Optimal. Leaf size=378 \[ -\frac{2 \sqrt{\sec (c+d x)} \left (a^2 (-(3 A+C))+a b B+2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (-5 a^2 b^2 (A+C)+2 a^3 b B+a^4 C+2 a b^3 B+A b^4\right )}{3 a b d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]
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Rubi [A] time = 1.03745, antiderivative size = 378, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4098, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (-5 a^2 b^2 (A+C)+2 a^3 b B+a^4 C+2 a b^3 B+A b^4\right )}{3 a b d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}-\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac{2 \sqrt{\sec (c+d x)} \left (a^2 (-(3 A+C))+a b B+2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
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Rule 4098
Rule 4100
Rule 4035
Rule 3856
Rule 2655
Rule 2653
Rule 3858
Rule 2663
Rule 2661
Rubi steps
\begin{align*} \int \frac{\sqrt{\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{\frac{1}{2} \left (-A b^2+a (b B-a C)\right )+\frac{3}{2} b (b B-a (A+C)) \sec (c+d x)+\frac{1}{2} \left (2 A b^2-2 a b B-a^2 C+3 b^2 C\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{-\frac{1}{4} b \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right )-\frac{1}{4} a b \left (4 a b B-a^2 (3 A+C)-b^2 (A+3 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a b \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )}-\frac{\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 a^2 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 \left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{3 a^2 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}-\frac{2 \left (A b^2-a (b B-a C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 7.61037, size = 5040, normalized size = 13.33 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.463, size = 5169, normalized size = 13.7 \begin{align*} \text{output too large to display} \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 \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\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{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, 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 \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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