Integrand size = 25, antiderivative size = 224 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}+\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}} \]
4/11*e^3/a^2/d/(e*sin(d*x+c))^(11/2)-2/11*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+ c))^(11/2)-2/11*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(11/2)-4/7*e/a^2/d/( e*sin(d*x+c))^(7/2)+16/77*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(7/2)-4/231*co s(d*x+c)/a^2/d/e/(e*sin(d*x+c))^(3/2)-4/231*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^ (1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2 ))*sin(d*x+c)^(1/2)/a^2/d/e^2/(e*sin(d*x+c))^(1/2)
Time = 1.65 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.50 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=-\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (52+97 \cos (c+d x)+4 \cos (2 (c+d x))+\cos (3 (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {11}{2}}(c+d x)\right )}{1848 a^2 d e^2 \sqrt {e \sin (c+d x)}} \]
-1/1848*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]^5*(52 + 97*Cos[c + d*x] + 4*Cos [2*(c + d*x)] + Cos[3*(c + d*x)] + Csc[(c + d*x)/2]^4*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(11/2)))/(a^2*d*e^2*Sqrt[e*Sin[c + d*x]])
Time = 0.95 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4360, 3042, 3354, 3042, 3352, 2009}
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.
| \(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 (e \sin (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2 (e \sin (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2 \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{13/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{13/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{13/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{13/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{13/2}}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \left (\frac {4 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{231 d e^6 \sqrt {e \sin (c+d x)}}-\frac {4 a^2 \cos (c+d x)}{231 d e^5 (e \sin (c+d x))^{3/2}}-\frac {4 a^2}{7 d e^3 (e \sin (c+d x))^{7/2}}+\frac {16 a^2 \cos (c+d x)}{77 d e^3 (e \sin (c+d x))^{7/2}}+\frac {4 a^2}{11 d e (e \sin (c+d x))^{11/2}}-\frac {2 a^2 \cos ^3(c+d x)}{11 d e (e \sin (c+d x))^{11/2}}-\frac {2 a^2 \cos (c+d x)}{11 d e (e \sin (c+d x))^{11/2}}\right )}{a^4}\) |
(e^4*((4*a^2)/(11*d*e*(e*Sin[c + d*x])^(11/2)) - (2*a^2*Cos[c + d*x])/(11* d*e*(e*Sin[c + d*x])^(11/2)) - (2*a^2*Cos[c + d*x]^3)/(11*d*e*(e*Sin[c + d *x])^(11/2)) - (4*a^2)/(7*d*e^3*(e*Sin[c + d*x])^(7/2)) + (16*a^2*Cos[c + d*x])/(77*d*e^3*(e*Sin[c + d*x])^(7/2)) - (4*a^2*Cos[c + d*x])/(231*d*e^5* (e*Sin[c + d*x])^(3/2)) + (4*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin [c + d*x]])/(231*d*e^6*Sqrt[e*Sin[c + d*x]])))/a^4
3.2.33.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 6.73 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.71
| method | result | size |
| default | \(\frac {\frac {4 e^{3} \left (11 \cos \left (d x +c \right )^{2}-4\right )}{77 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {11}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {13}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{7}+47 \sin \left (d x +c \right )^{5}-87 \sin \left (d x +c \right )^{3}+42 \sin \left (d x +c \right )\right )}{231 e^{2} a^{2} \sin \left (d x +c \right )^{6} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(160\) |
(4/77*e^3/a^2/(e*sin(d*x+c))^(11/2)*(11*cos(d*x+c)^2-4)-2/231/e^2*((-sin(d *x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(13/2)*EllipticF((-sin(d* x+c)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^7+47*sin(d*x+c)^5-87*sin(d*x+c)^3+ 42*sin(d*x+c))/a^2/sin(d*x+c)^6/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\frac {2 \, {\left ({\left (\sqrt {2} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} \cos \left (d x + c\right ) - \sqrt {2}\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + 47 \, \cos \left (d x + c\right ) + 24\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{231 \, {\left (a^{2} d e^{3} \cos \left (d x + c\right )^{4} + 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{3} - 2 \, a^{2} d e^{3} \cos \left (d x + c\right ) - a^{2} d e^{3}\right )}} \]
2/231*((sqrt(2)*cos(d*x + c)^4 + 2*sqrt(2)*cos(d*x + c)^3 - 2*sqrt(2)*cos( d*x + c) - sqrt(2))*sqrt(-I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I* sin(d*x + c)) + (sqrt(2)*cos(d*x + c)^4 + 2*sqrt(2)*cos(d*x + c)^3 - 2*sqr t(2)*cos(d*x + c) - sqrt(2))*sqrt(I*e)*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 + 47*cos(d*x + c) + 24)*sqrt(e*sin(d*x + c)))/(a^2*d*e^3*cos(d*x + c)^4 + 2*a^2*d*e^3*c os(d*x + c)^3 - 2*a^2*d*e^3*cos(d*x + c) - a^2*d*e^3)
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]