Integrand size = 23, antiderivative size = 109 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b} d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \]
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Time = 0.09 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3266, 472, 211} \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} d \sqrt {a+b}}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^5(c+d x)}{5 a d} \]
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Rule 211
Rule 472
Rule 3266
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^6 \left (a+(a+b) x^2\right )} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {1}{a x^6}+\frac {2 a-b}{a^2 x^4}+\frac {a^2-a b+b^2}{a^3 x^2}+\frac {b^3}{a^3 \left (-a-(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d}+\frac {b^3 \text {Subst}\left (\int \frac {1}{-a-(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{a^3 d} \\ & = -\frac {b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{7/2} \sqrt {a+b} d}-\frac {\left (a^2-a b+b^2\right ) \cot (c+d x)}{a^3 d}-\frac {(2 a-b) \cot ^3(c+d x)}{3 a^2 d}-\frac {\cot ^5(c+d x)}{5 a d} \\ \end{align*}
Time = 2.31 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.35 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {(2 a+b-b \cos (2 (c+d x))) \csc ^2(c+d x) \left (15 b^3 \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a} \sqrt {a+b} \cot (c+d x) \left (8 a^2-10 a b+15 b^2+a (4 a-5 b) \csc ^2(c+d x)+3 a^2 \csc ^4(c+d x)\right )\right )}{30 a^{7/2} \sqrt {a+b} d \left (b+a \csc ^2(c+d x)\right )} \]
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Time = 1.13 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {-\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}-a b +b^{2}}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(96\) |
default | \(\frac {-\frac {b^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}-\frac {1}{5 a \tan \left (d x +c \right )^{5}}-\frac {2 a -b}{3 a^{2} \tan \left (d x +c \right )^{3}}-\frac {a^{2}-a b +b^{2}}{a^{3} \tan \left (d x +c \right )}}{d}\) | \(96\) |
risch | \(-\frac {2 i \left (15 b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+30 a b \,{\mathrm e}^{6 i \left (d x +c \right )}-60 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+80 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-70 a b \,{\mathrm e}^{4 i \left (d x +c \right )}+90 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-40 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+50 a b \,{\mathrm e}^{2 i \left (d x +c \right )}-60 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+8 a^{2}-10 a b +15 b^{2}\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a^{2}+2 i a b -2 a \sqrt {-a^{2}-a b}-b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{2 \sqrt {-a^{2}-a b}\, d \,a^{3}}\) | \(346\) |
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Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (97) = 194\).
Time = 0.29 (sec) , antiderivative size = 595, normalized size of antiderivative = 5.46 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [-\frac {4 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 20 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \sin \left (d x + c\right ) + 60 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{60 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\right )} d\right )} \sin \left (d x + c\right )}, -\frac {2 \, {\left (8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} - 10 \, {\left (4 \, a^{4} - a^{3} b + a^{2} b^{2} + 6 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 15 \, {\left (b^{3} \cos \left (d x + c\right )^{4} - 2 \, b^{3} \cos \left (d x + c\right )^{2} + b^{3}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 30 \, {\left (a^{4} + a b^{3}\right )} \cos \left (d x + c\right )}{30 \, {\left ({\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} + a^{4} b\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{5} + a^{4} b\right )} d\right )} \sin \left (d x + c\right )}\right ] \]
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\[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\csc ^{6}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
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Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, b^{3} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {15 \, {\left (a^{2} - a b + b^{2}\right )} \tan \left (d x + c\right )^{4} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
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Time = 0.47 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.42 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} b^{3}}{\sqrt {a^{2} + a b} a^{3}} + \frac {15 \, a^{2} \tan \left (d x + c\right )^{4} - 15 \, a b \tan \left (d x + c\right )^{4} + 15 \, b^{2} \tan \left (d x + c\right )^{4} + 10 \, a^{2} \tan \left (d x + c\right )^{2} - 5 \, a b \tan \left (d x + c\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
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Time = 13.77 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87 \[ \int \frac {\csc ^6(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2-a\,b+b^2\right )+\frac {a^2}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a\,b}{3}-\frac {2\,a^2}{3}\right )}{a^3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^5}-\frac {b^3\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )}{a^{7/2}\,d\,\sqrt {a+b}} \]
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