Integrand size = 23, antiderivative size = 192 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^5}{4 a^3 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {b^4 (5 a+2 b)}{2 a^3 (a+b)^4 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(2 a+5 b) \csc ^2(e+f x)}{2 (a+b)^4 f}-\frac {\csc ^4(e+f x)}{4 (a+b)^3 f}+\frac {b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^5 f}+\frac {\left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))}{(a+b)^5 f} \]
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Time = 0.32 (sec) , antiderivative size = 192, 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 = {4223, 457, 90} \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {b^5}{4 a^3 f (a+b)^3 \left (a \cos ^2(e+f x)+b\right )^2}+\frac {b^4 (5 a+2 b)}{2 a^3 f (a+b)^4 \left (a \cos ^2(e+f x)+b\right )}+\frac {\left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))}{f (a+b)^5}+\frac {b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f (a+b)^5}-\frac {\csc ^4(e+f x)}{4 f (a+b)^3}+\frac {(2 a+5 b) \csc ^2(e+f x)}{2 f (a+b)^4} \]
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Rule 90
Rule 457
Rule 4223
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^{11}}{\left (1-x^2\right )^3 \left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {x^5}{(1-x)^3 (b+a x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {Subst}\left (\int \left (-\frac {1}{(a+b)^3 (-1+x)^3}+\frac {-2 a-5 b}{(a+b)^4 (-1+x)^2}+\frac {-a^2-5 a b-10 b^2}{(a+b)^5 (-1+x)}-\frac {b^5}{a^2 (a+b)^3 (b+a x)^3}+\frac {b^4 (5 a+2 b)}{a^2 (a+b)^4 (b+a x)^2}-\frac {b^3 \left (10 a^2+5 a b+b^2\right )}{a^2 (a+b)^5 (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {b^5}{4 a^3 (a+b)^3 f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {b^4 (5 a+2 b)}{2 a^3 (a+b)^4 f \left (b+a \cos ^2(e+f x)\right )}+\frac {(2 a+5 b) \csc ^2(e+f x)}{2 (a+b)^4 f}-\frac {\csc ^4(e+f x)}{4 (a+b)^3 f}+\frac {b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (b+a \cos ^2(e+f x)\right )}{2 a^3 (a+b)^5 f}+\frac {\left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))}{(a+b)^5 f} \\ \end{align*}
Time = 5.53 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (2 (a+b) (2 a+5 b) \csc ^2(e+f x)-(a+b)^2 \csc ^4(e+f x)+4 \left (a^2+5 a b+10 b^2\right ) \log (\sin (e+f x))+\frac {2 b^3 \left (10 a^2+5 a b+b^2\right ) \log \left (a+b-a \sin ^2(e+f x)\right )}{a^3}-\frac {b^5 (a+b)^2}{a^3 \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {2 b^4 (a+b) (5 a+2 b)}{a^3 \left (a+b-a \sin ^2(e+f x)\right )}\right )}{32 (a+b)^5 f \left (a+b \sec ^2(e+f x)\right )^3} \]
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Time = 53.50 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {-\frac {1}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +19 b}{16 \left (a +b \right )^{4} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}-\frac {1}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -19 b}{16 \left (a +b \right )^{4} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}+\frac {b^{3} \left (\frac {\left (10 a^{2}+5 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {b \left (5 a^{2}+7 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{5}}}{f}\) | \(254\) |
default | \(\frac {-\frac {1}{16 \left (a +b \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}-\frac {7 a +19 b}{16 \left (a +b \right )^{4} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}-\frac {1}{16 \left (a +b \right )^{3} \left (1+\cos \left (f x +e \right )\right )^{2}}-\frac {-7 a -19 b}{16 \left (a +b \right )^{4} \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (a^{2}+5 a b +10 b^{2}\right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{5}}+\frac {b^{3} \left (\frac {\left (10 a^{2}+5 a b +b^{2}\right ) \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{a^{3}}-\frac {b^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {b \left (5 a^{2}+7 a b +2 b^{2}\right )}{a^{3} \left (b +a \cos \left (f x +e \right )^{2}\right )}\right )}{2 \left (a +b \right )^{5}}}{f}\) | \(254\) |
risch | \(\text {Expression too large to display}\) | \(1676\) |
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Leaf count of result is larger than twice the leaf count of optimal. 859 vs. \(2 (182) = 364\).
