Integrand size = 33, antiderivative size = 118 \[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}{e \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]
2*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arcta n(a,c))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1 /2)/(b+(a^2+c^2)^(1/2)))^(1/2))*((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2 )^(1/2)))^(1/2)/e/cos(e*x+d)^(1/2)/(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)
Result contains complex when optimal does not.
Time = 16.02 (sec) , antiderivative size = 506, normalized size of antiderivative = 4.29 \[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {4 \left (i a-i b+c+\sqrt {a^2-b^2+c^2}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {\left (-i a+i b+c+\sqrt {a^2-b^2+c^2}\right ) (-\cos (d+e x)+i \sin (d+e x))}{i a-i b+c+\sqrt {a^2-b^2+c^2}}}\right ),\frac {b+i \sqrt {a^2-b^2+c^2}}{b-i \sqrt {a^2-b^2+c^2}}\right ) \sqrt {\frac {\left (-i a+i b+c+\sqrt {a^2-b^2+c^2}\right ) (-\cos (d+e x)+i \sin (d+e x))}{i a-i b+c+\sqrt {a^2-b^2+c^2}}} (\cos (d+e x)+i \sin (d+e x)) \sqrt {-\frac {i \left (-c+\sqrt {a^2-b^2+c^2}+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )\right )}{\left (-i a+i b-c+\sqrt {a^2-b^2+c^2}\right ) \left (-i+\tan \left (\frac {1}{2} (d+e x)\right )\right )}} \sqrt {-\frac {i \left (c+\sqrt {a^2-b^2+c^2}+(-a+b) \tan \left (\frac {1}{2} (d+e x)\right )\right )}{\left (i a-i b+c+\sqrt {a^2-b^2+c^2}\right ) \left (-i+\tan \left (\frac {1}{2} (d+e x)\right )\right )}}}{\left (a+i \left (i b+c+\sqrt {a^2-b^2+c^2}\right )\right ) e \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]
(4*(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])*EllipticF[ArcSin[Sqrt[(((-I)*a + I*b + c + Sqrt[a^2 - b^2 + c^2])*(-Cos[d + e*x] + I*Sin[d + e*x]))/(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])]], (b + I*Sqrt[a^2 - b^2 + c^2])/(b - I *Sqrt[a^2 - b^2 + c^2])]*Sqrt[(((-I)*a + I*b + c + Sqrt[a^2 - b^2 + c^2])* (-Cos[d + e*x] + I*Sin[d + e*x]))/(I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])] *(Cos[d + e*x] + I*Sin[d + e*x])*Sqrt[((-I)*(-c + Sqrt[a^2 - b^2 + c^2] + (a - b)*Tan[(d + e*x)/2]))/(((-I)*a + I*b - c + Sqrt[a^2 - b^2 + c^2])*(-I + Tan[(d + e*x)/2]))]*Sqrt[((-I)*(c + Sqrt[a^2 - b^2 + c^2] + (-a + b)*Ta n[(d + e*x)/2]))/((I*a - I*b + c + Sqrt[a^2 - b^2 + c^2])*(-I + Tan[(d + e *x)/2]))])/((a + I*(I*b + c + Sqrt[a^2 - b^2 + c^2]))*e*Sqrt[Cos[d + e*x]] *Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])
Time = 0.51 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3642, 3042, 3606, 3042, 3140}
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.
| \(\displaystyle \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}dx\) |
| \(\Big \downarrow \) 3642 |
| \(\displaystyle \frac {\sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
| \(\Big \downarrow \) 3606 |
| \(\displaystyle \frac {\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
(2*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2]) ])/(e*Sqrt[Cos[d + e*x]]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])
3.5.55.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sq rt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]] Int[1/Sqrt[a/(a + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - ArcTan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2 , 0] && NeQ[b^2 + c^2, 0] && !GtQ[a + Sqrt[b^2 + c^2], 0]
Int[cos[(d_.) + (e_.)*(x_)]^(n_)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + ( c_.)*tan[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[d + e*x]^n*((a + b*Sec[d + e*x] + c*Tan[d + e*x])^n/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n) Int[(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c, d , e}, x] && !IntegerQ[n]
Result contains complex when optimal does not.
