Optimal. Leaf size=228 \[ -\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{6 a^3 d}-\frac{(13 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(8 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]
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Rubi [A] time = 0.498514, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4020, 3787, 3771, 2639, 2641} \[ -\frac{(13 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{6 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{(49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{(8 A-3 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{15 a d (a \sec (c+d x)+a)^2}-\frac{(A-B) \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 4020
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\frac{1}{2} a (11 A-B)-\frac{5}{2} a (A-B) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{1}{2} a^2 (41 A-6 B)-\frac{3}{2} a^2 (8 A-3 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \frac{\frac{3}{4} a^3 (49 A-9 B)-\frac{5}{4} a^3 (13 A-3 B) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^6}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(49 A-9 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{20 a^3}-\frac{(13 A-3 B) \int \sqrt{\sec (c+d x)} \, dx}{12 a^3}\\ &=-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\left ((49 A-9 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{20 a^3}-\frac{\left ((13 A-3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac{(49 A-9 B) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{10 a^3 d}-\frac{(13 A-3 B) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{6 a^3 d}-\frac{(A-B) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(8 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{(13 A-3 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 6.5179, size = 364, normalized size = 1.6 \[ -\frac{e^{-i d x} \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) (A+B \sec (c+d x)) \left (i (49 A-9 B) e^{-\frac{3}{2} i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^5 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )+160 (13 A-3 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )+2 \cos (c+d x) \left (-30 i (49 A-9 B) \cos \left (\frac{1}{2} (c+d x)\right )-15 i (49 A-9 B) \cos \left (\frac{3}{2} (c+d x)\right )+142 A \sin \left (\frac{1}{2} (c+d x)\right )+205 A \sin \left (\frac{3}{2} (c+d x)\right )+87 A \sin \left (\frac{5}{2} (c+d x)\right )-147 i A \cos \left (\frac{5}{2} (c+d x)\right )-42 B \sin \left (\frac{1}{2} (c+d x)\right )-45 B \sin \left (\frac{3}{2} (c+d x)\right )-27 B \sin \left (\frac{5}{2} (c+d x)\right )+27 i B \cos \left (\frac{5}{2} (c+d x)\right )\right )\right )}{120 a^3 d (\sec (c+d x)+1)^3 (A \cos (c+d x)+B)} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.895, size = 451, normalized size = 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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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{\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{4} + 3 \, a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + a^{3} \sec \left (d x + c\right )}, 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 )}^{3} \sqrt{\sec \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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