Optimal. Leaf size=234 \[ \frac{7 a^2 \sqrt{\sin (c+d x)} \text{EllipticF}\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right ),2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}+\frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \]
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Rubi [A] time = 0.419121, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {3872, 2873, 2636, 2642, 2641, 2564, 325, 329, 212, 206, 203, 2570, 2571} \[ \frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{7 a^2 \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{3 d e^2 \sqrt{e \sin (c+d x)}}-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2873
Rule 2636
Rule 2642
Rule 2641
Rule 2564
Rule 325
Rule 329
Rule 212
Rule 206
Rule 203
Rule 2570
Rule 2571
Rubi steps
\begin{align*} \int \frac{(a+a \sec (c+d x))^2}{(e \sin (c+d x))^{5/2}} \, dx &=\int \frac{(-a-a \cos (c+d x))^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\\ &=\int \left (\frac{a^2}{(e \sin (c+d x))^{5/2}}+\frac{2 a^2 \sec (c+d x)}{(e \sin (c+d x))^{5/2}}+\frac{a^2 \sec ^2(c+d x)}{(e \sin (c+d x))^{5/2}}\right ) \, dx\\ &=a^2 \int \frac{1}{(e \sin (c+d x))^{5/2}} \, dx+a^2 \int \frac{\sec ^2(c+d x)}{(e \sin (c+d x))^{5/2}} \, dx+\left (2 a^2\right ) \int \frac{\sec (c+d x)}{(e \sin (c+d x))^{5/2}} \, dx\\ &=-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{a^2 \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{3 e^2}+\frac{\left (5 a^2\right ) \int \frac{\sec ^2(c+d x)}{\sqrt{e \sin (c+d x)}} \, dx}{3 e^2}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^{5/2} \left (1-\frac{x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e}\\ &=-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{e^2}\right )} \, dx,x,e \sin (c+d x)\right )}{d e^3}+\frac{\left (5 a^2\right ) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{6 e^2}+\frac{\left (a^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{3 e^2 \sqrt{e \sin (c+d x)}}\\ &=-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{2 a^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d e^2 \sqrt{e \sin (c+d x)}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{e^2}} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d e^3}+\frac{\left (5 a^2 \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{6 e^2 \sqrt{e \sin (c+d x)}}\\ &=-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{7 a^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d e^2 \sqrt{e \sin (c+d x)}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{e-x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d e^2}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{e+x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{d e^2}\\ &=\frac{2 a^2 \tan ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}+\frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{e \sin (c+d x)}}{\sqrt{e}}\right )}{d e^{5/2}}-\frac{4 a^2}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \cos (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}-\frac{2 a^2 \sec (c+d x)}{3 d e (e \sin (c+d x))^{3/2}}+\frac{7 a^2 F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{3 d e^2 \sqrt{e \sin (c+d x)}}+\frac{5 a^2 \sec (c+d x) \sqrt{e \sin (c+d x)}}{3 d e^3}\\ \end{align*}
Mathematica [C] time = 46.6692, size = 169, normalized size = 0.72 \[ -\frac{a^2 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \sin (c+d x)} \sec ^4\left (\frac{1}{2} \sin ^{-1}(\sin (c+d x))\right ) \left (3 \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},\sin ^2(c+d x)\right )+4 \sqrt{\cos ^2(c+d x)} \csc ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},\sin ^2(c+d x)\right )+4 \sqrt{\cos ^2(c+d x)} \csc ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{3}{2},\frac{1}{4},\sin ^2(c+d x)\right )+3\right )}{3 d e^3} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 2.245, size = 301, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}}{6\,\cos \left ( dx+c \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) d} \left ( 7\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ){e}^{7/2}-14\,{e}^{7/2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-8\,{e}^{7/2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+12\,{\it Artanh} \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) \left ( e\sin \left ( dx+c \right ) \right ) ^{3/2}{e}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+12\, \left ( e\sin \left ( dx+c \right ) \right ) ^{3/2}\arctan \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ){e}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}+20\,{e}^{7/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+8\,{e}^{7/2}\cos \left ( dx+c \right ) -12\,{\it Artanh} \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ) \left ( e\sin \left ( dx+c \right ) \right ) ^{3/2}{e}^{2}\cos \left ( dx+c \right ) -12\, \left ( e\sin \left ( dx+c \right ) \right ) ^{3/2}\arctan \left ({\frac{\sqrt{e\sin \left ( dx+c \right ) }}{\sqrt{e}}} \right ){e}^{2}\cos \left ( dx+c \right ) -6\,{e}^{7/2} \right ){e}^{-{\frac{9}{2}}} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \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] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}\right )} \sqrt{e \sin \left (d x + c\right )}}{{\left (e^{3} \cos \left (d x + c\right )^{2} - e^{3}\right )} \sin \left (d x + c\right )}, 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 \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \sin \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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