Optimal. Leaf size=203 \[ \frac{a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}-\frac{\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d} \]
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Rubi [A] time = 0.280267, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2691, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}-\frac{\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{d} \]
Antiderivative was successfully verified.
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Rule 2691
Rule 2753
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \sec ^2(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac{\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\int \sqrt{a+b \sin (c+d x)} \left (\frac{3 b^2}{2}+\frac{3}{2} a b \sin (c+d x)\right ) \, dx\\ &=\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac{2}{3} \int \frac{3 a b^2+\frac{3}{4} b \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}+\frac{1}{2} \left (a \left (a^2-b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{1}{2} \left (a^2+3 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx\\ &=\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac{\left (\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}}} \, dx}{2 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{a b \cos (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{\sec (c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{d}-\frac{\left (a^2+3 b^2\right ) E\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (c+d x)}}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{a \left (a^2-b^2\right ) F\left (\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}{d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.867851, size = 203, normalized size = 1. \[ \frac{-a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (a^2 b+a^3+3 a b^2+3 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+2 a^2 b \sec (c+d x)+a^2 b \sin (c+d x) \tan (c+d x)+a^3 \tan (c+d x)+3 a b^2 \tan (c+d x)+b^3 \sin (c+d x) \tan (c+d x)}{d \sqrt{a+b \sin (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.685, size = 1039, normalized size = 5.1 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{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 ({\left (2 \, a b \sec \left (d x + c\right )^{2} \sin \left (d x + c\right ) -{\left (b^{2} \cos \left (d x + c\right )^{2} - a^{2} - b^{2}\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, 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(-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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