3.243 \(\int \frac{(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=145 \[ \frac{6 e^5 \sin (c+d x) \sqrt{e \cos (c+d x)}}{7 a^2 d}+\frac{18 e^3 \sin (c+d x) (e \cos (c+d x))^{5/2}}{35 a^2 d}+\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )} \]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (6*e^5*Sqrt[e*Cos[c + d*
x]]*Sin[c + d*x])/(7*a^2*d) + (18*e^3*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(35*a^2*d) + (4*e*(e*Cos[c + d*x])^
(9/2))/(5*d*(a^2 + a^2*Sin[c + d*x]))

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Rubi [A]  time = 0.109562, antiderivative size = 145, 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 = {2680, 2635, 2642, 2641} \[ \frac{6 e^5 \sin (c+d x) \sqrt{e \cos (c+d x)}}{7 a^2 d}+\frac{18 e^3 \sin (c+d x) (e \cos (c+d x))^{5/2}}{35 a^2 d}+\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (6*e^5*Sqrt[e*Cos[c + d*
x]]*Sin[c + d*x])/(7*a^2*d) + (18*e^3*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(35*a^2*d) + (4*e*(e*Cos[c + d*x])^
(9/2))/(5*d*(a^2 + a^2*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{11/2}}{(a+a \sin (c+d x))^2} \, dx &=\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (9 e^2\right ) \int (e \cos (c+d x))^{7/2} \, dx}{5 a^2}\\ &=\frac{18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (9 e^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{7 a^2}\\ &=\frac{6 e^5 \sqrt{e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (3 e^6\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{7 a^2}\\ &=\frac{6 e^5 \sqrt{e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{\left (3 e^6 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{7 a^2 \sqrt{e \cos (c+d x)}}\\ &=\frac{6 e^6 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{7 a^2 d \sqrt{e \cos (c+d x)}}+\frac{6 e^5 \sqrt{e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac{18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac{4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.188916, size = 66, normalized size = 0.46 \[ -\frac{4 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \, _2F_1\left (-\frac{1}{4},\frac{13}{4};\frac{17}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{13 a^2 d e (\sin (c+d x)+1)^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^2,x]

[Out]

(-4*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[-1/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(13*a^2*d*e*(1
+ Sin[c + d*x])^(13/4))

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Maple [A]  time = 0.56, size = 203, normalized size = 1.4 \begin{align*} -{\frac{2\,{e}^{6}}{35\,{a}^{2}d} \left ( -80\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}+120\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +112\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-168\, \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 ) -20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +84\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-14\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \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))^(11/2)/(a+a*sin(d*x+c))^2,x)

[Out]

-2/35/a^2/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(-80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8+120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+112*sin(1/2*d*x+1/2*c)^7-168*sin(1/2*d*x+1/2*c)^5+15*(2*sin(1/
2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-20*sin(1/2*d*x+1/2*
c)^2*cos(1/2*d*x+1/2*c)+84*sin(1/2*d*x+1/2*c)^3-14*sin(1/2*d*x+1/2*c))/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{11}{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.

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(11/2)/(a*sin(d*x + c) + a)^2, 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^{5} \cos \left (d x + c\right )^{5}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e^5*cos(d*x + c)^5/(a^2*cos(d*x + c)^2 - 2*a^2*sin(d*x + c) - 2*a^2), 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))**(11/2)/(a+a*sin(d*x+c))**2,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{11}{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.

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(11/2)/(a*sin(d*x + c) + a)^2, x)