Optimal. Leaf size=148 \[ \frac{2 (a B+A b) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 (a B+A b) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 (3 a A+5 b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
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Rubi [A] time = 0.163442, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3996, 3787, 3769, 3771, 2641, 2639} \[ \frac{2 (a B+A b) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 (a B+A b) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 (3 a A+5 b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
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Rule 3996
Rule 3787
Rule 3769
Rule 3771
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+b \sec (c+d x)) (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-\frac{2}{5} \int \frac{-\frac{5}{2} (A b+a B)-\frac{1}{2} (3 a A+5 b B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}-(-A b-a B) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx-\frac{1}{5} (-3 a A-5 b B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{1}{3} (-A b-a B) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{5} \left ((-3 a A-5 b B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{2 (3 a A+5 b B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}-\frac{1}{3} \left ((-A b-a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (3 a A+5 b B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{2 (A b+a B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a A \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+a B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.604255, size = 108, normalized size = 0.73 \[ \frac{\sqrt{\sec (c+d x)} \left (10 (a B+A b) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+\sin (2 (c+d x)) (3 a A \cos (c+d x)+5 a B+5 A b)+6 (3 a A+5 b B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{15 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.806, size = 371, normalized size = 2.5 \begin{align*} -{\frac{2}{15\,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 ( -24\,Aa\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+ \left ( 24\,Aa+20\,Ab+20\,Ba \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) + \left ( -6\,Aa-10\,Ab-10\,Ba \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +5\,A\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) b-9\,A\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a+5\,B\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) a-15\,B\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) b \right ){\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 (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}}{\sec \left (d x + c\right )^{\frac{5}{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{B b \sec \left (d x + c\right )^{2} + A a +{\left (B a + A b\right )} \sec \left (d x + c\right )}{\sec \left (d x + c\right )^{\frac{5}{2}}}, x\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 + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )}{\sec ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx \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 (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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