3.258 \(\int \frac{(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=118 \[ \frac{6 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^3 d \sqrt{\cos (c+d x)}}+\frac{6 e (e \cos (c+d x))^{3/2}}{5 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a \sin (c+d x)+a)^2} \]

[Out]

(6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]) - (4*e*(e*Cos[c + d*x])^(3
/2))/(5*a*d*(a + a*Sin[c + d*x])^2) + (6*e*(e*Cos[c + d*x])^(3/2))/(5*d*(a^3 + a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.130675, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2680, 2683, 2640, 2639} \[ \frac{6 e^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^3 d \sqrt{\cos (c+d x)}}+\frac{6 e (e \cos (c+d x))^{3/2}}{5 d \left (a^3 \sin (c+d x)+a^3\right )}-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(6*e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^3*d*Sqrt[Cos[c + d*x]]) - (4*e*(e*Cos[c + d*x])^(3
/2))/(5*a*d*(a + a*Sin[c + d*x])^2) + (6*e*(e*Cos[c + d*x])^(3/2))/(5*d*(a^3 + a^3*Sin[c + d*x]))

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2683

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(g*Cos[e
+ f*x])^(p + 1))/(a*f*g*(p - 1)*(a + b*Sin[e + f*x])), x] + Dist[p/(a*(p - 1)), Int[(g*Cos[e + f*x])^p, x], x
] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] &&  !GeQ[p, 1] && IntegerQ[2*p]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(e \cos (c+d x))^{5/2}}{(a+a \sin (c+d x))^3} \, dx &=-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a+a \sin (c+d x))^2}-\frac{\left (3 e^2\right ) \int \frac{\sqrt{e \cos (c+d x)}}{a+a \sin (c+d x)} \, dx}{5 a^2}\\ &=-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a+a \sin (c+d x))^2}+\frac{6 e (e \cos (c+d x))^{3/2}}{5 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{\left (3 e^2\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 a^3}\\ &=-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a+a \sin (c+d x))^2}+\frac{6 e (e \cos (c+d x))^{3/2}}{5 d \left (a^3+a^3 \sin (c+d x)\right )}+\frac{\left (3 e^2 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^3 \sqrt{\cos (c+d x)}}\\ &=\frac{6 e^2 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{3/2}}{5 a d (a+a \sin (c+d x))^2}+\frac{6 e (e \cos (c+d x))^{3/2}}{5 d \left (a^3+a^3 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.0840857, size = 66, normalized size = 0.56 \[ -\frac{(e \cos (c+d x))^{7/2} \, _2F_1\left (\frac{7}{4},\frac{9}{4};\frac{11}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{7 \sqrt [4]{2} a^3 d e (\sin (c+d x)+1)^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

-((e*Cos[c + d*x])^(7/2)*Hypergeometric2F1[7/4, 9/4, 11/4, (1 - Sin[c + d*x])/2])/(7*2^(1/4)*a^3*d*e*(1 + Sin[
c + d*x])^(7/4))

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Maple [B]  time = 1.941, size = 330, normalized size = 2.8 \begin{align*}{\frac{2\,{e}^{3}}{5\,{a}^{3}d} \left ( 12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+3\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \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.

[In]

int((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^3,x)

[Out]

2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/
2)*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin
(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1
/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)+20*sin(1/2*d*x+1/2*c)^5+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),2^(1/2))+2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-20*sin(1/2*d*x+1/2*c)^3+sin(1/2*d*x+1/2*c))*
e^3/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(5/2)/(a*sin(d*x + c) + a)^3, x)

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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 )} e^{2} \cos \left (d x + c\right )^{2}}{3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e^2*cos(d*x + c)^2/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)
*sin(d*x + c)), x)

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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.

[In]

integrate((e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(5/2)/(a*sin(d*x + c) + a)^3, x)