Integrand size = 25, antiderivative size = 190 \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\frac {4 e^3}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos (c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {2 e^3 \cos ^3(c+d x)}{7 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 e}{3 a^2 d (e \sin (c+d x))^{3/2}}+\frac {16 e \cos (c+d x)}{21 a^2 d (e \sin (c+d x))^{3/2}}+\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 a^2 d \sqrt {e \sin (c+d x)}} \]
4/7*e^3/a^2/d/(e*sin(d*x+c))^(7/2)-2/7*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c)) ^(7/2)-2/7*e^3*cos(d*x+c)^3/a^2/d/(e*sin(d*x+c))^(7/2)-4/3*e/a^2/d/(e*sin( d*x+c))^(3/2)+16/21*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(3/2)-20/21*(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*sin(d*x+c))^(1/2)
Time = 1.97 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=-\frac {\csc ^3(c+d x) \left (16 (8+11 \cos (c+d x)) \sin ^4\left (\frac {1}{2} (c+d x)\right )+40 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right ) \sin ^{\frac {7}{2}}(c+d x)\right )}{42 a^2 d \sqrt {e \sin (c+d x)}} \]
-1/42*(Csc[c + d*x]^3*(16*(8 + 11*Cos[c + d*x])*Sin[(c + d*x)/2]^4 + 40*El lipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(7/2)))/(a^2*d*Sqrt[e*Sin[c + d*x]])
Time = 0.87 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^2 \sqrt {e \sin (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2 \sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2 \sqrt {e \sin (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x-\frac {\pi }{2}\right )-a\right )^2 \sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {e^4 \int \frac {\cos ^2(c+d x) (a-a \cos (c+d x))^2}{(e \sin (c+d x))^{9/2}}dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^4 \int \frac {\sin \left (c+d x-\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^2}{\left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{9/2}}dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {e^4 \int \left (\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{9/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{9/2}}+\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}}\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^4 \left (\frac {20 a^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{21 d e^4 \sqrt {e \sin (c+d x)}}-\frac {4 a^2}{3 d e^3 (e \sin (c+d x))^{3/2}}+\frac {16 a^2 \cos (c+d x)}{21 d e^3 (e \sin (c+d x))^{3/2}}+\frac {4 a^2}{7 d e (e \sin (c+d x))^{7/2}}-\frac {2 a^2 \cos ^3(c+d x)}{7 d e (e \sin (c+d x))^{7/2}}-\frac {2 a^2 \cos (c+d x)}{7 d e (e \sin (c+d x))^{7/2}}\right )}{a^4}\) |
(e^4*((4*a^2)/(7*d*e*(e*Sin[c + d*x])^(7/2)) - (2*a^2*Cos[c + d*x])/(7*d*e *(e*Sin[c + d*x])^(7/2)) - (2*a^2*Cos[c + d*x]^3)/(7*d*e*(e*Sin[c + d*x])^ (7/2)) - (4*a^2)/(3*d*e^3*(e*Sin[c + d*x])^(3/2)) + (16*a^2*Cos[c + d*x])/ (21*d*e^3*(e*Sin[c + d*x])^(3/2)) + (20*a^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(21*d*e^4*Sqrt[e*Sin[c + d*x]])))/a^4
3.2.31.3.1 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 = 5.89 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {\frac {4 e^{3} \left (7 \cos \left (d x +c \right )^{2}-4\right )}{21 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}-\frac {2 \left (5 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {9}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+11 \sin \left (d x +c \right )^{5}-17 \sin \left (d x +c \right )^{3}+6 \sin \left (d x +c \right )\right )}{21 a^{2} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(148\) |
(4/21/a^2*e^3/(e*sin(d*x+c))^(7/2)*(7*cos(d*x+c)^2-4)-2/21*(5*(-sin(d*x+c) +1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(9/2)*EllipticF((-sin(d*x+c)+1 )^(1/2),1/2*2^(1/2))+11*sin(d*x+c)^5-17*sin(d*x+c)^3+6*sin(d*x+c))/a^2/sin (d*x+c)^4/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 = 165, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\frac {2 \, {\left (5 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {i \, e} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \sqrt {e \sin \left (d x + c\right )} {\left (11 \, \cos \left (d x + c\right ) + 8\right )}\right )}}{21 \, {\left (a^{2} d e \cos \left (d x + c\right )^{2} + 2 \, a^{2} d e \cos \left (d x + c\right ) + a^{2} d e\right )}} \]
2/21*(5*(sqrt(2)*cos(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*sqrt(- I*e)*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2) *cos(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*sqrt(I*e)*weierstrassP Inverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) - sqrt(e*sin(d*x + c))*(11*co s(d*x + c) + 8))/(a^2*d*e*cos(d*x + c)^2 + 2*a^2*d*e*cos(d*x + c) + a^2*d* e)
\[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\frac {\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )} + 2 \sqrt {e \sin {\left (c + d x \right )}} \sec {\left (c + d x \right )} + \sqrt {e \sin {\left (c + d x \right )}}}\, dx}{a^{2}} \]
Integral(1/(sqrt(e*sin(c + d*x))*sec(c + d*x)**2 + 2*sqrt(e*sin(c + d*x))* sec(c + d*x) + sqrt(e*sin(c + d*x))), x)/a**2
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,\sqrt {e\,\sin \left (c+d\,x\right )}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]