Optimal. Leaf size=128 \[ -\frac{2 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (a^2 (A+2 C)+2 A b^2\right )}{2 a^3}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d} \]
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Rubi [A] time = 0.387222, antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4105, 4104, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (\frac{2 A b^2}{a^2}+A+2 C\right )}{2 a}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \sin (c+d x) \cos (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
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Rule 4105
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\int \frac{\cos (c+d x) \left (2 A b-a (A+2 C) \sec (c+d x)-A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a}\\ &=-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}+\frac{\int \frac{2 A b^2+a^2 (A+2 C)+a A b \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2}\\ &=\frac{\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\left (b \left (A b^2+a^2 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3}\\ &=\frac{\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\left (A b^2+a^2 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^3}\\ &=\frac{\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\left (2 \left (A b^2+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 )}{a^3 d}\\ &=\frac{\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac{2 b \left (A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 \sqrt{a-b} \sqrt{a+b} d}-\frac{A b \sin (c+d x)}{a^2 d}+\frac{A \cos (c+d x) \sin (c+d x)}{2 a d}\\ \end{align*}
Mathematica [A] time = 0.357908, size = 115, normalized size = 0.9 \[ \frac{2 (c+d x) \left (a^2 (A+2 C)+2 A b^2\right )+\frac{8 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a^2 A \sin (2 (c+d x))-4 a A b \sin (c+d x)}{4 a^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.114, size = 296, normalized size = 2.3 \begin{align*} -{\frac{A}{ad} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}Ab}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) b}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{A}{ad}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A{b}^{2}}{d{a}^{3}}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}}-2\,{\frac{A{b}^{3}}{d{a}^{3}\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{Cb}{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 ) } \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 = 0.56873, size = 848, normalized size = 6.62 \begin{align*} \left [\frac{{\left ({\left (A + 2 \, C\right )} a^{4} +{\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} d x +{\left (C a^{2} b + A b^{3}\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 ) -{\left (2 \, A a^{3} b - 2 \, A a b^{3} -{\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )} d}, \frac{{\left ({\left (A + 2 \, C\right )} a^{4} +{\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} d x - 2 \,{\left (C a^{2} b + A b^{3}\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 (2 \, A a^{3} b - 2 \, A a b^{3} -{\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{5} - a^{3} 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 + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\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 = 1.2499, size = 269, normalized size = 2.1 \begin{align*} \frac{\frac{{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )}{\left (d x + c\right )}}{a^{3}} - \frac{4 \,{\left (C a^{2} b + A b^{3}\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^{3}} - \frac{2 \,{\left (A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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