Optimal. Leaf size=122 \[ \frac {\left (3 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\sec ^2(c+d x) \left (\left (3 a^2 A-2 a b B+A b^2\right ) \sin (c+d x)+2 b (2 a A-b B)\right )}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{4 d} \]
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Rubi [A] time = 0.16, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {2837, 821, 778, 206} \[ \frac {\left (3 a^2 A-2 a b B-A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\sec ^2(c+d x) \left (\left (3 a^2 A-2 a b B+A b^2\right ) \sin (c+d x)+2 b (2 a A-b B)\right )}{8 d}+\frac {\sec ^4(c+d x) (a+b \sin (c+d x))^2 (A \sin (c+d x)+B)}{4 d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 778
Rule 821
Rule 2837
Rubi steps
\begin {align*} \int \sec ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {(a+x)^2 \left (A+\frac {B x}{b}\right )}{\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))^2}{4 d}-\frac {b^3 \operatorname {Subst}\left (\int \frac {(a+x) (-3 a A+2 b B-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))^2}{4 d}+\frac {\sec ^2(c+d x) \left (2 b (2 a A-b B)+\left (3 a^2 A+A b^2-2 a b B\right ) \sin (c+d x)\right )}{8 d}+\frac {\left (b \left (3 a^2 A-A b^2-2 a b B\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 A-A b^2-2 a b B\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))^2}{4 d}+\frac {\sec ^2(c+d x) \left (2 b (2 a A-b B)+\left (3 a^2 A+A b^2-2 a b B\right ) \sin (c+d x)\right )}{8 d}\\ \end {align*}
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Mathematica [A] time = 1.78, size = 186, normalized size = 1.52 \[ \frac {4 \left (b^2-a^2\right ) \sec ^4(c+d x) (a+b \sin (c+d x))^3 ((b B-a A) \sin (c+d x)-a B+A b)+\left (-3 a^2 A+2 a b B+A b^2\right ) \left (-2 \left (a^4-b^4\right ) \tan (c+d x) \sec (c+d x)+\left (4 a b^3-6 a^3 b\right ) \tan ^2(c+d x)+2 a^3 b \sec ^2(c+d x)+\left (a^2-b^2\right )^2 (\log (1-\sin (c+d x))-\log (\sin (c+d x)+1))\right )}{16 d \left (a^2-b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 173, normalized size = 1.42 \[ \frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, B b^{2} \cos \left (d x + c\right )^{2} + 4 \, B a^{2} + 8 \, A a b + 4 \, B b^{2} + 2 \, {\left (2 \, A a^{2} + 4 \, B a b + 2 \, A b^{2} + {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 187, normalized size = 1.53 \[ \frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \sin \left (d x + c\right )^{3} - 2 \, B a b \sin \left (d x + c\right )^{3} - A b^{2} \sin \left (d x + c\right )^{3} - 4 \, B b^{2} \sin \left (d x + c\right )^{2} - 5 \, A a^{2} \sin \left (d x + c\right ) - 2 \, B a b \sin \left (d x + c\right ) - A b^{2} \sin \left (d x + c\right ) - 2 \, B a^{2} - 4 \, A a b + 2 \, B b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.56, size = 299, normalized size = 2.45 \[ \frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {B \,a^{2}}{4 d \cos \left (d x +c \right )^{4}}+\frac {A a b}{2 d \cos \left (d x +c \right )^{4}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}+\frac {B a b \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}+\frac {B a b \sin \left (d x +c \right )}{4 d}-\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {A \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {A \,b^{2} \sin \left (d x +c \right )}{8 d}-\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {B \,b^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 171, normalized size = 1.40 \[ \frac {{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (4 \, B b^{2} \sin \left (d x + c\right )^{2} - {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 2 \, B a^{2} + 4 \, A a b - 2 \, B b^{2} + {\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 12.38, size = 181, normalized size = 1.48 \[ \frac {\sin \left (c+d\,x\right )\,\left (\frac {5\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,a^2}{4}-\frac {B\,b^2}{4}+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {A\,a\,b}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\frac {4\,\sin \left (c+d\,x\right )\,\left (-\frac {3\,A\,a^2}{16}+\frac {B\,a\,b}{8}+\frac {A\,b^2}{16}\right )}{-\frac {3\,A\,a^2}{4}+\frac {B\,a\,b}{2}+\frac {A\,b^2}{4}}\right )\,\left (-\frac {3\,A\,a^2}{8}+\frac {B\,a\,b}{4}+\frac {A\,b^2}{8}\right )}{d} \]
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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