Optimal. Leaf size=90 \[ \frac{(a-2 b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} f}+\frac{x}{a^2}-\frac{(a+b) \tan (e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]
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Rubi [A] time = 0.176449, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4141, 1975, 470, 522, 203, 205} \[ \frac{(a-2 b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} f}+\frac{x}{a^2}-\frac{(a+b) \tan (e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]
Antiderivative was successfully verified.
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Rule 4141
Rule 1975
Rule 470
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{a+b+(a-b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a b f}\\ &=-\frac{(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^2 f}+\frac{((a-2 b) (a+b)) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 b f}\\ &=\frac{x}{a^2}+\frac{(a-2 b) \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{3/2} f}-\frac{(a+b) \tan (e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 2.98309, size = 249, normalized size = 2.77 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\frac{\left (-a^2+a b+2 b^2\right ) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{b f \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}+2 x (a \cos (2 (e+f x))+a+2 b)+\frac{(a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}\right )}{8 a^2 \left (a+b \sec ^2(e+f x)\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.096, size = 168, normalized size = 1.9 \begin{align*}{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{2}}}-{\frac{\tan \left ( fx+e \right ) }{2\,fb \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{1}{2\,fb}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{1}{2\,fa}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}}-{\frac{\tan \left ( fx+e \right ) }{2\,fa \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{b}{f{a}^{2}}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.609202, size = 933, normalized size = 10.37 \begin{align*} \left [\frac{8 \, a b f x \cos \left (f x + e\right )^{2} + 8 \, b^{2} f x - 4 \,{\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt{-\frac{a + b}{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 b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt{-\frac{a + b}{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 )}{8 \,{\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}, \frac{4 \, a b f x \cos \left (f x + e\right )^{2} + 4 \, b^{2} f x - 2 \,{\left (a^{2} + a b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) -{\left ({\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b - 2 \, b^{2}\right )} \sqrt{\frac{a + b}{b}} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{a + b}{b}}}{2 \,{\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \,{\left (a^{3} b f \cos \left (f x + e\right )^{2} + a^{2} b^{2} f\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.25342, size = 170, normalized size = 1.89 \begin{align*} \frac{\frac{2 \,{\left (f x + e\right )}}{a^{2}} + \frac{{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )}{\left (a^{2} - a b - 2 \, b^{2}\right )}}{\sqrt{a b + b^{2}} a^{2} b} - \frac{a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a b}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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