Integrand size = 25, antiderivative size = 114 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {14 e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d \sqrt {\cos (c+d x)}}+\frac {14 e^3 (e \cos (c+d x))^{3/2} \sin (c+d x)}{15 a^2 d}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2+a^2 \sin (c+d x)\right )} \]
14/15*e^3*(e*cos(d*x+c))^(3/2)*sin(d*x+c)/a^2/d+4/3*e*(e*cos(d*x+c))^(7/2) /d/(a^2+a^2*sin(d*x+c))+14/5*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^2/d/co s(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 = 0.58 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=-\frac {2\ 2^{3/4} (e \cos (c+d x))^{11/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{4},\frac {15}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{11 a^2 d e (1+\sin (c+d x))^{11/4}} \]
(-2*2^(3/4)*(e*Cos[c + d*x])^(11/2)*Hypergeometric2F1[1/4, 11/4, 15/4, (1 - Sin[c + d*x])/2])/(11*a^2*d*e*(1 + Sin[c + d*x])^(11/4))
Time = 0.47 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3159, 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.
\(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle \frac {7 e^2 \int (e \cos (c+d x))^{5/2}dx}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}dx}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {7 e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \cos (c+d x)}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 e^2 \left (\frac {3}{5} e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {7 e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {7 e^2 \left (\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{5 \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {7 e^2 \left (\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{5 d \sqrt {\cos (c+d x)}}+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{3/2}}{5 d}\right )}{3 a^2}+\frac {4 e (e \cos (c+d x))^{7/2}}{3 d \left (a^2 \sin (c+d x)+a^2\right )}\) |
(4*e*(e*Cos[c + d*x])^(7/2))/(3*d*(a^2 + a^2*Sin[c + d*x])) + (7*e^2*((6*e ^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(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)))/(3*a^2)
3.3.44.3.1 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 _)])^(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] + Simp[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] && EqQ[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]
Time = 19.35 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.67
method | result | size |
default | \(-\frac {2 e^{5} \left (24 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-40 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+40 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(190\) |
-2/15/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^5*(24*s in(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x +1/2*c)-40*sin(1/2*d*x+1/2*c)^5+6*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)^3-10*sin(1/2*d*x+1/2*c) )/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.95 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\frac {21 i \, \sqrt {2} e^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} e^{\frac {9}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, e^{4} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 10 \, e^{4} \cos \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{15 \, a^{2} d} \]
1/15*(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*(3*e ^4*cos(d*x + c)*sin(d*x + c) - 10*e^4*cos(d*x + c))*sqrt(e*cos(d*x + c)))/ (a^2*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^2} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^2} \,d x \]