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

Optimal. Leaf size=101 \[ \frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a d \sqrt {\cos (c+d x)}}+\frac {2 e^3 \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 a d}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d} \]

[Out]

2/7*e*(e*cos(d*x+c))^(7/2)/a/d+2/5*e^3*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+6/5*e^4*(cos(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/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 101, 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 = {2682, 2635, 2640, 2639} \[ \frac {2 e^3 \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 a d}+\frac {6 e^4 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 a d \sqrt {\cos (c+d x)}}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}+\frac {e^2 \int (e \cos (c+d x))^{5/2} \, dx}{a}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}+\frac {2 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {\left (3 e^4\right ) \int \sqrt {e \cos (c+d x)} \, dx}{5 a}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}+\frac {2 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {\left (3 e^4 \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a \sqrt {\cos (c+d x)}}\\ &=\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}+\frac {6 e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)}}+\frac {2 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 66, normalized size = 0.65 \[ -\frac {4\ 2^{3/4} (e \cos (c+d x))^{11/2} \, _2F_1\left (-\frac {3}{4},\frac {11}{4};\frac {15}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{11 a d e (\sin (c+d x)+1)^{11/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \cos \left (d x + c\right )} e^{4} \cos \left (d x + c\right )^{4}}{a \sin \left (d x + c\right ) + a}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

integral(sqrt(e*cos(d*x + c))*e^4*cos(d*x + c)^4/(a*sin(d*x + c) + a), 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)),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.79, size = 216, normalized size = 2.14 \[ \frac {2 e^{5} \left (80 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+56 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-160 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-56 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+120 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

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)),x)

[Out]

2/35/a/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^5*(80*sin(1/2*d*x+1/2*c)^9+56*sin(1/2*d*x+1/2*
c)^6*cos(1/2*d*x+1/2*c)-160*sin(1/2*d*x+1/2*c)^7-56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+120*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)+
14*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-40*sin(1/2*d*x+1/2*c)^3+5*sin(1/2*d*x+1/2*c))/d

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

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)),x, algorithm="maxima")

[Out]

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

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

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)),x)

[Out]

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

[Out]

Timed out

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