Optimal. Leaf size=189 \[ \frac{2 b \left (3 a^2 A b-2 a^3 B+a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 A+a b B-2 A b^2\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{(2 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]
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Rubi [A] time = 0.673164, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3000, 3055, 3001, 3770, 2659, 205} \[ \frac{2 b \left (3 a^2 A b-2 a^3 B+a b^2 B-2 A b^3\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 A+a b B-2 A b^2\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{(2 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]
Antiderivative was successfully verified.
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Rule 3000
Rule 3055
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac{b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (a^2 A-2 A b^2+a b B-a (A b-a B) \cos (c+d x)+b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A-2 A b^2+a b B\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-\left (a^2-b^2\right ) (2 A b-a B)+a b (A b-a B) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (a^2 A-2 A b^2+a b B\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{(2 A b-a B) \int \sec (c+d x) \, dx}{a^3}+\frac{\left (b \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right )\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac{(2 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (2 b \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right )\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^3 \left (a^2-b^2\right ) d}\\ &=\frac{2 b \left (3 a^2 A b-2 A b^3-2 a^3 B+a b^2 B\right ) \tan ^{-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{(2 A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{\left (a^2 A-2 A b^2+a b B\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac{b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.7186, size = 240, normalized size = 1.27 \[ \frac{-\frac{2 b \left (-3 a^2 A b+2 a^3 B-a b^2 B+2 A b^3\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac{a b^2 (a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+a A \tan (c+d x)-a B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+a B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 A b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 A b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.174, size = 502, normalized size = 2.7 \begin{align*} -{\frac{A}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) Ab}{d{a}^{3}}}-{\frac{B}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) Ab}{d{a}^{3}}}+{\frac{B}{{a}^{2}d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) A}{{a}^{2}d \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 ) B}{da \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{A{b}^{2}}{da \left ( a-b \right ) \left ( a+b \right ) \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 ) }-4\,{\frac{A{b}^{4}}{d{a}^{3} \left ( a-b \right ) \left ( a+b \right ) \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 ) }-4\,{\frac{Bb}{d \left ( a-b \right ) \left ( a+b \right ) \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 ) }+2\,{\frac{B{b}^{3}}{{a}^{2}d \left ( a-b \right ) \left ( a+b \right ) \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 = 69.4434, size = 2419, normalized size = 12.8 \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 \cos{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos{\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 = 1.40503, size = 545, normalized size = 2.88 \begin{align*} \frac{\frac{2 \,{\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2} - B a b^{3} + 2 \, 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^{5} - a^{3} b^{2}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A 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} - B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A 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 ) + B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A 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 (B a - 2 \, A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{{\left (B a - 2 \, A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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