Optimal. Leaf size=105 \[ -\frac {a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{7/2}}-\frac {a^2 \cot (e+f x)}{f (a+b)^3}-\frac {\cot ^5(e+f x)}{5 f (a+b)}-\frac {(2 a+b) \cot ^3(e+f x)}{3 f (a+b)^2} \]
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Rubi [A] time = 0.14, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4132, 461, 205} \[ -\frac {a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{f (a+b)^{7/2}}-\frac {a^2 \cot (e+f x)}{f (a+b)^3}-\frac {\cot ^5(e+f x)}{5 f (a+b)}-\frac {(2 a+b) \cot ^3(e+f x)}{3 f (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 205
Rule 461
Rule 4132
Rubi steps
\begin {align*} \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(a+b) x^6}+\frac {2 a+b}{(a+b)^2 x^4}+\frac {a^2}{(a+b)^3 x^2}-\frac {a^2 b}{(a+b)^3 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f}-\frac {\left (a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{(a+b)^3 f}\\ &=-\frac {a^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{7/2} f}-\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f}\\ \end {align*}
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Mathematica [C] time = 1.72, size = 318, normalized size = 3.03 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\sqrt {a+b} \csc (e) \sqrt {b (\cos (e)-i \sin (e))^4} \csc ^5(e+f x) \left (10 \left (8 a^2+b^2\right ) \sin (f x)-40 a^2 \sin (2 e+3 f x)+8 a^2 \sin (4 e+5 f x)+30 a b \sin (2 e+3 f x)+15 a b \sin (4 e+3 f x)-9 a b \sin (4 e+5 f x)-30 b (3 a+b) \sin (2 e+f x)+10 b^2 \sin (2 e+3 f x)-2 b^2 \sin (4 e+5 f x)\right )+240 a^2 b (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )\right )}{480 f (a+b)^{7/2} \sqrt {b (\cos (e)-i \sin (e))^4} \left (a+b \sec ^2(e+f x)\right )} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.53, size = 587, normalized size = 5.59 \[ \left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \, a^{2} \cos \left (f x + e\right )}{60 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.30, size = 180, normalized size = 1.71 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a^{2} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 15 \, a b \tan \left (f x + e\right )^{2} + 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.01, size = 116, normalized size = 1.10 \[ -\frac {a^{2} b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{5 f \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{f \left (a +b \right )^{3} \tan \left (f x +e \right )}-\frac {2 a}{3 f \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {b}{3 f \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 137, normalized size = 1.30 \[ -\frac {\frac {15 \, a^{2} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.05, size = 112, normalized size = 1.07 \[ -\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+b\right )}{3\,{\left (a+b\right )}^2}+\frac {a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+b\right )}^3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5}-\frac {a^2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{f\,{\left (a+b\right )}^{7/2}} \]
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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