Optimal. Leaf size=165 \[ -\frac{(A-C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1) \sqrt{\sec (c+d x)}}+\frac{2 (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.394108, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4221, 3042, 2978, 2748, 2641, 2639} \[ -\frac{(A-C) \sin (c+d x)}{a^2 d (\cos (c+d x)+1) \sqrt{\sec (c+d x)}}+\frac{2 (A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{(A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A+C) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 4221
Rule 3042
Rule 2978
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sqrt{\sec (c+d x)}}{(a+a \cos (c+d x))^2} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} a (5 A-C)-\frac{1}{2} a (A-5 C) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(A-C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sqrt{\sec (c+d x)}}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}+\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{a^2 (A+C)+\frac{3}{2} a^2 (A-C) \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{(A-C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sqrt{\sec (c+d x)}}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}+\frac{\left ((A-C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}+\frac{\left ((A+C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a^2}\\ &=\frac{(A-C) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{a^2 d}+\frac{2 (A+C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 a^2 d}-\frac{(A-C) \sin (c+d x)}{a^2 d (1+\cos (c+d x)) \sqrt{\sec (c+d x)}}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 4.76956, size = 450, normalized size = 2.73 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left ((7 A-5 C) \cos \left (\frac{1}{2} (c-d x)\right )+2 (A-2 C) \cos \left (\frac{1}{2} (3 c+d x)\right )+3 (A-C) \cos \left (\frac{1}{2} (c+3 d x)\right )\right )}{2 \sqrt{\sec (c+d x)}}-2 \sqrt{2} A \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+8 A \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \sqrt{2} C \csc (c) e^{-i d x} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \left (\left (-1+e^{2 i c}\right ) e^{2 i d x} \, _2F_1\left (\frac{1}{2},\frac{3}{4};\frac{7}{4};-e^{2 i (c+d x)}\right )-3 \sqrt{1+e^{2 i (c+d x)}}\right )+8 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.344, size = 419, normalized size = 2.5 \begin{align*}{\frac{1}{6\,{a}^{2}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 12\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-4\,A\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+6\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -12\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-4\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-6\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -16\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+20\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+3\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-9\,C \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+A+C \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}, x\right ) \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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