Optimal. Leaf size=219 \[ -\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (5 a^2-20 a b+16 b^2\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
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Rubi [A] time = 0.23, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3663, 462, 453, 271, 192, 191} \[ -\frac {8 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {4 b \left (5 a^2-20 a b+16 b^2\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (5 a^2-20 a b+16 b^2\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 192
Rule 271
Rule 453
Rule 462
Rule 3663
Rubi steps
\begin {align*} \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {2 (5 a-4 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\left (-15 a^2+12 (5 a-4 b) b\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (4 b \left (-15 a^2+12 (5 a-4 b) b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\left (8 b \left (-15 a^2+12 (5 a-4 b) b\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{45 a^4 f}\\ &=-\frac {\left (5 a^2-4 (5 a-4 b) b\right ) \cot (e+f x)}{5 a^3 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {2 (5 a-4 b) \cot ^3(e+f x)}{15 a^2 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^4 f \left (a+b \tan ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2-4 (5 a-4 b) b\right ) \tan (e+f x)}{15 a^5 f \sqrt {a+b \tan ^2(e+f x)}}\\ \end {align*}
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Mathematica [A] time = 2.34, size = 174, normalized size = 0.79 \[ \frac {\sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac {5 b (b-a) \sin (2 (e+f x)) \left (\left (6 a^2-17 a b+11 b^2\right ) \cos (2 (e+f x))+6 a^2-7 a b-11 b^2\right )}{((a-b) \cos (2 (e+f x))+a+b)^2}-\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+2 a (2 a-7 b) \csc ^2(e+f x)-66 a b+73 b^2\right )\right )}{15 \sqrt {2} a^5 f} \]
Antiderivative was successfully verified.
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fricas [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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{6}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.10, size = 371, normalized size = 1.69 \[ -\frac {\left (8 \left (\cos ^{8}\left (f x +e \right )\right ) a^{4}-112 \left (\cos ^{8}\left (f x +e \right )\right ) a^{3} b +328 \left (\cos ^{8}\left (f x +e \right )\right ) a^{2} b^{2}-352 \left (\cos ^{8}\left (f x +e \right )\right ) a \,b^{3}+128 \left (\cos ^{8}\left (f x +e \right )\right ) b^{4}-20 \left (\cos ^{6}\left (f x +e \right )\right ) a^{4}+292 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3} b -976 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b^{2}+1216 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{3}-512 \left (\cos ^{6}\left (f x +e \right )\right ) b^{4}+15 \left (\cos ^{4}\left (f x +e \right )\right ) a^{4}-240 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3} b +1008 a^{2} b^{2} \left (\cos ^{4}\left (f x +e \right )\right )-1536 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{3}+768 \left (\cos ^{4}\left (f x +e \right )\right ) b^{4}+60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b -400 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}+832 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{3}-512 \left (\cos ^{2}\left (f x +e \right )\right ) b^{4}+40 a^{2} b^{2}-160 a \,b^{3}+128 b^{4}\right ) \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \left (\cos ^{5}\left (f x +e \right )\right )}{15 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{4} \sin \left (f x +e \right )^{5} a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.73, size = 337, normalized size = 1.54 \[ -\frac {\frac {40 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {20 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {160 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3}} + \frac {128 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{5}} + \frac {64 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{4}} + \frac {15}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )} - \frac {60 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )} + \frac {48 \, b^{2}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{3} \tan \left (f x + e\right )} + \frac {10}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{3}} - \frac {8 \, b}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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