Optimal. Leaf size=126 \[ \frac {\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}-\frac {a (4 a-3 b) \log (\sin (c+d x)+1)}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d} \]
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Rubi [A] time = 0.22, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2837, 12, 1805, 823, 801} \[ \frac {\sec ^4(c+d x) \left (a^2+2 a b \sin (c+d x)+b^2\right )}{4 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}-\frac {a (4 a-3 b) \log (\sin (c+d x)+1)}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 801
Rule 823
Rule 1805
Rule 2837
Rubi steps
\begin {align*} \int \csc (c+d x) \sec ^5(c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {b (a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^6 \operatorname {Subst}\left (\int \frac {(a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {b^4 \operatorname {Subst}\left (\int \frac {-4 a^2-6 a x}{x \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-8 a^2 b^2-6 a b^2 x}{x \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}-\frac {\operatorname {Subst}\left (\int \left (-\frac {a (4 a+3 b)}{b-x}-\frac {8 a^2}{x}+\frac {a (4 a-3 b)}{b+x}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac {a (4 a+3 b) \log (1-\sin (c+d x))}{8 d}+\frac {a^2 \log (\sin (c+d x))}{d}-\frac {a (4 a-3 b) \log (1+\sin (c+d x))}{8 d}+\frac {a \sec ^2(c+d x) (2 a+3 b \sin (c+d x))}{4 d}+\frac {\sec ^4(c+d x) \left (a^2+b^2+2 a b \sin (c+d x)\right )}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.92, size = 137, normalized size = 1.09 \[ \frac {16 a^2 \log (\sin (c+d x))-\frac {(a+b) (5 a+b)}{\sin (c+d x)-1}+\frac {(a-b) (5 a-b)}{\sin (c+d x)+1}+\frac {(a+b)^2}{(\sin (c+d x)-1)^2}+\frac {(a-b)^2}{(\sin (c+d x)+1)^2}-2 a (4 a+3 b) \log (1-\sin (c+d x))-2 a (4 a-3 b) \log (\sin (c+d x)+1)}{16 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 144, normalized size = 1.14 \[ \frac {8 \, a^{2} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2} + 2 \, {\left (3 \, a b \cos \left (d x + c\right )^{2} + 2 \, a b\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 134, normalized size = 1.06 \[ \frac {8 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \sin \left (d x + c\right )^{4} - 3 \, a b \sin \left (d x + c\right )^{3} - 8 \, a^{2} \sin \left (d x + c\right )^{2} + 5 \, a b \sin \left (d x + c\right ) + 6 \, a^{2} + b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.61, size = 125, normalized size = 0.99 \[ \frac {a^{2}}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{2}}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {a b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 a b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{4 d}+\frac {3 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {b^{2}}{4 d \cos \left (d x +c \right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 130, normalized size = 1.03 \[ \frac {8 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - {\left (4 \, a^{2} - 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (4 \, a^{2} + 3 \, a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a b \sin \left (d x + c\right )^{3} + 2 \, a^{2} \sin \left (d x + c\right )^{2} - 5 \, a b \sin \left (d x + c\right ) - 3 \, a^{2} - b^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{8 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.13, size = 131, normalized size = 1.04 \[ \frac {a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}+\frac {-\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {3\,a^2}{4}-\frac {3\,a\,b\,{\sin \left (c+d\,x\right )}^3}{4}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{4}+\frac {b^2}{4}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (4\,a+3\,b\right )}{8\,d}-\frac {a\,\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (4\,a-3\,b\right )}{8\,d} \]
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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