c+d x))-\frac {\sin (c+d x)+2}{(\sin (c+d x)+1)^2}}{4 a^2 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.74, size = 105, normalized size = 1.75 \[ \frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, \sin \left (d x + c\right ) + 4}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.75, size = 71, normalized size = 1.18 \[ \frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} - \frac {3 \, \sin \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) + 11}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 72, normalized size = 1.20 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8 a^{2} d}-\frac {1}{4 a^{2} d \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 a^{2} d \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8 a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.81, size = 72, normalized size = 1.20 \[ -\frac {\frac {2 \, {\left (\sin \left (d x + c\right ) + 2\right )}}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.52, size = 60, normalized size = 1.00 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,a^2\,d}-\frac {\frac {\sin \left (c+d\,x\right )}{4}+\frac {1}{2}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]
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 \[ \frac {\int \frac {\sec {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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