Time = 1.68 (sec) , antiderivative size = 859, normalized size of antiderivative = 4.47 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {3 \, a^{5} b^{2} + 12 \, a^{4} b^{3} + 9 \, a^{3} b^{4} + 9 \, a^{2} b^{5} + 12 \, a b^{6} + 3 \, b^{7} - 2 \, {\left (2 \, a^{7} + 7 \, a^{6} b + 5 \, a^{5} b^{2} - 5 \, a^{3} b^{4} - 7 \, a^{2} b^{5} - 2 \, a b^{6}\right )} \cos \left (f x + e\right )^{6} + {\left (3 \, a^{7} + 4 \, a^{6} b - 19 \, a^{5} b^{2} - 20 \, a^{4} b^{3} - 20 \, a^{3} b^{4} - 19 \, a^{2} b^{5} + 4 \, a b^{6} + 3 \, b^{7}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{6} b + 10 \, a^{5} b^{2} + 2 \, a^{4} b^{3} - 2 \, a^{2} b^{5} - 10 \, a b^{6} - 3 \, b^{7}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left ({\left (10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} \cos \left (f x + e\right )^{8} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7} - 2 \, {\left (10 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 4 \, a^{2} b^{5} - a b^{6}\right )} \cos \left (f x + e\right )^{6} + {\left (10 \, a^{4} b^{3} - 35 \, a^{3} b^{4} - 9 \, a^{2} b^{5} + a b^{6} + b^{7}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (10 \, a^{3} b^{4} - 5 \, a^{2} b^{5} - 4 \, a b^{6} - b^{7}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + 4 \, {\left ({\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2}\right )} \cos \left (f x + e\right )^{8} + a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} - 2 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} - 10 \, a^{4} b^{3}\right )} \cos \left (f x + e\right )^{6} + {\left (a^{7} + a^{6} b - 9 \, a^{5} b^{2} - 35 \, a^{4} b^{3} + 10 \, a^{3} b^{4}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 10 \, a^{3} b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{4 \, {\left ({\left (a^{10} + 5 \, a^{9} b + 10 \, a^{8} b^{2} + 10 \, a^{7} b^{3} + 5 \, a^{6} b^{4} + a^{5} b^{5}\right )} f \cos \left (f x + e\right )^{8} - 2 \, {\left (a^{10} + 4 \, a^{9} b + 5 \, a^{8} b^{2} - 5 \, a^{6} b^{4} - 4 \, a^{5} b^{5} - a^{4} b^{6}\right )} f \cos \left (f x + e\right )^{6} + {\left (a^{10} + a^{9} b - 9 \, a^{8} b^{2} - 25 \, a^{7} b^{3} - 25 \, a^{6} b^{4} - 9 \, a^{5} b^{5} + a^{4} b^{6} + a^{3} b^{7}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{9} b + 4 \, a^{8} b^{2} + 5 \, a^{7} b^{3} - 5 \, a^{5} b^{5} - 4 \, a^{4} b^{6} - a^{3} b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{8} b^{2} + 5 \, a^{7} b^{3} + 10 \, a^{6} b^{4} + 10 \, a^{5} b^{5} + 5 \, a^{4} b^{6} + a^{3} b^{7}\right )} f\right )}} \]
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Timed out. \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (182) = 364\).
Time = 0.23 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.36 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {2 \, {\left (10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{8} + 5 \, a^{7} b + 10 \, a^{6} b^{2} + 10 \, a^{5} b^{3} + 5 \, a^{4} b^{4} + a^{3} b^{5}} + \frac {2 \, {\left (a^{2} + 5 \, a b + 10 \, b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}} + \frac {2 \, {\left (2 \, a^{6} + 5 \, a^{5} b - 5 \, a^{2} b^{4} - 2 \, a b^{5}\right )} \sin \left (f x + e\right )^{6} - a^{6} - 3 \, a^{5} b - 3 \, a^{4} b^{2} - a^{3} b^{3} - {\left (9 \, a^{6} + 29 \, a^{5} b + 20 \, a^{4} b^{2} - 10 \, a^{2} b^{4} - 13 \, a b^{5} - 3 \, b^{6}\right )} \sin \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{6} + 11 \, a^{5} b + 13 \, a^{4} b^{2} + 5 \, a^{3} b^{3}\right )} \sin \left (f x + e\right )^{2}}{{\left (a^{9} + 4 \, a^{8} b + 6 \, a^{7} b^{2} + 4 \, a^{6} b^{3} + a^{5} b^{4}\right )} \sin \left (f x + e\right )^{8} - 2 \, {\left (a^{9} + 5 \, a^{8} b + 10 \, a^{7} b^{2} + 10 \, a^{6} b^{3} + 5 \, a^{5} b^{4} + a^{4} b^{5}\right )} \sin \left (f x + e\right )^{6} + {\left (a^{9} + 6 \, a^{8} b + 15 \, a^{7} b^{2} + 20 \, a^{6} b^{3} + 15 \, a^{5} b^{4} + 6 \, a^{4} b^{5} + a^{3} b^{6}\right )} \sin \left (f x + e\right )^{4}}}{4 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1912 vs. \(2 (182) = 364\).
Time = 0.53 (sec) , antiderivative size = 1912, normalized size of antiderivative = 9.96 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
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Time = 22.49 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.70 \[ \int \frac {\cot ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+5\,a\,b+10\,b^2\right )}{f\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )}-\frac {\frac {1}{4\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (a+3\,b\right )}{2\,{\left (a+b\right )}^2}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-4\,a^3\,b-15\,a^2\,b^2+9\,a\,b^3+2\,b^4\right )}{4\,a^2\,\left (a+b\right )\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (-a^3\,b^2-4\,a^2\,b^3+4\,a\,b^4+b^5\right )}{2\,a^2\,{\left (a+b\right )}^2\,\left (a^2+2\,a\,b+b^2\right )}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a^3\,f}+\frac {b^3\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )\,\left (10\,a^2+5\,a\,b+b^2\right )}{2\,a^3\,f\,{\left (a+b\right )}^5} \]
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