Integrand size = 25, antiderivative size = 132 \[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}+\frac {10 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 a d \sqrt {e \cos (c+d x)}}+\frac {10 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{21 a d}+\frac {2 e^3 (e \cos (c+d x))^{5/2} \sin (c+d x)}{7 a d} \]
2/9*e*(e*cos(d*x+c))^(9/2)/a/d+2/7*e^3*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a/d +10/21*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/d/(e*cos(d*x+c))^(1/2)+10/21*e^5 *sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.50 \[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=-\frac {8 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {13}{4},\frac {17}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{13 a d e (1+\sin (c+d x))^{13/4}} \]
(-8*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[-5/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(13*a*d*e*(1 + Sin[c + d*x])^(13/4))
Time = 0.55 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3161, 3042, 3115, 3042, 3115, 3042, 3121, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle \frac {e^2 \int (e \cos (c+d x))^{7/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \int (e \cos (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {e^2 \left (\frac {5}{7} e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{9/2}}{9 a d}\) |
(2*e*(e*Cos[c + d*x])^(9/2))/(9*a*d) + (e^2*((2*e*(e*Cos[c + d*x])^(5/2)*S in[c + d*x])/(7*d) + (5*e^2*((2*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x) /2, 2])/(3*d*Sqrt[e*Cos[c + d*x]]) + (2*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x ])/(3*d)))/7))/a
3.3.33.3.1 Defintions of rubi rules used
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
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] + Si mp[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]
Time = 4.25 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.90
method | result | size |
default | \(-\frac {2 e^{6} \left (224 \left (\sin ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+144 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-560 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-216 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+560 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+168 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-280 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+70 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{63 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(251\) |
-2/63/a/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(224*si n(1/2*d*x+1/2*c)^11+144*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-560*sin(1/ 2*d*x+1/2*c)^9-216*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+560*sin(1/2*d*x +1/2*c)^7+168*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-280*sin(1/2*d*x+1/2* c)^5-48*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 ))+70*sin(1/2*d*x+1/2*c)^3-7*sin(1/2*d*x+1/2*c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\frac {-15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (7 \, e^{5} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, e^{5} \cos \left (d x + c\right )^{2} + 5 \, e^{5}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{63 \, a d} \]
1/63*(-15*I*sqrt(2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c)) + 15*I*sqrt(2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(7*e^5*cos(d*x + c)^4 + 3*(3*e^5*cos(d*x + c)^2 + 5*e^5)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(a*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{a \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{a+a \sin (c+d x)} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{a+a\,\sin \left (c+d\,x\right )} \,d x \]