Optimal. Leaf size=150 \[ -\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{3 a^2 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 \sin (c+d x)}{3 a^2 d e \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e \left (a^2 \sin (c+d x)+a^2\right ) \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e (a \sin (c+d x)+a)^2 \sqrt{e \cos (c+d x)}} \]
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Rubi [A] time = 0.1574, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2681, 2683, 2636, 2640, 2639} \[ -\frac{2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{3 a^2 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 \sin (c+d x)}{3 a^2 d e \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e \left (a^2 \sin (c+d x)+a^2\right ) \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e (a \sin (c+d x)+a)^2 \sqrt{e \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2681
Rule 2683
Rule 2636
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int \frac{1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2} \, dx &=-\frac{2}{9 d e \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}+\frac{5 \int \frac{1}{(e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \, dx}{9 a}\\ &=-\frac{2}{9 d e \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\int \frac{1}{(e \cos (c+d x))^{3/2}} \, dx}{3 a^2}\\ &=\frac{2 \sin (c+d x)}{3 a^2 d e \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\int \sqrt{e \cos (c+d x)} \, dx}{3 a^2 e^2}\\ &=\frac{2 \sin (c+d x)}{3 a^2 d e \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}-\frac{\sqrt{e \cos (c+d x)} \int \sqrt{\cos (c+d x)} \, dx}{3 a^2 e^2 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 \sin (c+d x)}{3 a^2 d e \sqrt{e \cos (c+d x)}}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}-\frac{2}{9 d e \sqrt{e \cos (c+d x)} \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 0.0682538, size = 66, normalized size = 0.44 \[ \frac{\sqrt [4]{\sin (c+d x)+1} \, _2F_1\left (-\frac{1}{4},\frac{13}{4};\frac{3}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{2 \sqrt [4]{2} a^2 d e \sqrt{e \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 2.269, size = 488, normalized size = 3.3 \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 \frac{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (a \sin \left (d x + c\right ) + a\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 (-\frac{\sqrt{e \cos \left (d x + c\right )}}{a^{2} e^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} e^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} e^{2} \cos \left (d x + c\right )^{2}}, 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{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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