Time = 25.26 (sec) , antiderivative size = 1012, normalized size of antiderivative = 8.58
| method | result | size |
| default | \(\frac {4 \left (i \cos \left (e x +d \right ) a b +i a \sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-i \sqrt {a^{2}-b^{2}+c^{2}}\, \cos \left (e x +d \right ) c -i c \sqrt {a^{2}-b^{2}+c^{2}}+i a^{2}+i b c \sin \left (e x +d \right )-i \cos \left (e x +d \right ) b^{2}-i a c \sin \left (e x +d \right )-i a b -i b \sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )+i c^{2}+i \cos \left (e x +d \right ) c^{2}+\sqrt {a^{2}-b^{2}+c^{2}}\, \cos \left (e x +d \right ) b +a^{2} \sin \left (e x +d \right )-a c \cos \left (e x +d \right )-b^{2} \sin \left (e x +d \right )+a \sqrt {a^{2}-b^{2}+c^{2}}-c b \right ) \sqrt {\cos \left (e x +d \right )}\, \sqrt {-\frac {\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )+a \cos \left (e x +d \right )-b \cos \left (e x +d \right )+c \sin \left (e x +d \right )-a +b \right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-a \cos \left (e x +d \right )+b \cos \left (e x +d \right )-c \sin \left (e x +d \right )+a -b \right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}}\, \sqrt {\frac {\left (i \sin \left (e x +d \right )+\cos \left (e x +d \right )-1\right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}+c^{2}}}{\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-a \cos \left (e x +d \right )+b \cos \left (e x +d \right )-c \sin \left (e x +d \right )+a -b \right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}}\, \sqrt {\frac {\left (i \sin \left (e x +d \right )-\cos \left (e x +d \right )+1\right ) \left (a -b \right ) \sqrt {a^{2}-b^{2}+c^{2}}}{\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-a \cos \left (e x +d \right )+b \cos \left (e x +d \right )-c \sin \left (e x +d \right )+a -b \right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-\frac {\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )+a \cos \left (e x +d \right )-b \cos \left (e x +d \right )+c \sin \left (e x +d \right )-a +b \right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}{\left (\sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )-a \cos \left (e x +d \right )+b \cos \left (e x +d \right )-c \sin \left (e x +d \right )+a -b \right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}+c \right )}}, \sqrt {\frac {\left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}+c \right ) \left (i a -i b +\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}{\left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}-c \right )}}\right ) \sqrt {a +b \sec \left (e x +d \right )+c \tan \left (e x +d \right )}}{e \left (b +a \cos \left (e x +d \right )+c \sin \left (e x +d \right )\right ) \left (i a -i b -\sqrt {a^{2}-b^{2}+c^{2}}+c \right ) \sqrt {a^{2}-b^{2}+c^{2}}}\) | \(1012\) |
4/e*(I*cos(e*x+d)*a*b+I*a*(a^2-b^2+c^2)^(1/2)*sin(e*x+d)-I*(a^2-b^2+c^2)^( 1/2)*cos(e*x+d)*c-I*c*(a^2-b^2+c^2)^(1/2)+I*a^2+I*b*c*sin(e*x+d)-I*cos(e*x +d)*b^2-I*a*c*sin(e*x+d)-I*a*b-I*b*(a^2-b^2+c^2)^(1/2)*sin(e*x+d)+I*c^2+I* cos(e*x+d)*c^2+(a^2-b^2+c^2)^(1/2)*cos(e*x+d)*b+a^2*sin(e*x+d)-a*c*cos(e*x +d)-b^2*sin(e*x+d)+a*(a^2-b^2+c^2)^(1/2)-c*b)*cos(e*x+d)^(1/2)*(-((a^2-b^2 +c^2)^(1/2)*sin(e*x+d)+a*cos(e*x+d)-b*cos(e*x+d)+c*sin(e*x+d)-a+b)/((a^2-b ^2+c^2)^(1/2)*sin(e*x+d)-a*cos(e*x+d)+b*cos(e*x+d)-c*sin(e*x+d)+a-b)*(I*a- I*b-(a^2-b^2+c^2)^(1/2)+c)/(I*a-I*b+(a^2-b^2+c^2)^(1/2)+c))^(1/2)*((I*sin( e*x+d)+cos(e*x+d)-1)/((a^2-b^2+c^2)^(1/2)*sin(e*x+d)-a*cos(e*x+d)+b*cos(e* x+d)-c*sin(e*x+d)+a-b)*(a-b)*(a^2-b^2+c^2)^(1/2)/(I*a-I*b-(a^2-b^2+c^2)^(1 /2)-c))^(1/2)*((I*sin(e*x+d)-cos(e*x+d)+1)/((a^2-b^2+c^2)^(1/2)*sin(e*x+d) -a*cos(e*x+d)+b*cos(e*x+d)-c*sin(e*x+d)+a-b)*(a-b)*(a^2-b^2+c^2)^(1/2)/(I* a-I*b+(a^2-b^2+c^2)^(1/2)+c))^(1/2)*EllipticF((-((a^2-b^2+c^2)^(1/2)*sin(e *x+d)+a*cos(e*x+d)-b*cos(e*x+d)+c*sin(e*x+d)-a+b)/((a^2-b^2+c^2)^(1/2)*sin (e*x+d)-a*cos(e*x+d)+b*cos(e*x+d)-c*sin(e*x+d)+a-b)*(I*a-I*b-(a^2-b^2+c^2) ^(1/2)+c)/(I*a-I*b+(a^2-b^2+c^2)^(1/2)+c))^(1/2),((I*a-I*b+(a^2-b^2+c^2)^( 1/2)+c)*(I*a-I*b+(a^2-b^2+c^2)^(1/2)-c)/(I*a-I*b-(a^2-b^2+c^2)^(1/2)+c)/(I *a-I*b-(a^2-b^2+c^2)^(1/2)-c))^(1/2))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/ (b+a*cos(e*x+d)+c*sin(e*x+d))/(I*a-I*b-(a^2-b^2+c^2)^(1/2)+c)/(a^2-b^2+c^2 )^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 506, normalized size of antiderivative = 4.29 \[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {\sqrt {2} \sqrt {a - i \, c} {\left (-i \, a + c\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{2} c^{2} + 6 i \, a c^{3} - 3 \, c^{4} + 2 i \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} c\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )}}, \frac {8 \, {\left (9 \, a^{5} b - 8 \, a^{3} b^{3} - 27 \, a b c^{4} - 9 i \, b c^{5} + 2 i \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} c^{3} - 6 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} c^{2} + 3 i \, {\left (9 \, a^{4} b - 8 \, a^{2} b^{3}\right )} c\right )}}{27 \, {\left (a^{6} + 3 \, a^{4} c^{2} + 3 \, a^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b + 2 i \, b c + 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (-i \, a^{2} - i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (a^{2} + c^{2}\right )}}\right ) + \sqrt {2} \sqrt {a + i \, c} {\left (i \, a + c\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{2} c^{2} - 6 i \, a c^{3} - 3 \, c^{4} - 2 i \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} c\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )}}, \frac {8 \, {\left (9 \, a^{5} b - 8 \, a^{3} b^{3} - 27 \, a b c^{4} + 9 i \, b c^{5} - 2 i \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} c^{3} - 6 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} c^{2} - 3 i \, {\left (9 \, a^{4} b - 8 \, a^{2} b^{3}\right )} c\right )}}{27 \, {\left (a^{6} + 3 \, a^{4} c^{2} + 3 \, a^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b - 2 i \, b c + 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (i \, a^{2} + i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (a^{2} + c^{2}\right )}}\right )}{{\left (a^{2} + c^{2}\right )} e} \]
(sqrt(2)*sqrt(a - I*c)*(-I*a + c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2* b^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^ 2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9* a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2*b^3) *c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b + 2*I*b*c + 3*(a^2 + c ^2)*cos(e*x + d) - 3*(-I*a^2 - I*c^2)*sin(e*x + d))/(a^2 + c^2)) + sqrt(2) *sqrt(a + I*c)*(I*a + c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b ^2*c^2 - 6*I*a*c^3 - 3*c^4 - 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c ^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 + 9*I*b*c^5 - 2*I*(9*a^2*b + 4 *b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 - 3*I*(9*a^4*b - 8*a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b - 2*I*b*c + 3*(a^2 + c^2)*cos(e *x + d) - 3*(I*a^2 + I*c^2)*sin(e*x + d))/(a^2 + c^2)))/((a^2 + c^2)*e)
\[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int \frac {1}{\sqrt {a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}} \sqrt {\cos {\left (d + e x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}} \,d x } \]
\[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int { \frac {1}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a} \sqrt {\cos \left (e x + d\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {\cos (d+e x)} \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int \frac {1}{\sqrt {\cos \left (d+e\,x\right )}\,\sqrt {a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}}} \,d x \]