Optimal. Leaf size=177 \[ -\frac{2 \left (3 a^2 b^2 C+a^3 b B-2 a^4 C-2 a b^3 B+A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{C \tan (c+d x)}{b^2 d} \]
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Rubi [A] time = 0.64743, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4090, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{2 \left (3 a^2 b^2 C+a^3 b B-2 a^4 C-2 a b^3 B+A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{C \tan (c+d x)}{b^2 d} \]
Antiderivative was successfully verified.
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Rule 4090
Rule 4082
Rule 3998
Rule 3770
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (-b \left (A b^2-a (b B-a C)\right )+\left (a^2-b^2\right ) (b B-a C) \sec (c+d x)+b \left (a^2-b^2\right ) C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{C \tan (c+d x)}{b^2 d}+\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\sec (c+d x) \left (-b^2 \left (A b^2-a (b B-a C)\right )+b \left (a^2-b^2\right ) (b B-2 a C) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{C \tan (c+d x)}{b^2 d}+\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{(b B-2 a C) \int \sec (c+d x) \, dx}{b^3}-\frac{\left (A b^4+a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{C \tan (c+d x)}{b^2 d}+\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (A b^4+a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^4 \left (a^2-b^2\right )}\\ &=\frac{(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}+\frac{C \tan (c+d x)}{b^2 d}+\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 \left (A b^4+a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right )\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^4 \left (a^2-b^2\right ) d}\\ &=\frac{(b B-2 a C) \tanh ^{-1}(\sin (c+d x))}{b^3 d}-\frac{2 \left (A b^4+a^3 b B-2 a b^3 B-2 a^4 C+3 a^2 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-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac{C \tan (c+d x)}{b^2 d}+\frac{a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [B] time = 3.00144, size = 382, normalized size = 2.16 \[ \frac{2 (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{2 \left (a \left (a^2 b B-2 a^3 C+3 a b^2 C-2 b^3 B\right )+A b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{a b \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b)}-(b B-2 a C) (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+(b B-2 a C) (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{b C \sin \left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}\right )}{b^3 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.092, size = 630, normalized size = 3.6 \begin{align*} \text{result too large to display} \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 = 63.7846, size = 2514, normalized size = 14.2 \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] 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 ) \sec ^{2}{\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 [B] time = 1.38399, size = 598, normalized size = 3.38 \begin{align*} \frac{\frac{2 \,{\left (2 \, C a^{4} - B a^{3} b - 3 \, C a^{2} b^{2} + 2 \, B a b^{3} - A 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^{2} b^{3} - b^{5}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{2 \,{\left (2 \, C 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} - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A 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} + C b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C 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 ) - C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A 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 ) + C 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 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a + b\right )}{\left (a^{2} b^{2} - b^{4}\right )}} - \frac{{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac{{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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