Optimal. Leaf size=296 \[ \frac{2 (b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 b^2 d}+\frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 b^2 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 b d} \]
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Rubi [A] time = 1.11021, antiderivative size = 296, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.186, Rules used = {4102, 4106, 3849, 2805, 3787, 3771, 2639, 2641} \[ \frac{2 \sin (c+d x) \sqrt{\sec (c+d x)} \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{5 b^3 d}-\frac{2 a \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a+b)}+\frac{2 (b B-a C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{3 b^2 d}+\frac{2 (b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 b^2 d}+\frac{2 C \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 b d} \]
Antiderivative was successfully verified.
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Rule 4102
Rule 4106
Rule 3849
Rule 2805
Rule 3787
Rule 3771
Rule 2639
Rule 2641
Rubi steps
\begin{align*} \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{2 \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3 a C}{2}+\frac{1}{2} b (5 A+3 C) \sec (c+d x)+\frac{5}{2} (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{5 b}\\ &=\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\sqrt{\sec (c+d x)} \left (\frac{5}{4} a (b B-a C)+\frac{1}{4} b (5 b B+4 a C) \sec (c+d x)+\frac{3}{4} \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{15 b^2}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{8 \int \frac{-\frac{3}{8} a \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\frac{1}{8} b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \sec (c+d x)+\frac{5}{8} \left (3 a^2 b B+b^3 B-3 a^3 C-a b^2 (3 A+C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))} \, dx}{15 b^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{8 \int \frac{-\frac{3}{8} a^2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )-\left (-\frac{3}{8} a b \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right )+\frac{1}{8} a b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right )\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)}} \, dx}{15 a^2 b^3}-\frac{\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{a+b \sec (c+d x)} \, dx}{b^3}\\ &=\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{(b B-a C) \int \sqrt{\sec (c+d x)} \, dx}{3 b^2}-\frac{\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx}{5 b^3}-\frac{\left (a \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{b^3}\\ &=-\frac{2 a \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 (a+b) d}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}+\frac{\left ((b B-a C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 b^2}-\frac{\left (\left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 b^3}\\ &=-\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 b^3 d}+\frac{2 (b B-a C) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 b^2 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \sqrt{\cos (c+d x)} \Pi \left (\frac{2 a}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{b^3 (a+b) d}+\frac{2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sqrt{\sec (c+d x)} \sin (c+d x)}{5 b^3 d}+\frac{2 (b B-a C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 b^2 d}+\frac{2 C \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 b d}\\ \end{align*}
Mathematica [F] time = 80.3292, size = 0, normalized size = 0. \[ \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx \]
Verification is Not applicable to the result.
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Maple [B] time = 9.906, size = 800, normalized size = 2.7 \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(-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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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 )} \sec \left (d x + c\right )^{\frac{5}{2}}}{b \sec \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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