Integrand size = 33, antiderivative size = 118 \[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sqrt {\sec (d+e x)} \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}{e \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]
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))*sec(e*x+d)^(1/2)*((b+a*cos(e*x+d)+c*sin(e* x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)/e/(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 1.26 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.87 \[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {3}{2},\frac {b+\sqrt {1+\frac {a^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {a}{c}\right )\right )}{b-\sqrt {1+\frac {a^2}{c^2}} c},\frac {b+\sqrt {1+\frac {a^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {a}{c}\right )\right )}{b+\sqrt {1+\frac {a^2}{c^2}} c}\right ) \sqrt {\sec (d+e x)} \sec \left (d+e x+\arctan \left (\frac {a}{c}\right )\right ) \sqrt {b+a \cos (d+e x)+c \sin (d+e x)} \sqrt {-\frac {\sqrt {1+\frac {a^2}{c^2}} c \left (-1+\sin \left (d+e x+\arctan \left (\frac {a}{c}\right )\right )\right )}{b+\sqrt {1+\frac {a^2}{c^2}} c}} \sqrt {\frac {\sqrt {1+\frac {a^2}{c^2}} c \left (1+\sin \left (d+e x+\arctan \left (\frac {a}{c}\right )\right )\right )}{-b+\sqrt {1+\frac {a^2}{c^2}} c}} \sqrt {b+\sqrt {1+\frac {a^2}{c^2}} c \sin \left (d+e x+\arctan \left (\frac {a}{c}\right )\right )}}{\sqrt {1+\frac {a^2}{c^2}} c e \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \]
(2*AppellF1[1/2, 1/2, 1/2, 3/2, (b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + Arc Tan[a/c]])/(b - Sqrt[1 + a^2/c^2]*c), (b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(b + Sqrt[1 + a^2/c^2]*c)]*Sqrt[Sec[d + e*x]]*Sec[d + e*x + ArcTan[a/c]]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]*Sqrt[-((Sqrt[1 + a^2/c^2]*c*(-1 + Sin[d + e*x + ArcTan[a/c]]))/(b + Sqrt[1 + a^2/c^2]*c))] *Sqrt[(Sqrt[1 + a^2/c^2]*c*(1 + Sin[d + e*x + ArcTan[a/c]]))/(-b + Sqrt[1 + a^2/c^2]*c)]*Sqrt[b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]]])/( Sqrt[1 + a^2/c^2]*c*e*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])
Time = 0.50 (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, 3646, 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.
\(\displaystyle \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}dx\) |
\(\Big \downarrow \) 3646 |
\(\displaystyle \frac {\sqrt {\sec (d+e x)} \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (d+e x)} \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
\(\Big \downarrow \) 3606 |
\(\displaystyle \frac {\sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \frac {2 \sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}\) |
(2*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[Sec[d + e*x]]*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])/(e*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])
3.5.50.3.1 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[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(m_), x_Symbol] :> Simp[Sec[d + e*x]^n*((b + a*Cos[d + e*x] + c*Sin[d + e*x])^n/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n ) Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c , d, e}, x] && EqQ[m + n, 0] && !IntegerQ[n]
Result contains complex when optimal does not.
Time = 25.33 (sec) , antiderivative size = 1018, normalized size of antiderivative = 8.63
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 a^{2}-i \cos \left (e x +d \right ) c^{2}+i \cos \left (e x +d \right ) b^{2}+i a c \sin \left (e x +d \right )-i b c \sin \left (e x +d \right )+i a b -i c^{2}+i b \sqrt {a^{2}-b^{2}+c^{2}}\, \sin \left (e x +d \right )+i c \sqrt {a^{2}-b^{2}+c^{2}}-\sqrt {a^{2}-b^{2}+c^{2}}\, \cos \left (e x +d \right ) b +a c \cos \left (e x +d \right )-a^{2} \sin \left (e x +d \right )+b^{2} \sin \left (e x +d \right )-a \sqrt {a^{2}-b^{2}+c^{2}}+c b \right ) \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 )}\, \sqrt {\sec \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}}}\) | \(1018\) |
-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*a^2-I*cos(e*x+d)*c^2+I*cos(e*x+d)*b^2+I*a*c*sin(e*x+ d)-I*b*c*sin(e*x+d)+I*a*b-I*c^2+I*b*(a^2-b^2+c^2)^(1/2)*sin(e*x+d)+I*c*(a^ 2-b^2+c^2)^(1/2)-(a^2-b^2+c^2)^(1/2)*cos(e*x+d)*b+a*c*cos(e*x+d)-a^2*sin(e *x+d)+b^2*sin(e*x+d)-a*(a^2-b^2+c^2)^(1/2)+c*b)*cos(e*x+d)*(-((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*c os(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)*sec( 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 = 504, normalized size of antiderivative = 4.27 \[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\frac {\sqrt {2 \, a - 2 i \, c} {\left (-i \, a + c\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{2} c^{2} + 6 i \, a c^{3} - 3 \, c^{4} + 2 i \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} c\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )}}, \frac {8 \, {\left (9 \, a^{5} b - 8 \, a^{3} b^{3} - 27 \, a b c^{4} - 9 i \, b c^{5} + 2 i \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} c^{3} - 6 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} c^{2} + 3 i \, {\left (9 \, a^{4} b - 8 \, a^{2} b^{3}\right )} c\right )}}{27 \, {\left (a^{6} + 3 \, a^{4} c^{2} + 3 \, a^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b + 2 i \, b c + 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (-i \, a^{2} - i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (a^{2} + c^{2}\right )}}\right ) + \sqrt {2 \, a + 2 i \, c} {\left (i \, a + c\right )} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{2} c^{2} - 6 i \, a c^{3} - 3 \, c^{4} - 2 i \, {\left (3 \, a^{3} - 4 \, a b^{2}\right )} c\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )}}, \frac {8 \, {\left (9 \, a^{5} b - 8 \, a^{3} b^{3} - 27 \, a b c^{4} + 9 i \, b c^{5} - 2 i \, {\left (9 \, a^{2} b + 4 \, b^{3}\right )} c^{3} - 6 \, {\left (3 \, a^{3} b - 4 \, a b^{3}\right )} c^{2} - 3 i \, {\left (9 \, a^{4} b - 8 \, a^{2} b^{3}\right )} c\right )}}{27 \, {\left (a^{6} + 3 \, a^{4} c^{2} + 3 \, a^{2} c^{4} + c^{6}\right )}}, \frac {2 \, a b - 2 i \, b c + 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (e x + d\right ) - 3 \, {\left (i \, a^{2} + i \, c^{2}\right )} \sin \left (e x + d\right )}{3 \, {\left (a^{2} + c^{2}\right )}}\right )}{{\left (a^{2} + c^{2}\right )} e} \]
(sqrt(2*a - 2*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*a + 2*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), 8/2 7*(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)
\[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int \frac {\sqrt {\sec {\left (d + e x \right )}}}{\sqrt {a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int { \frac {\sqrt {\sec \left (e x + d\right )}}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int { \frac {\sqrt {\sec \left (e x + d\right )}}{\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\sec (d+e x)}}{\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (d+e\,x\right )}}}{\sqrt {a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}}} \,d x \]