Optimal. Leaf size=105 \[ -\frac {\sqrt {b} (a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f} \]
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Rubi [A] time = 0.12, 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 = {3663, 461, 205} \[ -\frac {\sqrt {b} (a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {\cot ^5(e+f x)}{5 a f} \]
Antiderivative was successfully verified.
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Rule 205
Rule 461
Rule 3663
Rubi steps
\begin {align*} \int \frac {\csc ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a x^6}+\frac {2 a-b}{a^2 x^4}+\frac {(a-b)^2}{a^3 x^2}-\frac {(a-b)^2 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f}-\frac {\left ((a-b)^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}\\ &=-\frac {(a-b)^2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f}-\frac {(a-b)^2 \cot (e+f x)}{a^3 f}-\frac {(2 a-b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\cot ^5(e+f x)}{5 a f}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 103, normalized size = 0.98 \[ \frac {-\sqrt {a} \cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+8 a^2+a (4 a-5 b) \csc ^2(e+f x)-25 a b+15 b^2\right )-15 \sqrt {b} (a-b)^2 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{15 a^{7/2} f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.46, size = 543, normalized size = 5.17 \[ \left [-\frac {4 \, {\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \, {\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 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 {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{60 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (4 \, a^{2} - 11 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 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 {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} 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 = 1.90, size = 151, normalized size = 1.44 \[ -\frac {\frac {15 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a^{3}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} - 30 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 5 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.55, size = 191, normalized size = 1.82 \[ -\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f a \sqrt {a b}}+\frac {2 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f \,a^{2} \sqrt {a b}}-\frac {b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{f \,a^{3} \sqrt {a b}}-\frac {1}{5 f a \tan \left (f x +e \right )^{5}}-\frac {2}{3 f a \tan \left (f x +e \right )^{3}}+\frac {b}{3 f \,a^{2} \tan \left (f x +e \right )^{3}}-\frac {1}{f a \tan \left (f x +e \right )}+\frac {2 b}{f \,a^{2} \tan \left (f x +e \right )}-\frac {b^{2}}{f \,a^{3} \tan \left (f x +e \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.80, size = 104, normalized size = 0.99 \[ -\frac {\frac {15 \, {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a^{2} - a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.39, size = 115, normalized size = 1.10 \[ -\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a-b\right )}{3\,a^2}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (a^2-2\,a\,b+b^2\right )}{a^3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,{\left (a-b\right )}^2}{\sqrt {a}\,\left (a^2-2\,a\,b+b^2\right )}\right )\,{\left (a-b\right )}^2}{a^{7/2}\,f} \]
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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