Integrand size = 25, antiderivative size = 112 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 a d e^2 \sqrt {e \cos (c+d x)}}+\frac {10 \sin (c+d x)}{21 a d e (e \cos (c+d x))^{3/2}}-\frac {2}{7 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))} \]
10/21*sin(d*x+c)/a/d/e/(e*cos(d*x+c))^(3/2)-2/7/d/e/(e*cos(d*x+c))^(3/2)/( a+a*sin(d*x+c))+10/21*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/a/d/e^2/(e*cos(d*x+c))^ (1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.59 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {11}{4},\frac {1}{4},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{3/4}}{3\ 2^{3/4} a d e (e \cos (c+d x))^{3/2}} \]
(Hypergeometric2F1[-3/4, 11/4, 1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x ])^(3/4))/(3*2^(3/4)*a*d*e*(e*Cos[c + d*x])^(3/2))
Time = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3162, 3042, 3116, 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.
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a) (e \cos (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (c+d x)+a) (e \cos (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3162 |
| \(\displaystyle \frac {5 \int \frac {1}{(e \cos (c+d x))^{5/2}}dx}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \int \frac {1}{\left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {e \cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\int \frac {1}{\sqrt {e \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d e^2 \sqrt {e \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 d e (e \cos (c+d x))^{3/2}}\right )}{7 a}-\frac {2}{7 d e (a \sin (c+d x)+a) (e \cos (c+d x))^{3/2}}\) |
-2/(7*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])) + (5*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (2*Si n[c + d*x])/(3*d*e*(e*Cos[c + d*x])^(3/2))))/(7*a)
3.3.41.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 _)]), x_Symbol] :> Simp[b*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p - 1)*(a + b*S in[e + f*x]))), x] + Simp[p/(a*(p - 1)) Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && !GeQ[p, 1] && Intege rQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(374\) vs. \(2(124)=248\).
Time = 4.24 (sec) , antiderivative size = 375, normalized size of antiderivative = 3.35
| method | result | size |
| default | \(-\frac {2 \left (40 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+40 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-60 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+30 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+16 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d}\) | \(375\) |
-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c) ^2-1)/a/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^2*(40*(si n(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2 *d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^6+40*sin(1/2*d*x+1/2*c)^6*cos(1/ 2*d*x+1/2*c)-60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c), 2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-40*sin(1/2* d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+30*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF( cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1 /2*c)^2+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*(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))-3*sin(1/2*d*x+1/2*c))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=-\frac {5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - i \, \sqrt {2} \cos \left (d x + c\right )^{2}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {e \cos \left (d x + c\right )} {\left (5 \, \cos \left (d x + c\right )^{2} - 5 \, \sin \left (d x + c\right ) - 2\right )}}{21 \, {\left (a d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d e^{3} \cos \left (d x + c\right )^{2}\right )}} \]
-1/21*(5*(I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) + I*sqrt(2)*cos(d*x + c)^2 )*sqrt(e)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(- I*sqrt(2)*cos(d*x + c)^2*sin(d*x + c) - I*sqrt(2)*cos(d*x + c)^2)*sqrt(e)* weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*sqrt(e*cos(d *x + c))*(5*cos(d*x + c)^2 - 5*sin(d*x + c) - 2))/(a*d*e^3*cos(d*x + c)^2* sin(d*x + c) + a*d*e^3*cos(d*x + c)^2)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}} \,d x } \]
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sin \left (d x + c\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (a+a\,\sin \left (c+d\,x\right )\right )} \,d x \]