Optimal. Leaf size=168 \[ -\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {b \sec ^4(c+d x) \left (\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}+2 a\right )}{4 d}+\frac {b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+3\right ) \sin (c+d x)+8 a\right )}{8 d}-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d} \]
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Rubi [A] time = 0.35, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 12, 1805, 1802} \[ -\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)}{16 d}+\frac {b \sec ^4(c+d x) \left (\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}+2 a\right )}{4 d}+\frac {b \sec ^2(c+d x) \left (b \left (\frac {7 a^2}{b^2}+3\right ) \sin (c+d x)+8 a\right )}{8 d}-\frac {a^2 \csc (c+d x)}{d}+\frac {2 a b \log (\sin (c+d x))}{d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 1802
Rule 1805
Rule 2837
Rubi steps
\begin {align*} \int \csc ^2(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^2 (a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^7 \operatorname {Subst}\left (\int \frac {(a+x)^2}{x^2 \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}-\frac {b^5 \operatorname {Subst}\left (\int \frac {-4 a^2-8 a x-3 \left (1+\frac {a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d}\\ &=\frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {8 a^2+16 a x+\left (3+\frac {7 a^2}{b^2}\right ) x^2}{x^2 \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}+\frac {b^3 \operatorname {Subst}\left (\int \left (\frac {15 a^2+16 a b+3 b^2}{2 b^3 (b-x)}+\frac {8 a^2}{b^2 x^2}+\frac {16 a}{b^2 x}+\frac {15 a^2-16 a b+3 b^2}{2 b^3 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=-\frac {a^2 \csc (c+d x)}{d}-\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {2 a b \log (\sin (c+d x))}{d}+\frac {\left (15 a^2-16 a b+3 b^2\right ) \log (1+\sin (c+d x))}{16 d}+\frac {b \sec ^2(c+d x) \left (8 a+\left (3+\frac {7 a^2}{b^2}\right ) b \sin (c+d x)\right )}{8 d}+\frac {b \sec ^4(c+d x) \left (2 a+\frac {\left (a^2+b^2\right ) \sin (c+d x)}{b}\right )}{4 d}\\ \end {align*}
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Mathematica [A] time = 2.83, size = 162, normalized size = 0.96 \[ -\frac {\left (15 a^2+16 a b+3 b^2\right ) \log (1-\sin (c+d x))-\left (15 a^2-16 a b+3 b^2\right ) \log (\sin (c+d x)+1)+16 a^2 \csc (c+d x)+\frac {(a+b) (7 a+3 b)}{\sin (c+d x)-1}+\frac {(7 a-3 b) (a-b)}{\sin (c+d x)+1}-\frac {(a+b)^2}{(\sin (c+d x)-1)^2}+\frac {(a-b)^2}{(\sin (c+d x)+1)^2}-32 a b \log (\sin (c+d x))}{16 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.47, size = 202, normalized size = 1.20 \[ \frac {32 \, a b \cos \left (d x + c\right )^{4} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, a^{2} + 4 \, b^{2} + 8 \, {\left (2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.33, size = 186, normalized size = 1.11 \[ \frac {32 \, a b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {16 \, {\left (2 \, a b \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )} + \frac {2 \, {\left (12 \, a b \sin \left (d x + c\right )^{4} - 7 \, a^{2} \sin \left (d x + c\right )^{3} - 3 \, b^{2} \sin \left (d x + c\right )^{3} - 32 \, a b \sin \left (d x + c\right )^{2} + 9 \, a^{2} \sin \left (d x + c\right ) + 5 \, b^{2} \sin \left (d x + c\right ) + 24 \, a b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.60, size = 195, normalized size = 1.16 \[ \frac {a^{2}}{4 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5 a^{2}}{8 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15 a^{2}}{8 d \sin \left (d x +c \right )}+\frac {15 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a b}{2 d \cos \left (d x +c \right )^{4}}+\frac {a b}{d \cos \left (d x +c \right )^{2}}+\frac {2 a b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{8 d}+\frac {3 b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 163, normalized size = 0.97 \[ \frac {32 \, a b \log \left (\sin \left (d x + c\right )\right ) + {\left (15 \, a^{2} - 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (15 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (8 \, a b \sin \left (d x + c\right )^{3} + 3 \, {\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{4} - 12 \, a b \sin \left (d x + c\right ) - 5 \, {\left (5 \, a^{2} + b^{2}\right )} \sin \left (d x + c\right )^{2} + 8 \, a^{2}\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.87, size = 169, normalized size = 1.01 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {15\,a^2}{16}-a\,b+\frac {3\,b^2}{16}\right )}{d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {15\,a^2}{16}+a\,b+\frac {3\,b^2}{16}\right )}{d}-\frac {a^2+{\sin \left (c+d\,x\right )}^4\,\left (\frac {15\,a^2}{8}+\frac {3\,b^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {25\,a^2}{8}+\frac {5\,b^2}{8}\right )-\frac {3\,a\,b\,\sin \left (c+d\,x\right )}{2}+a\,b\,{\sin \left (c+d\,x\right )}^3}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^3+\sin \left (c+d\,x\right )\right )}+\frac {2\,a\,b\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]
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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