Optimal. Leaf size=92 \[ -\frac{a^2 \tan (d+e x)}{b e \left (a^2 \sec (d+e x)+a b\right )}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b^2 e}+\frac{a x}{b^2} \]
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Rubi [A] time = 0.303862, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4172, 3923, 3919, 3831, 2659, 205} \[ -\frac{a^2 \tan (d+e x)}{b e \left (a^2 \sec (d+e x)+a b\right )}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b^2 e}+\frac{a x}{b^2} \]
Antiderivative was successfully verified.
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Rule 4172
Rule 3923
Rule 3919
Rule 3831
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{a+b \sec (d+e x)}{b^2+2 a b \sec (d+e x)+a^2 \sec ^2(d+e x)} \, dx &=\left (4 a^2\right ) \int \frac{a+b \sec (d+e x)}{\left (2 a b+2 a^2 \sec (d+e x)\right )^2} \, dx\\ &=-\frac{a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}+\frac{\int \frac{4 a^3 \left (a^2-b^2\right )+4 a^2 b \left (a^2-b^2\right ) \sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{2 a b \left (a^2-b^2\right )}\\ &=\frac{a x}{b^2}-\frac{a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (2 a \left (a^2-b^2\right )\right ) \int \frac{\sec (d+e x)}{2 a b+2 a^2 \sec (d+e x)} \, dx}{b^2}\\ &=\frac{a x}{b^2}-\frac{a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (a^2-b^2\right ) \int \frac{1}{1+\frac{b \cos (d+e x)}{a}} \, dx}{a b^2}\\ &=\frac{a x}{b^2}-\frac{a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}-\frac{\left (2 \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{b}{a}+\left (1-\frac{b}{a}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (d+e x)\right )\right )}{a b^2 e}\\ &=\frac{a x}{b^2}-\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{a+b}}\right )}{b^2 e}-\frac{a^2 \tan (d+e x)}{b e \left (a b+a^2 \sec (d+e x)\right )}\\ \end{align*}
Mathematica [A] time = 0.38539, size = 97, normalized size = 1.05 \[ \frac{2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (d+e x)\right )}{\sqrt{b^2-a^2}}\right )+\frac{a (a d+a e x-b \sin (d+e x)+b (d+e x) \cos (d+e x))}{a+b \cos (d+e x)}}{b^2 e} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.093, size = 163, normalized size = 1.8 \begin{align*} 2\,{\frac{a\arctan \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) }{{b}^{2}e}}-2\,{\frac{a\tan \left ( d/2+1/2\,ex \right ) }{be \left ( a \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}-b \left ( \tan \left ( d/2+1/2\,ex \right ) \right ) ^{2}+a+b \right ) }}-2\,{\frac{{a}^{2}}{{b}^{2}e\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( d/2+1/2\,ex \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+2\,{\frac{1}{e\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( d/2+1/2\,ex \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \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 [A] time = 2.01331, size = 670, normalized size = 7.28 \begin{align*} \left [\frac{2 \, a b e x \cos \left (e x + d\right ) + 2 \, a^{2} e x - 2 \, a b \sin \left (e x + d\right ) + \sqrt{-a^{2} + b^{2}}{\left (b \cos \left (e x + d\right ) + a\right )} \log \left (\frac{2 \, a b \cos \left (e x + d\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (e x + d\right ) + b\right )} \sin \left (e x + d\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (e x + d\right )^{2} + 2 \, a b \cos \left (e x + d\right ) + a^{2}}\right )}{2 \,{\left (b^{3} e \cos \left (e x + d\right ) + a b^{2} e\right )}}, \frac{a b e x \cos \left (e x + d\right ) + a^{2} e x - a b \sin \left (e x + d\right ) - \sqrt{a^{2} - b^{2}}{\left (b \cos \left (e x + d\right ) + a\right )} \arctan \left (-\frac{a \cos \left (e x + d\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (e x + d\right )}\right )}{b^{3} e \cos \left (e x + d\right ) + a b^{2} e}\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{a + b \sec{\left (d + e x \right )}}{\left (a \sec{\left (d + e x \right )} + b\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.23847, size = 196, normalized size = 2.13 \begin{align*}{\left (\frac{{\left (x e + d\right )} a}{b^{2}} - \frac{2 \, a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{{\left (a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )^{2} + a + b\right )} b} - \frac{2 \,{\left (\pi \left \lfloor \frac{x e + d}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right ) - b \tan \left (\frac{1}{2} \, x e + \frac{1}{2} \, d\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} \sqrt{a^{2} - b^{2}}}{b^{2}}\right )} e^{\left (-1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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