Optimal. Leaf size=180 \[ \frac{\left (a^2 B+a b C-2 b^2 B\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{2 b \left (3 a^2 b B-2 a^3 C+a b^2 C-2 b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 b B-a C)}{a^3} \]
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Rubi [A] time = 0.632915, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4030, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (a^2 B+a b C-2 b^2 B\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{2 b \left (3 a^2 b B-2 a^3 C+a b^2 C-2 b^3 B\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac{x (2 b B-a C)}{a^3} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4030
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\int \frac{\cos (c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\\ &=\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos (c+d x) \left (-a^2 B+2 b^2 B-a b C+a (b B-a C) \sec (c+d x)-b (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 B-2 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{-\left (a^2-b^2\right ) (2 b B-a C)+a b (b B-a C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{(2 b B-a C) x}{a^3}+\frac{\left (a^2 B-2 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 b B-a C) x}{a^3}+\frac{\left (a^2 B-2 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 b B-a C) x}{a^3}+\frac{\left (a^2 B-2 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^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 \left (a^2-b^2\right ) d}\\ &=-\frac{(2 b B-a C) x}{a^3}+\frac{2 b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^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 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (a^2 B-2 b^2 B+a b C\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [A] time = 0.778907, size = 147, normalized size = 0.82 \[ \frac{\frac{2 b \left (-3 a^2 b B+2 a^3 C-a b^2 C+2 b^3 B\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b^2 (a C-b B) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}+(c+d x) (a C-2 b B)+a B \sin (c+d x)}{a^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.121, size = 453, normalized size = 2.5 \begin{align*} 2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) b}{d{a}^{3}}}+2\,{\frac{C\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) B}{d{a}^{2} \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{{b}^{2}\tan \left ( 1/2\,dx+c/2 \right ) C}{ad \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 ) }}+6\,{\frac{B{b}^{2}}{ad \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 ) }-4\,{\frac{B{b}^{4}}{d{a}^{3} \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 ) }-4\,{\frac{Cb}{d \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 ) }+2\,{\frac{C{b}^{3}}{d{a}^{2} \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 ) } \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 = 0.680988, size = 1715, normalized size = 9.53 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.22539, size = 505, normalized size = 2.81 \begin{align*} -\frac{\frac{2 \,{\left (2 \, C a^{3} b - 3 \, B a^{2} b^{2} - C a b^{3} + 2 \, B b^{4}\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 )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \,{\left (B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}{\left (a^{4} - a^{2} b^{2}\right )}} - \frac{{\left (C a - 2 \, B b\right )}{\left (d x + c\right )}}{a^{3}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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