Optimal. Leaf size=165 \[ \frac {a^3 \log (\sin (c+d x))}{d}+\frac {\sec ^4(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d}-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d} \]
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Rubi [A] time = 0.25, antiderivative size = 165, 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} \[ -\frac {\left (9 a^2 b+8 a^3-b^3\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (-9 a^2 b+8 a^3+b^3\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {\sec ^4(c+d x) \left (b \left (3 a^2+b^2\right ) \sin (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d}+\frac {\sec ^2(c+d x) \left (b \left (9 a^2-b^2\right ) \sin (c+d x)+4 a^3\right )}{8 d}+\frac {a^3 \log (\sin (c+d x))}{d} \]
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))^3 \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {b (a+x)^3}{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)^3}{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 \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {b^4 \operatorname {Subst}\left (\int \frac {-4 a^3-\left (9 a^2-b^2\right ) x}{x \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {\operatorname {Subst}\left (\int \frac {-8 a^3 b^2-b^2 \left (9 a^2-b^2\right ) x}{x \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {\operatorname {Subst}\left (\int \left (\frac {-8 a^3-9 a^2 b+b^3}{2 (b-x)}-\frac {8 a^3}{x}+\frac {8 a^3-9 a^2 b+b^3}{2 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac {\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))}{16 d}+\frac {a^3 \log (\sin (c+d x))}{d}-\frac {\left (8 a^3-9 a^2 b+b^3\right ) \log (1+\sin (c+d x))}{16 d}+\frac {\sec ^2(c+d x) \left (4 a^3+b \left (9 a^2-b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^4(c+d x) \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \sin (c+d x)\right )}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.57, size = 157, normalized size = 0.95 \[ \frac {16 a^3 \log (\sin (c+d x))-\left (8 a^3+9 a^2 b-b^3\right ) \log (1-\sin (c+d x))-\left (8 a^3-9 a^2 b+b^3\right ) \log (\sin (c+d x)+1)-\frac {(5 a-b) (a+b)^2}{\sin (c+d x)-1}+\frac {(a-b)^2 (5 a+b)}{\sin (c+d x)+1}+\frac {(a+b)^3}{(\sin (c+d x)-1)^2}+\frac {(a-b)^3}{(\sin (c+d x)+1)^2}}{16 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.46, size = 173, normalized size = 1.05 \[ \frac {16 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 8 \, a^{3} \cos \left (d x + c\right )^{2} + 4 \, a^{3} + 12 \, a b^{2} + 2 \, {\left (6 \, a^{2} b + 2 \, b^{3} + {\left (9 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.35, size = 175, normalized size = 1.06 \[ \frac {16 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, a^{3} \sin \left (d x + c\right )^{4} - 9 \, a^{2} b \sin \left (d x + c\right )^{3} + b^{3} \sin \left (d x + c\right )^{3} - 16 \, a^{3} \sin \left (d x + c\right )^{2} + 15 \, a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + 12 \, a^{3} + 6 \, a b^{2}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.68, size = 216, normalized size = 1.31 \[ \frac {a^{3}}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3}}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 a^{2} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {9 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 a \,b^{2}}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {b^{3} \sin \left (d x +c \right )}{8 d}-\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.33, size = 160, normalized size = 0.97 \[ \frac {16 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - {\left (8 \, a^{3} - 9 \, a^{2} b + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, a^{3} + 9 \, a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (4 \, a^{3} \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 6 \, a^{3} - 6 \, a b^{2} - {\left (15 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.91, size = 169, normalized size = 1.02 \[ \frac {a^3\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {a^3}{2}-\frac {9\,a^2\,b}{16}+\frac {b^3}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {a^3}{2}+\frac {9\,a^2\,b}{16}-\frac {b^3}{16}\right )}{d}+\frac {\frac {3\,a\,b^2}{4}-{\sin \left (c+d\,x\right )}^3\,\left (\frac {9\,a^2\,b}{8}-\frac {b^3}{8}\right )+\frac {3\,a^3}{4}+\sin \left (c+d\,x\right )\,\left (\frac {15\,a^2\,b}{8}+\frac {b^3}{8}\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )} \]
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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