\(\int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx\) [243]

Optimal result

Integrand size = 25, antiderivative size = 145 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac {18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )} \]

[Out]

18/35*e^3*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^2/d+4/5*e*(e*cos(d*x+c))^(9/2)/d/(a^2+a^2*sin(d*x+c))+6/7*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^2/d/(e*co
s(d*x+c))^(1/2)+6/7*e^5*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a^2/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2759, 2715, 2721, 2720} \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sin (c+d x) \sqrt {e \cos (c+d x)}}{7 a^2 d}+\frac {18 e^3 \sin (c+d x) (e \cos (c+d x))^{5/2}}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )} \]

[In]

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

[Out]

(6*e^6*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(7*a^2*d*Sqrt[e*Cos[c + d*x]]) + (6*e^5*Sqrt[e*Cos[c + d*
x]]*Sin[c + d*x])/(7*a^2*d) + (18*e^3*(e*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(35*a^2*d) + (4*e*(e*Cos[c + d*x])^
(9/2))/(5*d*(a^2 + a^2*Sin[c + d*x]))

Rule 2715

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] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2759

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*} \text {integral}& = \frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (9 e^2\right ) \int (e \cos (c+d x))^{7/2} \, dx}{5 a^2} \\ & = \frac {18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (9 e^4\right ) \int (e \cos (c+d x))^{3/2} \, dx}{7 a^2} \\ & = \frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac {18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (3 e^6\right ) \int \frac {1}{\sqrt {e \cos (c+d x)}} \, dx}{7 a^2} \\ & = \frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac {18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {\left (3 e^6 \sqrt {\cos (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{7 a^2 \sqrt {e \cos (c+d x)}} \\ & = \frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^2 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^2 d}+\frac {18 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^2 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2+a^2 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {13}{4},\frac {17}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{13 a^2 d e (1+\sin (c+d x))^{13/4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 20.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.40

method	result	size
default	\(\frac {2 e^{6} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-112 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-84 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\)	\(203\)

[In]

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

[Out]

2/35/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^6*(80*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
8-120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-112*sin(1/2*d*x+1/2*c)^7+168*sin(1/2*d*x+1/2*c)^5+20*sin(1/2*d*x
+1/2*c)^2*cos(1/2*d*x+1/2*c)-15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/
2*d*x+1/2*c),2^(1/2))-84*sin(1/2*d*x+1/2*c)^3+14*sin(1/2*d*x+1/2*c))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.78 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {-15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (14 \, e^{5} \cos \left (d x + c\right )^{2} - 5 \, {\left (e^{5} \cos \left (d x + c\right )^{2} - 3 \, e^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{35 \, a^{2} d} \]

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/35*(-15*I*sqrt(2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 15*I*sqrt(2)*e^(11/2)
*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(14*e^5*cos(d*x + c)^2 - 5*(e^5*cos(d*x + c)^2
- 3*e^5)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(a^2*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(11/2)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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