Optimal. Leaf size=242 \[ -\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{11/2}}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 f (a+b)^5 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 f (a+b)^5}-\frac {(10 a+b) \cot ^3(e+f x)}{15 f (a+b)^4}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.37, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 462, 456, 1259, 1261, 205} \[ -\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{11/2}}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 f (a+b)^5 \left (a+b \tan ^2(e+f x)+b\right )}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 f (a+b)^4 \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 f (a+b)^5}-\frac {(10 a+b) \cot ^3(e+f x)}{15 f (a+b)^4}-\frac {\cot ^5(e+f x)}{5 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 205
Rule 456
Rule 462
Rule 1259
Rule 1261
Rule 4132
Rubi steps
\begin {align*} \int \frac {\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {10 a+b+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {4 (10 a+b)}{b (a+b)}-\frac {4 \left (5 a^2+4 b^2\right ) x^2}{b (a+b)^2}+\frac {3 \left (5 a^2+4 b^2\right ) x^4}{(a+b)^3}}{x^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{20 (a+b) f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-8 b (a+b) (10 a+b)-8 b \left (5 a^2-10 a b+3 b^2\right ) x^2+\frac {b^2 \left (35 a^2-40 a b+24 b^2\right ) x^4}{a+b}}{x^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{40 b (a+b)^4 f}\\ &=-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (-\frac {8 b (10 a+b)}{x^4}-\frac {8 b \left (5 a^2-20 a b+2 b^2\right )}{(a+b) x^2}+\frac {5 b^2 \left (15 a^2-40 a b+8 b^2\right )}{(a+b) \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{40 b (a+b)^4 f}\\ &=-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 (a+b)^5 f}-\frac {(10 a+b) \cot ^3(e+f x)}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\left (b \left (15 a^2-40 a b+8 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^5 f}\\ &=-\frac {\sqrt {b} \left (15 a^2-40 a b+8 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{11/2} f}-\frac {\left (5 a^2-20 a b+2 b^2\right ) \cot (e+f x)}{5 (a+b)^5 f}-\frac {(10 a+b) \cot ^3(e+f x)}{15 (a+b)^4 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \tan (e+f x)}{20 (a+b)^4 f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{40 (a+b)^5 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [C] time = 5.77, size = 479, normalized size = 1.98 \[ \frac {\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (8 \left (8 a^2-59 a b+23 b^2\right ) \csc (e) \sin (f x) \csc (e+f x) (a \cos (2 (e+f x))+a+2 b)^2+15 b \sec (2 e) \left (\left (9 a^2+16 a b-8 b^2\right ) \sin (2 e)+3 a (2 b-3 a) \sin (2 f x)\right ) (a \cos (2 (e+f x))+a+2 b)+\frac {15 b \left (15 a^2-40 a b+8 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}-60 b^2 (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))-24 (a+b)^2 \cot (e) \csc ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2-8 (4 a-11 b) (a+b) \cot (e) \csc ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)^2+24 (a+b)^2 \csc (e) \sin (f x) \csc ^5(e+f x) (a \cos (2 (e+f x))+a+2 b)^2+8 (4 a-11 b) (a+b) \csc (e) \sin (f x) \csc ^3(e+f x) (a \cos (2 (e+f x))+a+2 b)^2\right )}{960 f (a+b)^5 \left (a+b \sec ^2(e+f x)\right )^3} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.67, size = 1423, normalized size = 5.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 1.09, size = 382, normalized size = 1.58 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 8 \, a b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) + a^{2} b^{2} \tan \left (f x + e\right ) - 8 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 60 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 5 \, a b \tan \left (f x + e\right )^{2} - 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.18, size = 411, normalized size = 1.70 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right ) a^{2}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \left (\tan ^{3}\left (f x +e \right )\right ) a}{f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right ) a^{3}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {b^{2} \tan \left (f x +e \right ) a^{2}}{8 f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {b^{3} \tan \left (f x +e \right ) a}{f \left (a +b \right )^{5} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a^{2}}{8 f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right ) a}{f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}-\frac {b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{5} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 f \left (a +b \right )^{3} \tan \left (f x +e \right )^{5}}-\frac {2 a}{3 f \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}+\frac {b}{3 f \left (a +b \right )^{4} \tan \left (f x +e \right )^{3}}-\frac {a^{2}}{f \left (a +b \right )^{5} \tan \left (f x +e \right )}+\frac {4 a b}{f \left (a +b \right )^{5} \tan \left (f x +e \right )}-\frac {b^{2}}{f \left (a +b \right )^{5} \tan \left (f x +e \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.45, size = 434, normalized size = 1.79 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, {\left (15 \, a^{2} b^{2} - 40 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{8} + 25 \, {\left (15 \, a^{3} b - 25 \, a^{2} b^{2} - 32 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{6} + 8 \, {\left (15 \, a^{4} - 10 \, a^{3} b - 57 \, a^{2} b^{2} - 24 \, a b^{3} + 8 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + 24 \, a^{4} + 96 \, a^{3} b + 144 \, a^{2} b^{2} + 96 \, a b^{3} + 24 \, b^{4} + 8 \, {\left (10 \, a^{4} + 31 \, a^{3} b + 33 \, a^{2} b^{2} + 13 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}}{{\left (a^{5} b^{2} + 5 \, a^{4} b^{3} + 10 \, a^{3} b^{4} + 10 \, a^{2} b^{5} + 5 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{9} + 2 \, {\left (a^{6} b + 6 \, a^{5} b^{2} + 15 \, a^{4} b^{3} + 20 \, a^{3} b^{4} + 15 \, a^{2} b^{5} + 6 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{7} + {\left (a^{7} + 7 \, a^{6} b + 21 \, a^{5} b^{2} + 35 \, a^{4} b^{3} + 35 \, a^{3} b^{4} + 21 \, a^{2} b^{5} + 7 \, a b^{6} + b^{7}\right )} \tan \left (f x + e\right )^{5}}}{120 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.49, size = 267, normalized size = 1.10 \[ -\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a+b\right )}{15\,{\left (a+b\right )}^2}+\frac {5\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (15\,a^2\,b-40\,a\,b^2+8\,b^3\right )}{24\,{\left (a+b\right )}^4}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{15\,{\left (a+b\right )}^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^8\,\left (15\,a^2\,b^2-40\,a\,b^3+8\,b^4\right )}{8\,{\left (a+b\right )}^5}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5\,\left (a^2+2\,a\,b+b^2\right )+{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^5+5\,a^4\,b+10\,a^3\,b^2+10\,a^2\,b^3+5\,a\,b^4+b^5\right )}{{\left (a+b\right )}^{11/2}}\right )\,\left (15\,a^2-40\,a\,b+8\,b^2\right )}{8\,f\,{\left (a+b\right )}^{11/2}} \]
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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