Integrand size = 25, antiderivative size = 132 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {18 e^3 (e \cos (c+d x))^{5/2}}{5 a^3 d}+\frac {6 e^6 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^3 d \sqrt {e \cos (c+d x)}}+\frac {6 e^5 \sqrt {e \cos (c+d x)} \sin (c+d x)}{a^3 d}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a+a \sin (c+d x))^2} \]
18/5*e^3*(e*cos(d*x+c))^(5/2)/a^3/d+4*e*(e*cos(d*x+c))^(9/2)/a/d/(a+a*sin( d*x+c))^2+6*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^3/d/(e*cos(d*x+c))^(1/2)+6* e^5*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a^3/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.50 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{13/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {13}{4},\frac {17}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{13 a^3 d e (1+\sin (c+d x))^{13/4}} \]
(-2*2^(1/4)*(e*Cos[c + d*x])^(13/2)*Hypergeometric2F1[3/4, 13/4, 17/4, (1 - Sin[c + d*x])/2])/(13*a^3*d*e*(1 + Sin[c + d*x])^(13/4))
Time = 0.63 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3159, 3042, 3161, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{11/2}}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {9 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{\sin (c+d x) a+a}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9 e^2 \int \frac {(e \cos (c+d x))^{7/2}}{\sin (c+d x) a+a}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3161 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \int (e \cos (c+d x))^{3/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {9 e^2 \left (\frac {e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )}{a}+\frac {2 e (e \cos (c+d x))^{5/2}}{5 a d}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{a d (a \sin (c+d x)+a)^2}\) |
(4*e*(e*Cos[c + d*x])^(9/2))/(a*d*(a + a*Sin[c + d*x])^2) + (9*e^2*((2*e*( e*Cos[c + d*x])^(5/2))/(5*a*d) + (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]]*S in[c + d*x])/(3*d)))/a))/a^2
3.3.55.3.1 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 _)])^(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]
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 = 172.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.37
| method | result | size |
| default | \(\frac {2 e^{6} \left (8 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-12 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-15 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-34 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+19 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\) | \(181\) |
2/5/a^3/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^6*(8*sin( 1/2*d*x+1/2*c)^7+20*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x +1/2*c)^5-10*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))-34*sin(1/2*d*x+1/2*c)^3+19*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 = 102, normalized size of antiderivative = 0.77 \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\frac {-15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 i \, \sqrt {2} e^{\frac {11}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (e^{5} \cos \left (d x + c\right )^{2} + 5 \, e^{5} \sin \left (d x + c\right ) - 20 \, e^{5}\right )} \sqrt {e \cos \left (d x + c\right )}}{5 \, a^{3} d} \]
1/5*(-15*I*sqrt(2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*si n(d*x + c)) + 15*I*sqrt(2)*e^(11/2)*weierstrassPInverse(-4, 0, cos(d*x + c ) - I*sin(d*x + c)) - 2*(e^5*cos(d*x + c)^2 + 5*e^5*sin(d*x + c) - 20*e^5) *sqrt(e*cos(d*x + c)))/(a^3*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{11/2}}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]