3.264 \(\int \frac{(e \cos (c+d x))^{13/2}}{(a+a \sin (c+d x))^4} \, dx\)

Optimal. Leaf size=149 \[ -\frac{154 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^4 d}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{154 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a \sin (c+d x)+a)^3} \]

[Out]

(-154*e^6*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]) - (154*e^5*(e*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(15*a^4*d) - (4*e*(e*Cos[c + d*x])^(11/2))/(a*d*(a + a*Sin[c + d*x])^3) - (44*e^3*(e*
Cos[c + d*x])^(7/2))/(3*d*(a^4 + a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.153125, antiderivative size = 149, 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, 2640, 2639} \[ -\frac{154 e^5 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^4 d}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4 \sin (c+d x)+a^4\right )}-\frac{154 e^6 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a \sin (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

(-154*e^6*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^4*d*Sqrt[Cos[c + d*x]]) - (154*e^5*(e*Cos[c + d
*x])^(3/2)*Sin[c + d*x])/(15*a^4*d) - (4*e*(e*Cos[c + d*x])^(11/2))/(a*d*(a + a*Sin[c + d*x])^3) - (44*e^3*(e*
Cos[c + d*x])^(7/2))/(3*d*(a^4 + a^4*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 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))^{13/2}}{(a+a \sin (c+d x))^4} \, dx &=-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac{\left (11 e^2\right ) \int \frac{(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx}{a^2}\\ &=-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac{\left (77 e^4\right ) \int (e \cos (c+d x))^{5/2} \, dx}{3 a^4}\\ &=-\frac{154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac{\left (77 e^6\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 a^4}\\ &=-\frac{154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}-\frac{\left (77 e^6 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^4 \sqrt{\cos (c+d x)}}\\ &=-\frac{154 e^6 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{154 e^5 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^4 d}-\frac{4 e (e \cos (c+d x))^{11/2}}{a d (a+a \sin (c+d x))^3}-\frac{44 e^3 (e \cos (c+d x))^{7/2}}{3 d \left (a^4+a^4 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.219982, size = 66, normalized size = 0.44 \[ -\frac{2^{3/4} (e \cos (c+d x))^{15/2} \, _2F_1\left (\frac{5}{4},\frac{15}{4};\frac{19}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{15 a^4 d e (\sin (c+d x)+1)^{15/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(13/2)/(a + a*Sin[c + d*x])^4,x]

[Out]

-(2^(3/4)*(e*Cos[c + d*x])^(15/2)*Hypergeometric2F1[5/4, 15/4, 19/4, (1 - Sin[c + d*x])/2])/(15*a^4*d*e*(1 + S
in[c + d*x])^(15/4))

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Maple [A]  time = 1.056, size = 190, normalized size = 1.3 \begin{align*} -{\frac{2\,{e}^{7}}{15\,{a}^{4}d} \left ( -24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +80\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+231\,\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 ) -246\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -80\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+140\,\sin \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x)

[Out]

-2/15/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)/a^4*(-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+
24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+80*sin(1/2*d*x+1/2*c)^5+231*(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))-246*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-80*si
n(1/2*d*x+1/2*c)^3+140*sin(1/2*d*x+1/2*c))*e^7/d

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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.

[In]

integrate((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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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^{6} \cos \left (d x + c\right )^{6}}{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 )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*e^6*cos(d*x + c)^6/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*c
os(d*x + c)^2 - 2*a^4)*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))**(13/2)/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [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))^(13/2)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

Timed out