Optimal. Leaf size=100 \[ \frac{2 (a B-b C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b B-a C) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))} \]
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Rubi [A] time = 0.125723, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {4060, 12, 3831, 2659, 208} \[ \frac{2 (a B-b C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{d (a-b)^{3/2} (a+b)^{3/2}}-\frac{(b B-a C) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))} \]
Antiderivative was successfully verified.
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Rule 4060
Rule 12
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=-\frac{(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{a (a B-b C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{(a B-b C) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2-b^2}\\ &=-\frac{(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{(a B-b C) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{(2 (a B-b C)) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b \left (a^2-b^2\right ) d}\\ &=\frac{2 (a B-b C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{(a-b)^{3/2} (a+b)^{3/2} d}-\frac{(b B-a C) \tan (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.338846, size = 97, normalized size = 0.97 \[ \frac{\frac{(a C-b B) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}-\frac{2 (a B-b C) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}}{d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.085, size = 132, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( 2\,{\frac{ \left ( Bb-aC \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{Ba-Cb}{ \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \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 = 0.542636, size = 861, normalized size = 8.61 \begin{align*} \left [\frac{{\left (B a b - C b^{2} +{\left (B a^{2} - C a b\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \,{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d\right )}}, \frac{{\left (B a b - C b^{2} +{\left (B a^{2} - C a b\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) +{\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d}\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{\left (B + C \sec{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.17938, size = 235, normalized size = 2.35 \begin{align*} \frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}{\left (B a - C b\right )}}{{\left (a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}{\left (a^{2} - b^{2}\right )}}\right )}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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