Optimal. Leaf size=65 \[ \frac{(5 a+4 b) \tan ^3(e+f x)}{15 f}+\frac{(5 a+4 b) \tan (e+f x)}{5 f}+\frac{b \tan (e+f x) \sec ^4(e+f x)}{5 f} \]
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Rubi [A] time = 0.0432743, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4046, 3767} \[ \frac{(5 a+4 b) \tan ^3(e+f x)}{15 f}+\frac{(5 a+4 b) \tan (e+f x)}{5 f}+\frac{b \tan (e+f x) \sec ^4(e+f x)}{5 f} \]
Antiderivative was successfully verified.
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Rule 4046
Rule 3767
Rubi steps
\begin{align*} \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac{1}{5} (5 a+4 b) \int \sec ^4(e+f x) \, dx\\ &=\frac{b \sec ^4(e+f x) \tan (e+f x)}{5 f}-\frac{(5 a+4 b) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 f}\\ &=\frac{(5 a+4 b) \tan (e+f x)}{5 f}+\frac{b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac{(5 a+4 b) \tan ^3(e+f x)}{15 f}\\ \end{align*}
Mathematica [A] time = 0.194999, size = 61, normalized size = 0.94 \[ \frac{a \left (\frac{1}{3} \tan ^3(e+f x)+\tan (e+f x)\right )}{f}+\frac{b \left (\frac{1}{5} \tan ^5(e+f x)+\frac{2}{3} \tan ^3(e+f x)+\tan (e+f x)\right )}{f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.032, size = 58, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -a \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \tan \left ( fx+e \right ) -b \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 ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.01184, size = 58, normalized size = 0.89 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \,{\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{3} + 15 \,{\left (a + b\right )} \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 = 0.46819, size = 140, normalized size = 2.15 \begin{align*} \frac{{\left (2 \,{\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{4} +{\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{4}{\left (e + f x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.29146, size = 84, normalized size = 1.29 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 10 \, b \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right ) + 15 \, b \tan \left (f x + e\right )}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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