Optimal. Leaf size=211 \[ \frac{5 (2 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 a^2 d}+\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}-\frac{(7 A-4 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.356715, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4020, 3787, 3769, 3771, 2641, 2639} \[ \frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (\sec (c+d x)+1)}+\frac{5 (2 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{(7 A-4 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4020
Rule 3787
Rule 3769
Rule 3771
Rule 2641
Rule 2639
Rubi steps
\begin{align*} \int \frac{A+B \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{3}{2} a (3 A-B)-\frac{5}{2} a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{3 a^2}\\ &=-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\int \frac{\frac{15}{2} a^2 (2 A-B)-\frac{3}{2} a^2 (7 A-4 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac{(7 A-4 B) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{2 a^2}+\frac{(5 (2 A-B)) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx}{2 a^2}\\ &=\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{(5 (2 A-B)) \int \sqrt{\sec (c+d x)} \, dx}{6 a^2}-\frac{\left ((7 A-4 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 a^2}\\ &=-\frac{(7 A-4 B) \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{5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}+\frac{\left (5 (2 A-B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac{(7 A-4 B) \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{5 (2 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 a^2 d}+\frac{5 (2 A-B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)}}-\frac{(7 A-4 B) \sin (c+d x)}{3 a^2 d \sqrt{\sec (c+d x)} (1+\sec (c+d x))}-\frac{(A-B) \sin (c+d x)}{3 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^2}\\ \end{align*}
Mathematica [C] time = 6.80139, size = 899, normalized size = 4.26 \[ \frac{7 \sqrt{2} A 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)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\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 ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac{4 \sqrt{2} B 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)}} \csc \left (\frac{c}{2}\right ) \left (e^{2 i d x} \left (-1+e^{2 i c}\right ) \text{Hypergeometric2F1}\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 ) \sec \left (\frac{c}{2}\right ) \sec (c+d x) (A+B \sec (c+d x)) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{20 A \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}-\frac{10 B \sqrt{\cos (c+d x)} \csc \left (\frac{c}{2}\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right ) \sec \left (\frac{c}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)) \sin (c) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d (B+A \cos (c+d x)) (\sec (c+d x) a+a)^2}+\frac{\sec ^{\frac{3}{2}}(c+d x) (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 \sec \left (\frac{c}{2}\right ) \left (7 B \sin \left (\frac{d x}{2}\right )-10 A \sin \left (\frac{d x}{2}\right )\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}-\frac{2 (-2 \cos (2 c) A-5 A+3 B+B \cos (2 c)) \cos (d x) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right )}{d}+\frac{4 A \cos (2 d x) \sin (2 c)}{3 d}+\frac{8 (B-2 A) \cos (c) \sin (d x)}{d}+\frac{4 A \cos (2 c) \sin (2 d x)}{3 d}+\frac{4 (7 B-10 A) \tan \left (\frac{c}{2}\right )}{3 d}\right ) \cos ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{(B+A \cos (c+d x)) (\sec (c+d x) a+a)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A] time = 1.951, size = 435, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
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^{2} \sec \left (d x + c\right )^{4} + 2 \, a^{2} \sec \left (d x + c\right )^{3} + a^{2} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
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} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]