Optimal. Leaf size=135 \[ -\frac {4}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\csc ^5(c+d x)}{5 a^4 d}+\frac {\csc ^4(c+d x)}{a^4 d}-\frac {8 \csc ^3(c+d x)}{3 a^4 d}+\frac {6 \csc ^2(c+d x)}{a^4 d}-\frac {16 \csc (c+d x)}{a^4 d}-\frac {20 \log (\sin (c+d x))}{a^4 d}+\frac {20 \log (\sin (c+d x)+1)}{a^4 d} \]
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Rubi [A] time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2836, 12, 88} \[ -\frac {4}{d \left (a^4 \sin (c+d x)+a^4\right )}-\frac {\csc ^5(c+d x)}{5 a^4 d}+\frac {\csc ^4(c+d x)}{a^4 d}-\frac {8 \csc ^3(c+d x)}{3 a^4 d}+\frac {6 \csc ^2(c+d x)}{a^4 d}-\frac {16 \csc (c+d x)}{a^4 d}-\frac {20 \log (\sin (c+d x))}{a^4 d}+\frac {20 \log (\sin (c+d x)+1)}{a^4 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 88
Rule 2836
Rubi steps
\begin {align*} \int \frac {\cot ^5(c+d x) \csc (c+d x)}{(a+a \sin (c+d x))^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a^6 (a-x)^2}{x^6 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {a \operatorname {Subst}\left (\int \frac {(a-x)^2}{x^6 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a \operatorname {Subst}\left (\int \left (\frac {1}{x^6}-\frac {4}{a x^5}+\frac {8}{a^2 x^4}-\frac {12}{a^3 x^3}+\frac {16}{a^4 x^2}-\frac {20}{a^5 x}+\frac {4}{a^4 (a+x)^2}+\frac {20}{a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {16 \csc (c+d x)}{a^4 d}+\frac {6 \csc ^2(c+d x)}{a^4 d}-\frac {8 \csc ^3(c+d x)}{3 a^4 d}+\frac {\csc ^4(c+d x)}{a^4 d}-\frac {\csc ^5(c+d x)}{5 a^4 d}-\frac {20 \log (\sin (c+d x))}{a^4 d}+\frac {20 \log (1+\sin (c+d x))}{a^4 d}-\frac {4}{d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.28, size = 91, normalized size = 0.67 \[ -\frac {\frac {60}{\sin (c+d x)+1}+3 \csc ^5(c+d x)-15 \csc ^4(c+d x)+40 \csc ^3(c+d x)-90 \csc ^2(c+d x)+240 \csc (c+d x)+300 \log (\sin (c+d x))-300 \log (\sin (c+d x)+1)}{15 a^4 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.76, size = 283, normalized size = 2.10 \[ \frac {150 \, \cos \left (d x + c\right )^{4} - 325 \, \cos \left (d x + c\right )^{2} - 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 300 \, {\left (\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (150 \, \cos \left (d x + c\right )^{4} - 275 \, \cos \left (d x + c\right )^{2} + 119\right )} \sin \left (d x + c\right ) + 178}{15 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d - {\left (a^{4} d \cos \left (d x + c\right )^{4} - 2 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 248, normalized size = 1.84 \[ \frac {\frac {19200 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {9600 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {1920 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{2}} + \frac {21920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4350 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 840 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}} - \frac {3 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 30 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 175 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 840 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4350 \, a^{16} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{20}}}{480 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.68, size = 131, normalized size = 0.97 \[ -\frac {1}{5 d \,a^{4} \sin \left (d x +c \right )^{5}}+\frac {1}{d \,a^{4} \sin \left (d x +c \right )^{4}}-\frac {8}{3 d \,a^{4} \sin \left (d x +c \right )^{3}}+\frac {6}{d \,a^{4} \sin \left (d x +c \right )^{2}}-\frac {16}{d \,a^{4} \sin \left (d x +c \right )}-\frac {20 \ln \left (\sin \left (d x +c \right )\right )}{a^{4} d}-\frac {4}{d \,a^{4} \left (1+\sin \left (d x +c \right )\right )}+\frac {20 \ln \left (1+\sin \left (d x +c \right )\right )}{a^{4} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.78, size = 110, normalized size = 0.81 \[ -\frac {\frac {300 \, \sin \left (d x + c\right )^{5} + 150 \, \sin \left (d x + c\right )^{4} - 50 \, \sin \left (d x + c\right )^{3} + 25 \, \sin \left (d x + c\right )^{2} - 12 \, \sin \left (d x + c\right ) + 3}{a^{4} \sin \left (d x + c\right )^{6} + a^{4} \sin \left (d x + c\right )^{5}} - \frac {300 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {300 \, \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{15 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.91, size = 266, normalized size = 1.97 \[ \frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^4\,d}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{16\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{160\,a^4\,d}-\frac {20\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {34\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+524\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {569\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}-\frac {104\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {118\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {1}{5}}{d\,\left (32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+64\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+32\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}+\frac {40\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^4\,d}-\frac {145\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,a^4\,d} \]
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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