Optimal. Leaf size=119 \[ -\frac{(3 a-2 b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{5/2} f}-\frac{x}{a^2}+\frac{(3 a+b) \tan (e+f x)}{2 a b^2 f}-\frac{(a+b) \tan ^3(e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]
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Rubi [A] time = 0.266504, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4141, 1975, 470, 582, 522, 203, 205} \[ -\frac{(3 a-2 b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{5/2} f}-\frac{x}{a^2}+\frac{(3 a+b) \tan (e+f x)}{2 a b^2 f}-\frac{(a+b) \tan ^3(e+f x)}{2 a b f \left (a+b \tan ^2(e+f x)+b\right )} \]
Antiderivative was successfully verified.
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Rule 4141
Rule 1975
Rule 470
Rule 582
Rule 522
Rule 203
Rule 205
Rubi steps
\begin{align*} \int \frac{\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^6}{\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^6}{\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 ^3(e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 (a+b)+(3 a+b) x^2\right )}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a b f}\\ &=\frac{(3 a+b) \tan (e+f x)}{2 a b^2 f}-\frac{(a+b) \tan ^3(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) (3 a+b)+\left (3 a^2+4 a b-b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a b^2 f}\\ &=\frac{(3 a+b) \tan (e+f x)}{2 a b^2 f}-\frac{(a+b) \tan ^3(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{\left ((3 a-2 b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 b^2 f}\\ &=-\frac{x}{a^2}-\frac{(3 a-2 b) (a+b)^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^2 b^{5/2} f}+\frac{(3 a+b) \tan (e+f x)}{2 a b^2 f}-\frac{(a+b) \tan ^3(e+f x)}{2 a b f \left (a+b+b \tan ^2(e+f x)\right )}\\ \end{align*}
Mathematica [C] time = 4.8728, size = 286, normalized size = 2.4 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-\frac{(a+b)^2 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a^2 b^2 f (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}+\frac{(3 a-2 b) (a+b)^{3/2} (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{a^2 b^2 f \sqrt{b (\cos (e)-i \sin (e))^4}}-\frac{2 x (a \cos (2 (e+f x))+a+2 b)}{a^2}+\frac{2 \sec (e) \sin (f x) \sec (e+f x) (a \cos (2 (e+f x))+a+2 b)}{b^2 f}\right )}{8 \left (a+b \sec ^2(e+f x)\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.09, size = 242, normalized size = 2. \begin{align*}{\frac{\tan \left ( fx+e \right ) }{{b}^{2}f}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f{a}^{2}}}+{\frac{a\tan \left ( fx+e \right ) }{2\,{b}^{2}f \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( fx+e \right ) }{fb \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( fx+e \right ) }{2\,af \left ( a+b+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,a}{2\,{b}^{2}f}\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}}}}-2\,{\frac{1}{fb\sqrt{ \left ( a+b \right ) b}}\arctan \left ({\frac{\tan \left ( fx+e \right ) b}{\sqrt{ \left ( a+b \right ) b}}} \right ) }+{\frac{1}{2\,af}\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{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 = 0.637105, size = 1197, normalized size = 10.06 \begin{align*} \left [-\frac{8 \, a b^{2} f x \cos \left (f x + e\right )^{3} + 8 \, b^{3} f x \cos \left (f x + e\right ) +{\left ({\left (3 \, a^{3} + a^{2} b - 2 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )\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 ) - 4 \,{\left (2 \, a^{2} b +{\left (3 \, a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \,{\left (a^{3} b^{2} f \cos \left (f x + e\right )^{3} + a^{2} b^{3} f \cos \left (f x + e\right )\right )}}, -\frac{4 \, a b^{2} f x \cos \left (f x + e\right )^{3} + 4 \, b^{3} f x \cos \left (f x + e\right ) -{\left ({\left (3 \, a^{3} + a^{2} b - 2 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )\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 ) - 2 \,{\left (2 \, a^{2} b +{\left (3 \, a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{4 \,{\left (a^{3} b^{2} f \cos \left (f x + e\right )^{3} + a^{2} b^{3} f \cos \left (f x + e\right )\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 ^{6}{\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 = 3.98214, size = 220, normalized size = 1.85 \begin{align*} -\frac{\frac{2 \,{\left (f x + e\right )}}{a^{2}} - \frac{2 \, \tan \left (f x + e\right )}{b^{2}} + \frac{{\left (3 \, a^{3} + 4 \, a^{2} b - a b^{2} - 2 \, b^{3}\right )}{\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 )}}{\sqrt{a b + b^{2}} a^{2} b^{2}} - \frac{a^{2} \tan \left (f x + e\right ) + 2 \, a b \tan \left (f x + e\right ) + b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a b^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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