Optimal. Leaf size=88 \[ \frac{2 b^2 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{b B \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d} \]
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Rubi [A] time = 0.144806, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {21, 2802, 12, 2747, 3770, 2659, 205} \[ \frac{2 b^2 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{b B \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d} \]
Antiderivative was successfully verified.
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Rule 21
Rule 2802
Rule 12
Rule 2747
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{(a B+b B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=B \int \frac{\sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx\\ &=\frac{B \tan (c+d x)}{a d}-\frac{B \int \frac{b \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a}\\ &=\frac{B \tan (c+d x)}{a d}-\frac{(b B) \int \frac{\sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a}\\ &=\frac{B \tan (c+d x)}{a d}-\frac{(b B) \int \sec (c+d x) \, dx}{a^2}+\frac{\left (b^2 B\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^2}\\ &=-\frac{b B \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d}+\frac{\left (2 b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 d}\\ &=\frac{2 b^2 B \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 \sqrt{a-b} \sqrt{a+b} d}-\frac{b B \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{B \tan (c+d x)}{a d}\\ \end{align*}
Mathematica [A] time = 0.351039, size = 116, normalized size = 1.32 \[ \frac{B \left (-\frac{2 b^2 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+a \tan (c+d x)+b \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{a^2 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.138, size = 139, normalized size = 1.6 \begin{align*} -{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{Bb}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{Bb}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+2\,{\frac{{b}^{2}B}{{a}^{2}d\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \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 [B] time = 1.99455, size = 925, normalized size = 10.51 \begin{align*} \left [-\frac{\sqrt{-a^{2} + b^{2}} B b^{2} \cos \left (d x + c\right ) \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) +{\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac{2 \, \sqrt{a^{2} - b^{2}} B b^{2} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) -{\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (B a^{2} b - B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B a^{3} - B a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\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*} B \int \frac{\sec ^{2}{\left (c + d x \right )}}{a + b \cos{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.5398, size = 209, normalized size = 2.38 \begin{align*} \frac{\frac{2 \,{\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 )} B b^{2}}{\sqrt{a^{2} - b^{2}} a^{2}} - \frac{B b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac{B b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} a}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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