Optimal. Leaf size=48 \[ \frac{a \tan ^3(e+f x)}{3 f}-\frac{a \tan (e+f x)}{f}+a x+\frac{b \tan ^5(e+f x)}{5 f} \]
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Rubi [A] time = 0.0565195, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4141, 1802, 203} \[ \frac{a \tan ^3(e+f x)}{3 f}-\frac{a \tan (e+f x)}{f}+a x+\frac{b \tan ^5(e+f x)}{5 f} \]
Antiderivative was successfully verified.
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Rule 4141
Rule 1802
Rule 203
Rubi steps
\begin{align*} \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b \left (1+x^2\right )\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a+a x^2+b x^4+\frac{a}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{a \tan (e+f x)}{f}+\frac{a \tan ^3(e+f x)}{3 f}+\frac{b \tan ^5(e+f x)}{5 f}+\frac{a \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=a x-\frac{a \tan (e+f x)}{f}+\frac{a \tan ^3(e+f x)}{3 f}+\frac{b \tan ^5(e+f x)}{5 f}\\ \end{align*}
Mathematica [A] time = 0.0206226, size = 57, normalized size = 1.19 \[ \frac{a \tan ^3(e+f x)}{3 f}+\frac{a \tan ^{-1}(\tan (e+f x))}{f}-\frac{a \tan (e+f x)}{f}+\frac{b \tan ^5(e+f x)}{5 f} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.048, size = 50, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( a \left ({\frac{ \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{3}}-\tan \left ( fx+e \right ) +fx+e \right ) +{\frac{b \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 = 1.47521, size = 61, normalized size = 1.27 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \,{\left (f x + e\right )} a - 15 \, a \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 = 0.495778, size = 177, normalized size = 3.69 \begin{align*} \frac{15 \, a f x \cos \left (f x + e\right )^{5} -{\left ({\left (20 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} -{\left (5 \, a - 6 \, 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 [A] time = 2.9966, size = 54, normalized size = 1.12 \begin{align*} a \left (\begin{cases} x + \frac{\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac{\tan{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \tan ^{4}{\left (e \right )} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} x \tan ^{4}{\left (e \right )} \sec ^{2}{\left (e \right )} & \text{for}\: f = 0 \\\frac{\tan ^{5}{\left (e + f x \right )}}{5 f} & \text{otherwise} \end{cases}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.26045, size = 66, normalized size = 1.38 \begin{align*} \frac{3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \,{\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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