Optimal. Leaf size=124 \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{7/2}}-\frac {15 \cot (e+f x)}{8 f (a+b)^3}+\frac {5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.11, antiderivative size = 124, 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 = {4132, 290, 325, 205} \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 f (a+b)^{7/2}}-\frac {15 \cot (e+f x)}{8 f (a+b)^3}+\frac {5 \cot (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\cot (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 205
Rule 290
Rule 325
Rule 4132
Rubi steps
\begin {align*} \int \frac {\csc ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a+b) f}\\ &=\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {15 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^2 f}\\ &=-\frac {15 \cot (e+f x)}{8 (a+b)^3 f}+\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^3 f}\\ &=-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 (a+b)^{7/2} f}-\frac {15 \cot (e+f x)}{8 (a+b)^3 f}+\frac {\cot (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {5 \cot (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end {align*}
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Mathematica [C] time = 6.82, size = 987, normalized size = 7.96 \[ \frac {(\cos (2 e+2 f x) a+a+2 b)^3 \left (\frac {15 b \tan ^{-1}\left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {15 i b \tan ^{-1}\left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) \sec ^6(e+f x)}{(a+b)^3 \left (b \sec ^2(e+f x)+a\right )^3}+\frac {(\cos (2 e+2 f x) a+a+2 b) \csc (e) \csc (e+f x) \sec (2 e) \left (-32 \sin (f x) a^4+32 \sin (3 f x) a^4-48 \sin (2 e-f x) a^4+48 \sin (2 e+f x) a^4-32 \sin (4 e+f x) a^4-8 \sin (2 e+3 f x) a^4+32 \sin (4 e+3 f x) a^4-8 \sin (6 e+3 f x) a^4+8 \sin (2 e+5 f x) a^4+8 \sin (6 e+5 f x) a^4-64 b \sin (f x) a^3+46 b \sin (3 f x) a^3-128 b \sin (2 e-f x) a^3+146 b \sin (2 e+f x) a^3-82 b \sin (4 e+f x) a^3+18 b \sin (2 e+3 f x) a^3+73 b \sin (4 e+3 f x) a^3-9 b \sin (6 e+3 f x) a^3-9 b \sin (2 e+5 f x) a^3+9 b \sin (4 e+5 f x) a^3+22 b^2 \sin (f x) a^2-54 b^2 \sin (3 f x) a^2-106 b^2 \sin (2 e-f x) a^2+182 b^2 \sin (2 e+f x) a^2-54 b^2 \sin (4 e+f x) a^2+54 b^2 \sin (2 e+3 f x) a^2+24 b^2 \sin (4 e+3 f x) a^2-24 b^2 \sin (6 e+3 f x) a^2-2 b^2 \sin (2 e+5 f x) a^2+2 b^2 \sin (4 e+5 f x) a^2+80 b^3 \sin (f x) a-8 b^3 \sin (3 f x) a+80 b^3 \sin (2 e-f x) a+80 b^3 \sin (2 e+f x) a-80 b^3 \sin (4 e+f x) a+8 b^3 \sin (2 e+3 f x) a+8 b^3 \sin (4 e+3 f x) a-8 b^3 \sin (6 e+3 f x) a+16 b^4 \sin (f x)+16 b^4 \sin (2 e-f x)+16 b^4 \sin (2 e+f x)-16 b^4 \sin (4 e+f x)\right ) \sec ^6(e+f x)}{512 a^2 (a+b)^3 f \left (b \sec ^2(e+f x)+a\right )^3} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.61, size = 615, normalized size = 4.96 \[ \left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (5 \, 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 b \cos \left (f x + e\right )^{2} + b^{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 \, b^{2} \cos \left (f x + e\right )}{32 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\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 (5 \, 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 b \cos \left (f x + e\right )^{2} + b^{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 \, b^{2} \cos \left (f x + e\right )}{16 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 3 \, a^{3} b^{2} + 3 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\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 = 0.50, size = 184, normalized size = 1.48 \[ -\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {7 \, b^{2} \tan \left (f x + e\right )^{3} + 9 \, a b \tan \left (f x + e\right ) + 9 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} + \frac {8}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 1.07, size = 157, normalized size = 1.27 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \left (a +b \right )^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 a b \tan \left (f x +e \right )}{8 \left (a +b \right )^{3} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b^{2} \tan \left (f x +e \right )}{8 f \left (a +b \right )^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}-\frac {1}{f \left (a +b \right )^{3} \tan \left (f x +e \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 219, normalized size = 1.77 \[ -\frac {\frac {15 \, b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, b^{2} \tan \left (f x + e\right )^{4} + 25 \, {\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{5} + 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{3} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.11, size = 146, normalized size = 1.18 \[ -\frac {\frac {1}{a+b}+\frac {25\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,{\left (a+b\right )}^2}+\frac {15\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{8\,{\left (a+b\right )}^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3\,\left (2\,b^2+2\,a\,b\right )+\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+2\,a\,b+b^2\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{8\,f\,{\left (a+b\right )}^{7/2}} \]
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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