Integrand size = 25, antiderivative size = 101 \[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}+\frac {6 e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d \sqrt {\cos (c+d x)}}+\frac {2 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]
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)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.07 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=-\frac {4\ 2^{3/4} (e \cos (c+d x))^{11/2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {11}{4},\frac {15}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{11 a d e (1+\sin (c+d x))^{11/4}} \]
(-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))
Time = 0.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3161, 3042, 3115, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3161 |
\(\displaystyle \frac {e^2 \int (e \cos (c+d x))^{5/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \cos (c+d x)}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {e^2 \left (\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{7/2}}{7 a d}\) |
(2*e*(e*Cos[c + d*x])^(7/2))/(7*a*d) + (e^2*((6*e^2*Sqrt[e*Cos[c + d*x]]*E llipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]) + (2*e*(e*Cos[c + d*x]) ^(3/2)*Sin[c + d*x])/(5*d)))/a
3.3.34.3.1 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[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 = 3.68 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.14
method | result | size |
default | \(\frac {2 e^{5} \left (80 \left (\sin ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+56 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-160 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-56 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+120 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-40 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(216\) |
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+14*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+21*(sin(1/2*d*x+1/2*c)^2)^ (1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 ))-40*sin(1/2*d*x+1/2*c)^3+5*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 = 110, normalized size of antiderivative = 1.09 \[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\frac {21 i \, \sqrt {2} e^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} e^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (5 \, e^{4} \cos \left (d x + c\right )^{3} + 7 \, e^{4} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{35 \, a d} \]
1/35*(21*I*sqrt(2)*e^(9/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*I*sqrt(2)*e^(9/2)*weierstrassZeta( -4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(5*e ^4*cos(d*x + c)^3 + 7*e^4*cos(d*x + c)*sin(d*x + c))*sqrt(e*cos(d*x + c))) /(a*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{a \sin \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{a \sin \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{a+a \sin (c+d x)} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{a+a\,\sin \left (c+d\,x\right )} \,d x \]