Integrand size = 25, antiderivative size = 120 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {42 e^4 \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^4 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a+a \sin (c+d x))^3}+\frac {28 e^3 (e \cos (c+d x))^{3/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )} \]
-4/5*e*(e*cos(d*x+c))^(7/2)/a/d/(a+a*sin(d*x+c))^3+28/5*e^3*(e*cos(d*x+c)) ^(3/2)/d/(a^4+a^4*sin(d*x+c))+42/5*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^ 4/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 = 0.55 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {(e \cos (c+d x))^{11/2} \operatorname {Hypergeometric2F1}\left (\frac {9}{4},\frac {11}{4},\frac {15}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{11 \sqrt [4]{2} a^4 d e (1+\sin (c+d x))^{11/4}} \]
-1/11*((e*Cos[c + d*x])^(11/2)*Hypergeometric2F1[9/4, 11/4, 15/4, (1 - Sin [c + d*x])/2])/(2^(1/4)*a^4*d*e*(1 + Sin[c + d*x])^(11/4))
Time = 0.53 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04, 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, 3159, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{9/2}}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(\sin (c+d x) a+a)^2}dx}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \int \frac {(e \cos (c+d x))^{5/2}}{(\sin (c+d x) a+a)^2}dx}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {3 e^2 \int \sqrt {e \cos (c+d x)}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {3 e^2 \int \sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a^2 \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {3 e^2 \sqrt {e \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2 \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {7 e^2 \left (-\frac {6 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{a^2 d \sqrt {\cos (c+d x)}}-\frac {4 e (e \cos (c+d x))^{3/2}}{d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{5 a^2}-\frac {4 e (e \cos (c+d x))^{7/2}}{5 a d (a \sin (c+d x)+a)^3}\) |
(-4*e*(e*Cos[c + d*x])^(7/2))/(5*a*d*(a + a*Sin[c + d*x])^3) - (7*e^2*((-6 *e^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(a^2*d*Sqrt[Cos[c + d *x]]) - (4*e*(e*Cos[c + d*x])^(3/2))/(d*(a^2 + a^2*Sin[c + d*x]))))/(5*a^2 )
3.3.66.3.1 Defintions of rubi rules used
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]
Leaf count of result is larger than twice the leaf count of optimal. \(331\) vs. \(2(132)=264\).
Time = 5.69 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.77
\[-\frac {2 \left (128 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-84 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-128 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+84 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-80 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-21 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+80 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{5}}{5 \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\]
-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/a^4/sin(1/2*d*x+1/2 *c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*(128*sin(1/2*d*x+1/2*c)^6*cos(1/2* d*x+1/2*c)-84*(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4-128*sin(1/2*d *x+1/2*c)^4*cos(1/2*d*x+1/2*c)+84*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti cE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/ 2*c)^2-80*sin(1/2*d*x+1/2*c)^5+16*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))+80*sin(1/2*d*x+1/2*c)^3-12*sin(1/2*d*x+1/2*c) )*e^5/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 317, normalized size of antiderivative = 2.64 \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {21 \, {\left (i \, \sqrt {2} e^{4} \cos \left (d x + c\right )^{2} - i \, \sqrt {2} e^{4} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} e^{4} + {\left (-i \, \sqrt {2} e^{4} \cos \left (d x + c\right ) - 2 i \, \sqrt {2} e^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (-i \, \sqrt {2} e^{4} \cos \left (d x + c\right )^{2} + i \, \sqrt {2} e^{4} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} e^{4} + {\left (i \, \sqrt {2} e^{4} \cos \left (d x + c\right ) + 2 i \, \sqrt {2} e^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 8 \, {\left (4 \, e^{4} \cos \left (d x + c\right )^{2} + 3 \, e^{4} \cos \left (d x + c\right ) - e^{4} + {\left (4 \, e^{4} \cos \left (d x + c\right ) + e^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, {\left (a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d \cos \left (d x + c\right ) - 2 \, a^{4} d - {\left (a^{4} d \cos \left (d x + c\right ) + 2 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]
1/5*(21*(I*sqrt(2)*e^4*cos(d*x + c)^2 - I*sqrt(2)*e^4*cos(d*x + c) - 2*I*s qrt(2)*e^4 + (-I*sqrt(2)*e^4*cos(d*x + c) - 2*I*sqrt(2)*e^4)*sin(d*x + c)) *sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(-I*sqrt(2)*e^4*cos(d*x + c)^2 + I*sqrt(2)*e^4*cos(d *x + c) + 2*I*sqrt(2)*e^4 + (I*sqrt(2)*e^4*cos(d*x + c) + 2*I*sqrt(2)*e^4) *sin(d*x + c))*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(d*x + c) - I*sin(d*x + c))) - 8*(4*e^4*cos(d*x + c)^2 + 3*e^4*cos(d*x + c) - e^4 + (4*e^4*cos(d*x + c) + e^4)*sin(d*x + c))*sqrt(e*cos(d*x + c))) /(a^4*d*cos(d*x + c)^2 - a^4*d*cos(d*x + c) - 2*a^4*d - (a^4*d*cos(d*x + c ) + 2*a^4*d)*sin(d*x + c))
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {9}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{9/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]