Optimal. Leaf size=121 \[ \frac{(A+2 B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{B \sin (c+d x) \sqrt{\cos (c+d x)}}{a^2 d (\cos (c+d x)+1)}+\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]
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Rubi [A] time = 0.355074, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2954, 2978, 2748, 2641, 2639} \[ \frac{(A+2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{B \sin (c+d x) \sqrt{\cos (c+d x)}}{a^2 d (\cos (c+d x)+1)}+\frac{(A-B) \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
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Rule 2954
Rule 2978
Rule 2748
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=\int \frac{B+A \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx\\ &=\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a (A+5 B)+\frac{1}{2} a (A-B) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=-\frac{B \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}+\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (A+2 B)+\frac{3}{2} a^2 B \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{B \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}+\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{B \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}+\frac{(A+2 B) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{(A+2 B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{B \sqrt{\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}+\frac{(A-B) \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 6.51282, size = 921, normalized size = 7.61 \[ \frac{i B \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \left (\frac{2 e^{2 i d x} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt{e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt{e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)+1}}{3 i d \left (1+e^{2 i d x}\right ) \cos (c)-3 d \left (-1+e^{2 i d x}\right ) \sin (c)}-\frac{2 \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i d x} (\cos (c)+i \sin (c))^2\right ) \sqrt{e^{-i d x} \left (2 \left (1+e^{2 i d x}\right ) \cos (c)+2 i \left (-1+e^{2 i d x}\right ) \sin (c)\right )} \sqrt{e^{2 i d x} \cos (2 c)+i e^{2 i d x} \sin (2 c)+1}}{d \left (-1+e^{2 i d x}\right ) \sin (c)-i d \left (1+e^{2 i d x}\right ) \cos (c)}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{(A+B \sec (c+d x)) \left (-\frac{2 \sec \left (\frac{c}{2}\right ) \left (B \sin \left (\frac{d x}{2}\right )-A \sin \left (\frac{d x}{2}\right )\right ) \sec ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{2 (B-A) \tan \left (\frac{c}{2}\right ) \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{4 B \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{4 B \csc (c)}{d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{\sqrt{\cos (c+d x)} (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac{2 A \csc \left (\frac{c}{2}\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) \sqrt{\cot ^2(c)+1} (\sec (c+d x) a+a)^2}-\frac{4 B \csc \left (\frac{c}{2}\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \sqrt{1-\sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{-\sqrt{\cot ^2(c)+1} \sin (c) \sin \left (d x-\tan ^{-1}(\cot (c))\right )} \sqrt{\sin \left (d x-\tan ^{-1}(\cot (c))\right )+1} \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) \sqrt{\cot ^2(c)+1} (\sec (c+d x) a+a)^2} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 1.891, size = 350, normalized size = 2.9 \begin{align*} -{\frac{1}{6\,d{a}^{2}}\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 ( 2\,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}-12\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+4\,B\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\,B \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 ) +2\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+16\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-3\,A \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-3\,B \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+A-B \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(-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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )^{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{B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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