Integrand size = 25, antiderivative size = 180 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {78 e^8 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{7 a^4 d \sqrt {e \cos (c+d x)}}+\frac {78 e^7 \sqrt {e \cos (c+d x)} \sin (c+d x)}{7 a^4 d}+\frac {234 e^5 (e \cos (c+d x))^{5/2} \sin (c+d x)}{35 a^4 d}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a+a \sin (c+d x))^3}+\frac {52 e^3 (e \cos (c+d x))^{9/2}}{5 d \left (a^4+a^4 \sin (c+d x)\right )} \]
234/35*e^5*(e*cos(d*x+c))^(5/2)*sin(d*x+c)/a^4/d+4*e*(e*cos(d*x+c))^(13/2) /a/d/(a+a*sin(d*x+c))^3+52/5*e^3*(e*cos(d*x+c))^(9/2)/d/(a^4+a^4*sin(d*x+c ))+78/7*e^8*(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^4/d/(e*cos(d*x+c))^(1/2)+78/7*e ^7*sin(d*x+c)*(e*cos(d*x+c))^(1/2)/a^4/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.37 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=-\frac {2 \sqrt [4]{2} (e \cos (c+d x))^{17/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {17}{4},\frac {21}{4},\frac {1}{2} (1-\sin (c+d x))\right )}{17 a^4 d e (1+\sin (c+d x))^{17/4}} \]
(-2*2^(1/4)*(e*Cos[c + d*x])^(17/2)*Hypergeometric2F1[3/4, 17/4, 21/4, (1 - Sin[c + d*x])/2])/(17*a^4*d*e*(1 + Sin[c + d*x])^(17/4))
Time = 0.79 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3159, 3042, 3159, 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.
| \(\displaystyle \int \frac {(e \cos (c+d x))^{15/2}}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(e \cos (c+d x))^{15/2}}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {13 e^2 \int \frac {(e \cos (c+d x))^{11/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 e^2 \int \frac {(e \cos (c+d x))^{11/2}}{(\sin (c+d x) a+a)^2}dx}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \int (e \cos (c+d x))^{7/2}dx}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}dx}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \int (e \cos (c+d x))^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \cos (c+d x)}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {13 e^2 \left (\frac {9 e^2 \left (\frac {5}{7} e^2 \left (\frac {2 e^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {e \cos (c+d x)}}+\frac {2 e \sin (c+d x) \sqrt {e \cos (c+d x)}}{3 d}\right )+\frac {2 e \sin (c+d x) (e \cos (c+d x))^{5/2}}{7 d}\right )}{5 a^2}+\frac {4 e (e \cos (c+d x))^{9/2}}{5 d \left (a^2 \sin (c+d x)+a^2\right )}\right )}{a^2}+\frac {4 e (e \cos (c+d x))^{13/2}}{a d (a \sin (c+d x)+a)^3}\) |
(4*e*(e*Cos[c + d*x])^(13/2))/(a*d*(a + a*Sin[c + d*x])^3) + (13*e^2*((4*e *(e*Cos[c + d*x])^(9/2))/(5*d*(a^2 + a^2*Sin[c + d*x])) + (9*e^2*((2*e*(e* Cos[c + d*x])^(5/2)*Sin[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))/(5*a^2)))/a^2
3.3.63.3.1 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]
Time = 3.89 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.25
\[-\frac {2 e^{8} \left (80 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-224 \left (\sin ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-280 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+336 \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+160 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+195 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+392 \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-252 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 a^{4} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, d}\]
-2/35/a^4/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*e^8*(80*c os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-120*sin(1/2*d*x+1/2*c)^6*cos(1/2*d* x+1/2*c)-224*sin(1/2*d*x+1/2*c)^7-280*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2 *c)+336*sin(1/2*d*x+1/2*c)^5+160*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1 95*(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))+392*sin(1/2*d*x+1/2*c)^3-252*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 = 118, normalized size of antiderivative = 0.66 \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\frac {-195 i \, \sqrt {2} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 195 i \, \sqrt {2} e^{\frac {15}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 2 \, {\left (28 \, e^{7} \cos \left (d x + c\right )^{2} - 280 \, e^{7} - 5 \, {\left (e^{7} \cos \left (d x + c\right )^{2} - 17 \, e^{7}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{35 \, a^{4} d} \]
1/35*(-195*I*sqrt(2)*e^(15/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I* sin(d*x + c)) + 195*I*sqrt(2)*e^(15/2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 2*(28*e^7*cos(d*x + c)^2 - 280*e^7 - 5*(e^7*cos(d *x + c)^2 - 17*e^7)*sin(d*x + c))*sqrt(e*cos(d*x + c)))/(a^4*d)
Timed out. \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]
\[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
\[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]
Timed out. \[ \int \frac {(e \cos (c+d x))^{15/2}}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{15/2}}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]