3.255 \(\int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=132 \[ \frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sin (c+d x) \sqrt {e \cos (c+d x)}}{a^3 d}+\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2} \]

[Out]

18/5*e^3*(e*cos(d*x+c))^(5/2)/a^3/d+4*e*(e*cos(d*x+c))^(9/2)/a/d/(a+a*sin(d*x+c))^2+6*e^6*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a^3/d/(e*cos(d*x+c))^(1/2)+
6*e^5*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a^3/d

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Rubi [A]  time = 0.15, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2680, 2682, 2635, 2642, 2641} \[ \frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 \sin (c+d x) \sqrt {e \cos (c+d x)}}{a^3 d}+\frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt {e \cos (c+d x)}}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

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

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 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]

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 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]

Rubi steps

\begin {align*} \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx &=\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2}+\frac {\left (9 e^2\right ) \int \frac {(e \cos (c+d x))^{7/2}}{a+a \sin (c+d x)} \, dx}{a^2}\\ &=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2}+\frac {\left (9 e^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{a^3}\\ &=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2}+\frac {\left (3 e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{a^3}\\ &=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2}+\frac {\left (3 e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{a^3 \sqrt {e \cos (c+d x)}}\\ &=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^6 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 66, normalized size = 0.50 \[ -\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \, _2F_1\left (\frac {3}{4},\frac {13}{4};\frac {17}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{13 a^3 d e (\sin (c+d x)+1)^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(11/2)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-2*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[3/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(13*a^3*d*e*(1 +
Sin[c + d*x])^(13/4))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \cos \left (d x + c\right )} e^{5} \cos \left (d x + c\right )^{5}}{3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}, x\right ) \]

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))^3,x, algorithm="fricas")

[Out]

integral(-sqrt(e*cos(d*x + c))*e^5*cos(d*x + c)^5/(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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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))^(11/2)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.93, size = 181, normalized size = 1.37 \[ -\frac {2 e^{6} \left (-8 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}+10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+34 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-19 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^3,x)

[Out]

-2/5/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(-8*sin(1/2*d*x+1/2*c)^7-20*sin(1/2*d*x+1/
2*c)^4*cos(1/2*d*x+1/2*c)+12*sin(1/2*d*x+1/2*c)^5+15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)+10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+34*sin(1/2*d*x+1/2*c)^3-
19*sin(1/2*d*x+1/2*c))/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

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))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(11/2)/(a*sin(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(c + d*x))^(11/2)/(a + a*sin(c + d*x))^3,x)

[Out]

int((e*cos(c + d*x))^(11/2)/(a + a*sin(c + d*x))^3, 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))**(11/2)/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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