Optimal. Leaf size=51 \[ \frac{a+b}{2 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f} \]
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Rubi [A] time = 0.0823833, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 444, 43} \[ \frac{a+b}{2 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f} \]
Antiderivative was successfully verified.
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Rule 4138
Rule 444
Rule 43
Rubi steps
\begin{align*} \int \frac{\tan ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x \left (1-x^2\right )}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1-x}{(b+a x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{a+b}{a (b+a x)^2}-\frac{1}{a (b+a x)}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a+b}{2 a^2 f \left (b+a \cos ^2(e+f x)\right )}+\frac{\log \left (b+a \cos ^2(e+f x)\right )}{2 a^2 f}\\ \end{align*}
Mathematica [A] time = 0.731966, size = 81, normalized size = 1.59 \[ \frac{(a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+a \cos (2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+2 (a+b)}{2 a^2 f (a \cos (2 (e+f x))+a+2 b)} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.079, size = 68, normalized size = 1.3 \begin{align*}{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,{a}^{2}f}}+{\frac{1}{2\,fa \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{b}{2\,{a}^{2}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0.994416, size = 80, normalized size = 1.57 \begin{align*} -\frac{\frac{a + b}{a^{3} \sin \left (f x + e\right )^{2} - a^{3} - a^{2} b} - \frac{\log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0.524043, size = 131, normalized size = 2.57 \begin{align*} \frac{{\left (a \cos \left (f x + e\right )^{2} + b\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + a + b}{2 \,{\left (a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f\right )}} \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 [B] time = 1.86911, size = 458, normalized size = 8.98 \begin{align*} \frac{\frac{\log \left (a + b + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{2 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{2}} - \frac{2 \, \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} - \frac{a + b + \frac{6 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{2 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{{\left (a + b + \frac{2 \, a{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{2 \, b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} a^{2}}}{2 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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