Optimal. Leaf size=53 \[ \frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^5(e+f x)}{5 f}+\frac{2 a (a+b) \tan ^3(e+f x)}{3 f} \]
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Rubi [A] time = 0.0572108, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3191, 194} \[ \frac{a^2 \tan (e+f x)}{f}+\frac{(a+b)^2 \tan ^5(e+f x)}{5 f}+\frac{2 a (a+b) \tan ^3(e+f x)}{3 f} \]
Antiderivative was successfully verified.
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Rule 3191
Rule 194
Rubi steps
\begin{align*} \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+(a+b) x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2+2 a (a+b) x^2+(a+b)^2 x^4\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{a^2 \tan (e+f x)}{f}+\frac{2 a (a+b) \tan ^3(e+f x)}{3 f}+\frac{(a+b)^2 \tan ^5(e+f x)}{5 f}\\ \end{align*}
Mathematica [A] time = 0.353802, size = 67, normalized size = 1.26 \[ \frac{\tan (e+f x) \left (\left (4 a^2-2 a b-6 b^2\right ) \sec ^2(e+f x)+8 a^2+3 (a+b)^2 \sec ^4(e+f x)-4 a b+3 b^2\right )}{15 f} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.072, size = 101, normalized size = 1.9 \begin{align*}{\frac{1}{f} \left ( -{a}^{2} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{15}} \right ) \tan \left ( fx+e \right ) +2\,ab \left ( 1/5\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}} \right ) +{\frac{{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5}}{5\, \left ( \cos \left ( fx+e \right ) \right ) ^{5}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.00522, size = 74, normalized size = 1.4 \begin{align*} \frac{3 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 10 \,{\left (a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.81191, size = 194, normalized size = 3.66 \begin{align*} \frac{{\left ({\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (2 \, a^{2} - a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.16325, size = 116, normalized size = 2.19 \begin{align*} \frac{3 \, a^{2} \tan \left (f x + e\right )^{5} + 6 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 10 \, a^{2} \tan \left (f x + e\right )^{3} + 10 \, a b \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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