Optimal. Leaf size=99 \[ \frac {\left (3 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\sec ^2(c+d x) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{8 d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{4 d} \]
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Rubi [A] time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2668, 739, 639, 206} \[ \frac {\left (3 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\sec ^2(c+d x) \left (\left (3 a^2-b^2\right ) \sin (c+d x)+2 a b\right )}{8 d}+\frac {\sec ^4(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{4 d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 639
Rule 739
Rule 2668
Rubi steps
\begin {align*} \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {(a+x)^2}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{4 d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {-3 a^2+b^2-2 a x}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\left (b \left (3 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {\left (3 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\sec ^4(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \left (2 a b+\left (3 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}\\ \end {align*}
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Mathematica [A] time = 0.74, size = 166, normalized size = 1.68 \[ \frac {4 \left (b^2-a^2\right ) \sec ^4(c+d x) (b-a \sin (c+d x)) (a+b \sin (c+d x))^3+\left (b^2-3 a^2\right ) \left (-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+2 a^3 b \sec ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))\right )}{16 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 118, normalized size = 1.19 \[ \frac {{\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a b + 2 \, {\left ({\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.97, size = 118, normalized size = 1.19 \[ \frac {{\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{3} - b^{2} \sin \left (d x + c\right )^{3} - 5 \, a^{2} \sin \left (d x + c\right ) - b^{2} \sin \left (d x + c\right ) - 4 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.25, size = 165, normalized size = 1.67 \[ \frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a b}{2 d \cos \left (d x +c \right )^{4}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{8 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 115, normalized size = 1.16 \[ \frac {{\left (3 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (3 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{3} - 4 \, a b - {\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.10, size = 93, normalized size = 0.94 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (\frac {3\,a^2}{8}-\frac {b^2}{8}\right )}{d}+\frac {\left (\frac {b^2}{8}-\frac {3\,a^2}{8}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,a^2}{8}+\frac {b^2}{8}\right )\,\sin \left (c+d\,x\right )+\frac {a\,b}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )} \]
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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