3.256 \(\int \frac{(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=103 \[ -\frac{14 e^3 (e \cos (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{a^3 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a \sin (c+d x)+a)^2} \]

[Out]

(-14*e^3*(e*Cos[c + d*x])^(3/2))/(3*a^3*d) - (14*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^3*d*Sq
rt[Cos[c + d*x]]) - (4*e*(e*Cos[c + d*x])^(7/2))/(a*d*(a + a*Sin[c + d*x])^2)

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Rubi [A]  time = 0.154847, antiderivative size = 103, 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, 2682, 2640, 2639} \[ -\frac{14 e^3 (e \cos (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{a^3 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(9/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-14*e^3*(e*Cos[c + d*x])^(3/2))/(3*a^3*d) - (14*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^3*d*Sq
rt[Cos[c + d*x]]) - (4*e*(e*Cos[c + d*x])^(7/2))/(a*d*(a + a*Sin[c + d*x])^2)

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 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[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))^{9/2}}{(a+a \sin (c+d x))^3} \, dx &=-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a+a \sin (c+d x))^2}-\frac{\left (7 e^2\right ) \int \frac{(e \cos (c+d x))^{5/2}}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=-\frac{14 e^3 (e \cos (c+d x))^{3/2}}{3 a^3 d}-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a+a \sin (c+d x))^2}-\frac{\left (7 e^4\right ) \int \sqrt{e \cos (c+d x)} \, dx}{a^3}\\ &=-\frac{14 e^3 (e \cos (c+d x))^{3/2}}{3 a^3 d}-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a+a \sin (c+d x))^2}-\frac{\left (7 e^4 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{a^3 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 e^3 (e \cos (c+d x))^{3/2}}{3 a^3 d}-\frac{14 e^4 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt{\cos (c+d x)}}-\frac{4 e (e \cos (c+d x))^{7/2}}{a d (a+a \sin (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 0.0975292, size = 66, normalized size = 0.64 \[ -\frac{2^{3/4} (e \cos (c+d x))^{11/2} \, _2F_1\left (\frac{5}{4},\frac{11}{4};\frac{15}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{11 a^3 d e (\sin (c+d x)+1)^{11/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(9/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

-(2^(3/4)*(e*Cos[c + d*x])^(11/2)*Hypergeometric2F1[5/4, 11/4, 15/4, (1 - Sin[c + d*x])/2])/(11*a^3*d*e*(1 + S
in[c + d*x])^(11/4))

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Maple [A]  time = 0.987, size = 146, normalized size = 1.4 \begin{align*} -{\frac{2\,{e}^{5}}{3\,{a}^{3}d} \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+21\,\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 ) -24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+13\,\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))^(9/2)/(a+a*sin(d*x+c))^3,x)

[Out]

-2/3/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)/a^3*(4*sin(1/2*d*x+1/2*c)^5+21*(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))-24*sin(1/2*d*x+1/2*c)^2*cos(1
/2*d*x+1/2*c)-4*sin(1/2*d*x+1/2*c)^3+13*sin(1/2*d*x+1/2*c))*e^5/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{9}{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))^(9/2)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(9/2)/(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^{4} \cos \left (d x + c\right )^{4}}{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))^(9/2)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e^4*cos(d*x + c)^4/(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))**(9/2)/(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{9}{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))^(9/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(9/2)/(a*sin(d*x + c) + a)^3, x)