Optimal. Leaf size=116 \[ -\frac {\sqrt {b} (3 a-5 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {b (a-b) \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f} \]
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Rubi [A] time = 0.15, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3663, 456, 1261, 205} \[ -\frac {\sqrt {b} (3 a-5 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {b (a-b) \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f} \]
Antiderivative was successfully verified.
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Rule 205
Rule 456
Rule 1261
Rule 3663
Rubi steps
\begin {align*} \int \frac {\csc ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1+x^2}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {2}{a b}-\frac {2 (a-b) x^2}{a^2 b}+\frac {(a-b) x^4}{a^3}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \operatorname {Subst}\left (\int \left (-\frac {2}{a^2 b x^4}-\frac {2 (a-2 b)}{a^3 b x^2}+\frac {3 a-5 b}{a^3 \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}-\frac {((3 a-5 b) b) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^3 f}\\ &=-\frac {(3 a-5 b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{7/2} f}-\frac {(a-2 b) \cot (e+f x)}{a^3 f}-\frac {\cot ^3(e+f x)}{3 a^2 f}-\frac {(a-b) b \tan (e+f x)}{2 a^3 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.99, size = 112, normalized size = 0.97 \[ \frac {3 \sqrt {b} (5 b-3 a) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+\sqrt {a} \left (\frac {3 b (b-a) \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}-2 \cot (e+f x) \left (a \csc ^2(e+f x)+2 a-6 b\right )\right )}{6 a^{7/2} f} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.65, size = 587, normalized size = 5.06 \[ \left [-\frac {4 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, 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 ) - 12 \, {\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (4 \, a^{2} - 19 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - 14 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - 8 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (3 \, a^{2} - 11 \, a b + 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, a b + 5 \, 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 ) - 6 \, {\left (3 \, a b - 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\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 = 3.34, size = 142, normalized size = 1.22 \[ -\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b - 5 \, b^{2}\right )}}{\sqrt {a b} a^{3}} + \frac {3 \, {\left (a b \tan \left (f x + e\right ) - b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} a^{3}} + \frac {2 \, {\left (3 \, a \tan \left (f x + e\right )^{2} - 6 \, b \tan \left (f x + e\right )^{2} + a\right )}}{a^{3} \tan \left (f x + e\right )^{3}}}{6 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.62, size = 169, normalized size = 1.46 \[ -\frac {b \tan \left (f x +e \right )}{2 f \,a^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{2} \tan \left (f x +e \right )}{2 f \,a^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{2} \sqrt {a b}}+\frac {5 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \,a^{3} \sqrt {a b}}-\frac {1}{3 f \,a^{2} \tan \left (f x +e \right )^{3}}-\frac {1}{f \,a^{2} \tan \left (f x +e \right )}+\frac {2 b}{f \,a^{3} \tan \left (f x +e \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 115, normalized size = 0.99 \[ -\frac {\frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} - 5 \, a b\right )} \tan \left (f x + e\right )^{2} + 2 \, a^{2}}{a^{3} b \tan \left (f x + e\right )^{5} + a^{4} \tan \left (f x + e\right )^{3}} + \frac {3 \, {\left (3 \, a b - 5 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}}}{6 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 11.37, size = 108, normalized size = 0.93 \[ -\frac {\frac {1}{3\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (3\,a-5\,b\right )}{3\,a^2}+\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (3\,a-5\,b\right )}{2\,a^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^5+a\,{\mathrm {tan}\left (e+f\,x\right )}^3\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (3\,a-5\,b\right )}{2\,a^{7/2}\,f} \]
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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