Optimal. Leaf size=98 \[ \frac{2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{x (A b-a B)}{a^2}+\frac{A \sin (c+d x)}{a d} \]
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Rubi [A] time = 0.216397, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {4104, 3919, 3831, 2659, 208} \[ \frac{2 \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d \sqrt{a-b} \sqrt{a+b}}-\frac{x (A b-a B)}{a^2}+\frac{A \sin (c+d x)}{a d} \]
Antiderivative was successfully verified.
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Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{A \sin (c+d x)}{a d}-\frac{\int \frac{A b-a B-a C \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a}\\ &=-\frac{(A b-a B) x}{a^2}+\frac{A \sin (c+d x)}{a d}+\left (\frac{b (A b-a B)}{a^2}+C\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx\\ &=-\frac{(A b-a B) x}{a^2}+\frac{A \sin (c+d x)}{a d}+\frac{\left (\frac{b (A b-a B)}{a^2}+C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b}\\ &=-\frac{(A b-a B) x}{a^2}+\frac{A \sin (c+d x)}{a d}+\frac{\left (2 \left (\frac{b (A b-a B)}{a^2}+C\right )\right ) \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 d}\\ &=-\frac{(A b-a B) x}{a^2}+\frac{2 \left (\frac{b (A b-a B)}{a^2}+C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} \sqrt{a+b} d}+\frac{A \sin (c+d x)}{a d}\\ \end{align*}
Mathematica [A] time = 0.232455, size = 92, normalized size = 0.94 \[ \frac{-\frac{2 \left (a (a C-b B)+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+(c+d x) (a B-A b)+a A \sin (c+d x)}{a^2 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.11, size = 216, normalized size = 2.2 \begin{align*} 2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{Ab\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{ad}}+2\,{\frac{A{b}^{2}}{d{a}^{2}\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 ) }-2\,{\frac{Bb}{ad\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 ) }+2\,{\frac{C}{d\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 ) } \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 = 0.56765, size = 724, normalized size = 7.39 \begin{align*} \left [\frac{2 \,{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x +{\left (C a^{2} - B a b + A b^{2}\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 (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d}, \frac{{\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} d x +{\left (C a^{2} - B a b + A b^{2}\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 (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{4} - a^{2} b^{2}\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 (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )}}{a + b \sec{\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 = 1.24374, size = 197, normalized size = 2.01 \begin{align*} \frac{\frac{{\left (B a - A b\right )}{\left (d x + c\right )}}{a^{2}} + \frac{2 \, A \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} + \frac{2 \,{\left (C a^{2} - B a b + A b^{2}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a^{2}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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