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

Optimal. Leaf size=114 \[ \frac {14 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^2 d \sqrt {\cos (c+d x)}}+\frac {14 e^3 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^2 d}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )} \]

[Out]

14/15*e^3*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a^2/d+4/3*e*(e*cos(d*x+c))^(7/2)/d/(a^2+a^2*sin(d*x+c))+14/5*e^4*(co
s(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/a^2/d/
cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2680, 2635, 2640, 2639} \[ \frac {14 e^3 \sin (c+d x) (e \cos (c+d x))^{3/2}}{15 a^2 d}+\frac {14 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a^2 d \sqrt {\cos (c+d x)}}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(14*e^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*a^2*d*Sqrt[Cos[c + d*x]]) + (14*e^3*(e*Cos[c + d*x]
)^(3/2)*Sin[c + d*x])/(15*a^2*d) + (4*e*(e*Cos[c + d*x])^(7/2))/(3*d*(a^2 + a^2*Sin[c + d*x]))

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 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]

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 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]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx &=\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (7 e^2\right ) \int (e \cos (c+d x))^{5/2} \, dx}{3 a^2}\\ &=\frac {14 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^2 d}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (7 e^4\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 a^2}\\ &=\frac {14 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^2 d}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (7 e^4 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^2 \sqrt {\cos (c+d x)}}\\ &=\frac {14 e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)}}+\frac {14 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^2 d}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2+a^2 \sin (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 66, normalized size = 0.58 \[ -\frac {2\ 2^{3/4} (e \cos (c+d x))^{11/2} \, _2F_1\left (\frac {1}{4},\frac {11}{4};\frac {15}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{11 a^2 d e (\sin (c+d x)+1)^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )} e^{4} \cos \left (d x + c\right )^{4}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.00, size = 190, normalized size = 1.67 \[ \frac {2 e^{5} \left (-24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+40 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}-6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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))^2,x)

[Out]

2/15/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^5*(-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)+40*sin(1/2*d*x+1/2*c)^5+21*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-40*si
n(1/2*d*x+1/2*c)^3+10*sin(1/2*d*x+1/2*c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**2,x)

[Out]

Timed out

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