Optimal. Leaf size=152 \[ \frac{12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )}+\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac{10 a^4 \sin (c+d x) \sqrt{e \cos (c+d x)}}{d e^3}-\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt{e \cos (c+d x)}} \]
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Rubi [A] time = 0.223586, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2670, 2680, 2635, 2642, 2641} \[ \frac{12 a^8 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^4-a^4 \sin (c+d x)\right )}+\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac{10 a^4 \sin (c+d x) \sqrt{e \cos (c+d x)}}{d e^3}-\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt{e \cos (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2670
Rule 2680
Rule 2635
Rule 2642
Rule 2641
Rubi steps
\begin{align*} \int \frac{(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx &=\frac{a^8 \int \frac{(e \cos (c+d x))^{11/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8}\\ &=\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}-\frac{\left (3 a^6\right ) \int \frac{(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^2} \, dx}{e^6}\\ &=\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac{12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac{\left (15 a^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{e^4}\\ &=-\frac{10 a^4 \sqrt{e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac{12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac{\left (5 a^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{e^2}\\ &=-\frac{10 a^4 \sqrt{e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac{12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}-\frac{\left (5 a^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{e^2 \sqrt{e \cos (c+d x)}}\\ &=-\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d e^2 \sqrt{e \cos (c+d x)}}-\frac{10 a^4 \sqrt{e \cos (c+d x)} \sin (c+d x)}{d e^3}+\frac{4 a^7 (e \cos (c+d x))^{9/2}}{3 d e^7 (a-a \sin (c+d x))^3}+\frac{12 a^6 (e \cos (c+d x))^{5/2}}{d e^5 \left (a^2-a^2 \sin (c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 0.0749828, size = 66, normalized size = 0.43 \[ \frac{16 \sqrt [4]{2} a^4 (\sin (c+d x)+1)^{3/4} \, _2F_1\left (-\frac{9}{4},-\frac{3}{4};\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{3 d e (e \cos (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.776, size = 263, normalized size = 1.7 \begin{align*}{\frac{2\,{a}^{4}}{3\,d{e}^{2}} \left ( -8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +30\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -48\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-15\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -18\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +48\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-20\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-1} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \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 (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\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 (a^{4} \cos \left (d x + c\right )^{4} - 8 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} - 4 \,{\left (a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}}{e^{3} \cos \left (d x + c\right )^{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 (a \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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