Optimal. Leaf size=106 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {b} f (a+b)^{5/2}}+\frac {3 \tan (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\tan (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
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Rubi [A] time = 0.08, antiderivative size = 106, 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 = {4146, 199, 205} \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {b} f (a+b)^{5/2}}+\frac {3 \tan (e+f x)}{8 f (a+b)^2 \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\tan (e+f x)}{4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2} \]
Antiderivative was successfully verified.
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Rule 199
Rule 205
Rule 4146
Rubi steps
\begin {align*} \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 (a+b) f}\\ &=\frac {\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {3 \tan (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 (a+b)^2 f}\\ &=\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {b} (a+b)^{5/2} f}+\frac {\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {3 \tan (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 = 2.54, size = 265, normalized size = 2.50 \[ \frac {\sec ^6(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac {\sec (2 e) \left (a (5 a+2 b) \sin (2 f x)-\left (5 a^2+16 a b+8 b^2\right ) \sin (2 e)\right ) (a \cos (2 (e+f x))+a+2 b)}{a^2}+\frac {4 b (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a^2}-\frac {3 (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b)^2 \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{64 f (a+b)^2 \left (a+b \sec ^2(e+f x)\right )^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.96, size = 580, normalized size = 5.47 \[ \left [-\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a b - b^{2}} \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 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \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 ) - 4 \, {\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 130, normalized size = 1.23 \[ \frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {3 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) + 5 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.79, size = 97, normalized size = 0.92 \[ \frac {\tan \left (f x +e \right )}{4 \left (a +b \right ) f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {3 \tan \left (f x +e \right )}{8 \left (a +b \right )^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {3 \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{8 f \left (a +b \right )^{2} \sqrt {\left (a +b \right ) b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.46, size = 156, normalized size = 1.47 \[ \frac {\frac {3 \, b \tan \left (f x + e\right )^{3} + 5 \, {\left (a + b\right )} \tan \left (f x + e\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {3 \, \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b}}}{8 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 5.00, size = 112, normalized size = 1.06 \[ \frac {\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{8\,\left (a+b\right )}+\frac {3\,b\,{\mathrm {tan}\left (e+f\,x\right )}^3}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{8\,\sqrt {b}\,f\,{\left (a+b\right )}^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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