Integrand size = 25, antiderivative size = 167 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \]
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Time = 0.38 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 486, 597, 12, 385, 209} \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)^2}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {\cot ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{5 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{15 f (a+b)} \]
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Rule 12
Rule 209
Rule 385
Rule 486
Rule 597
Rule 2000
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\sqrt {a+b \left (1+x^2\right )}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {a+b+b x^2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}+\frac {\text {Subst}\left (\int \frac {b-5 (a+b)-4 b x^2}{x^4 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 f} \\ & = -\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {\text {Subst}\left (\int \frac {-15 a^2-25 a b-8 b^2+2 b (b-5 (a+b)) x^2}{x^2 \left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b) f} \\ & = -\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}+\frac {\text {Subst}\left (\int -\frac {15 a (a+b)^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f} \\ & = -\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {a \text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f}-\frac {a \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f} \\ & = -\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {\left (15 a^2+25 a b+8 b^2\right ) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b)^2 f}-\frac {(b-5 (a+b)) \cot ^3(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{15 (a+b) f}-\frac {\cot ^5(e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{5 f} \\ \end{align*}
Time = 2.88 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.07 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=-\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right ) \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f \sqrt {a+2 b+a \cos (2 e+2 f x)}}-\frac {\cot (e+f x) \left (23 a^2+40 a b+15 b^2-\left (11 a^2+21 a b+10 b^2\right ) \csc ^2(e+f x)+3 (a+b)^2 \csc ^4(e+f x)\right ) \sqrt {a+b \sec ^2(e+f x)}}{15 (a+b)^2 f} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(1015\) vs. \(2(149)=298\).
Time = 5.90 (sec) , antiderivative size = 1016, normalized size of antiderivative = 6.08
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Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (149) = 298\).
Time = 2.30 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.08 \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) - 8 \, {\left ({\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 59 \, a b + 20 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 25 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}, \frac {15 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (35 \, a^{2} + 59 \, a b + 20 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 25 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
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\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int \sqrt {a + b \sec ^{2}{\left (e + f x \right )}} \cot ^{6}{\left (e + f x \right )}\, dx \]
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\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \]
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\[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int { \sqrt {b \sec \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )^{6} \,d x } \]
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Timed out. \[ \int \cot ^6(e+f x) \sqrt {a+b \sec ^2(e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^6\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]
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