Optimal. Leaf size=120 \[ -\frac{2 e \cos ^2(c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{1-\frac{m}{2}} F_1\left (\frac{5}{2};\frac{1-m}{2},\frac{4-m}{2};\frac{7}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{5 a d \sqrt{a \sec (c+d x)+a}} \]
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Rubi [A] time = 0.373547, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3876, 2886, 135, 133} \[ -\frac{2 e \cos ^2(c+d x) (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{1-\frac{m}{2}} F_1\left (\frac{5}{2};\frac{1-m}{2},\frac{4-m}{2};\frac{7}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{5 a d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
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Rule 3876
Rule 2886
Rule 135
Rule 133
Rubi steps
\begin{align*} \int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx &=\frac{\sqrt{-a-a \cos (c+d x)} \int \frac{(-\cos (c+d x))^{3/2} (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^{3/2}} \, dx}{\sqrt{-\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=-\frac{\left (e (-a-a \cos (c+d x))^{\frac{1}{2}+\frac{1-m}{2}} (-a+a \cos (c+d x))^{\frac{1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int (-x)^{3/2} (-a-a x)^{-\frac{3}{2}+\frac{1}{2} (-1+m)} (-a+a x)^{\frac{1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{d \sqrt{-\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{\left (e (1+\cos (c+d x))^{1-\frac{m}{2}} (-a-a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} (-a+a \cos (c+d x))^{\frac{1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int (-x)^{3/2} (1+x)^{-\frac{3}{2}+\frac{1}{2} (-1+m)} (-a+a x)^{\frac{1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a d \sqrt{-\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{\left (e (1-\cos (c+d x))^{\frac{1}{2}-\frac{m}{2}} (1+\cos (c+d x))^{1-\frac{m}{2}} (-a-a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} (-a+a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-1+m)} (-x)^{3/2} (1+x)^{-\frac{3}{2}+\frac{1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a d \sqrt{-\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=-\frac{2 e F_1\left (\frac{5}{2};\frac{1-m}{2},\frac{4-m}{2};\frac{7}{2};\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac{1-m}{2}} \cos ^2(c+d x) (1+\cos (c+d x))^{1-\frac{m}{2}} (e \sin (c+d x))^{-1+m}}{5 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [B] time = 2.86349, size = 484, normalized size = 4.03 \[ \frac{4 (m+3) \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (F_1\left (\frac{m+1}{2};-\frac{1}{2},m;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 F_1\left (\frac{m+1}{2};-\frac{1}{2},m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right ) (e \sin (c+d x))^m}{d (m+1) (a (\sec (c+d x)+1))^{3/2} \left (-4 (m+3) \cos ^2\left (\frac{1}{2} (c+d x)\right ) F_1\left (\frac{m+1}{2};-\frac{1}{2},m+1;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+(\cos (c+d x)-1) \left (2 m F_1\left (\frac{m+3}{2};-\frac{1}{2},m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-4 (m+1) F_1\left (\frac{m+3}{2};-\frac{1}{2},m+2;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{m+3}{2};\frac{1}{2},m;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 F_1\left (\frac{m+3}{2};\frac{1}{2},m+1;\frac{m+5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )+(m+3) (\cos (c+d x)+1) F_1\left (\frac{m+1}{2};-\frac{1}{2},m;\frac{m+3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.175, size = 0, normalized size = 0. \begin{align*} \int{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \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 (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{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{a \sec \left (d x + c\right ) + a} \left (e \sin \left (d x + c\right )\right )^{m}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{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{\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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