Integrand size = 33, antiderivative size = 118 \[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\frac {2 E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{e \sqrt {\sec (d+e x)} \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}} \]
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))*EllipticE(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))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(1/2)/e/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)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 6.80 (sec) , antiderivative size = 1580, normalized size of antiderivative = 13.39 \[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx =\text {Too large to display} \]
(2*a*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(c*e*Sqrt[Sec[d + e*x]]) + (2*b*AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x + ArcTan[a/c]]*Sqrt[(c*Sqrt[(a ^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + A rcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(Sqrt[1 + a^2/c^2]*c*e*Sqrt[Sec[d + e*x]]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]) + (a^2*(-((c*AppellF1[-1/2, -1/2, -1/2, 1/2, -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c ^2/a^2]*(1 - b/(a*Sqrt[1 + c^2/a^2])))), -((b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*(-1 - b/(a*Sqrt[1 + c^2/a^2]))) )]*Sin[d + e*x - ArcTan[c/a]])/(a*Sqrt[1 + c^2/a^2]*Sqrt[(a*Sqrt[(a^2 + c^ 2)/a^2] - a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(b + a*Sqrt[ (a^2 + c^2)/a^2])]*Sqrt[b + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c /a]]]*Sqrt[(a*Sqrt[(a^2 + c^2)/a^2] + a*Sqrt[(a^2 + c^2)/a^2]*Cos[d + e*x - ArcTan[c/a]])/(-b + a*Sqrt[(a^2 + c^2)/a^2])])) - ((2*a*(b + a*Sqrt[1 + c^2/a^2]*Cos[d + e*x - ArcTan[c/a]]))/(a^2 + c^2) - (c*Sin[d + e*x - Ar...
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, 3646, 3042, 3598, 3042, 3132}
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.
| \(\displaystyle \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}}dx\) |
| \(\Big \downarrow \) 3646 |
| \(\displaystyle \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{\sqrt {\sec (d+e x)} \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx}{\sqrt {\sec (d+e x)} \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}\) |
| \(\Big \downarrow \) 3598 |
| \(\displaystyle \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 \sqrt {a+b \sec (d+e x)+c \tan (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\sec (d+e x)} \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\) |
(2*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[a + b*Sec[d + e*x] + c*Tan[d + e*x]])/(e*Sqrt[Sec[d + e*x]] *Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])])
3.5.49.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(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[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]]/Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])] Int[Sqrt[a/(a + S qrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - Arc Tan[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 higher order function than in optimal. Order 9 vs. order 4.
Time = 26.93 (sec) , antiderivative size = 2660, normalized size of antiderivative = 22.54
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(2660\) |
| default | \(\text {Expression too large to display}\) | \(12922\) |
I*(-I*exp(I*(e*x+d))^2*c+a*exp(I*(e*x+d))^2+2*b*exp(I*(e*x+d))+I*c+a)/e/(e xp(I*(e*x+d))*(-I*exp(I*(e*x+d))^2*c+a*exp(I*(e*x+d))^2+2*b*exp(I*(e*x+d)) +I*c+a))^(1/2)/(exp(RootOf(_Z^2+1,index=1)*(e*x+d))/(exp(RootOf(_Z^2+1,ind ex=1)*(e*x+d))^2+1))^(1/2)*(-(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,ind ex=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-2*b*exp(RootOf( _Z^2+1,index=1)*(e*x+d))-RootOf(_Z^2+1,index=1)*c-a)/(exp(RootOf(_Z^2+1,in dex=1)*(e*x+d))^2+1))^(1/2)*(-(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,in dex=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-2*b*exp(RootOf (_Z^2+1,index=1)*(e*x+d))-RootOf(_Z^2+1,index=1)*c-a)*(exp(RootOf(_Z^2+1,i ndex=1)*(e*x+d))^2+1))^(1/2)/(RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,ind ex=1)*(e*x+d))^2*c-exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a-2*b*exp(RootOf( _Z^2+1,index=1)*(e*x+d))-RootOf(_Z^2+1,index=1)*c-a)*(exp(RootOf(_Z^2+1,in dex=1)*(e*x+d))*(-RootOf(_Z^2+1,index=1)*exp(RootOf(_Z^2+1,index=1)*(e*x+d ))^2*c+exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*a+2*b*exp(RootOf(_Z^2+1,index =1)*(e*x+d))+RootOf(_Z^2+1,index=1)*c+a))^(1/2)/((-RootOf(_Z^2+1,index=1)* exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2*c+exp(RootOf(_Z^2+1,index=1)*(e*x+d) )^2*a+2*b*exp(RootOf(_Z^2+1,index=1)*(e*x+d))+RootOf(_Z^2+1,index=1)*c+a)* (exp(RootOf(_Z^2+1,index=1)*(e*x+d))^2+1))^(1/2)*2^(1/2)-I/e*(-2*b*(-b+(-a ^2+b^2-c^2)^(1/2))/(I*c-a)*((exp(I*(e*x+d))+(-b+(-a^2+b^2-c^2)^(1/2))/(I*c -a))/(-b+(-a^2+b^2-c^2)^(1/2))*(I*c-a))^(1/2)*((exp(I*(e*x+d))-(b+(-a^2...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 1367, normalized size of antiderivative = 11.58 \[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\text {Too large to display} \]
1/3*((-I*a*b + b*c)*sqrt(2*a - 2*I*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)) + (I* a*b + b*c)*sqrt(2*a + 2*I*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)*c os(e*x + d) - 3*(I*a^2 + I*c^2)*sin(e*x + d))/(a^2 + c^2)) - 3*(-I*a^2 - I *c^2)*sqrt(2*a - 2*I*c)*weierstrassZeta(-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), 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)/...
\[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\int \frac {\sqrt {a + b \sec {\left (d + e x \right )} + c \tan {\left (d + e x \right )}}}{\sqrt {\sec {\left (d + e x \right )}}}\, dx \]
\[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\int { \frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}{\sqrt {\sec \left (e x + d\right )}} \,d x } \]
\[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\int { \frac {\sqrt {b \sec \left (e x + d\right ) + c \tan \left (e x + d\right ) + a}}{\sqrt {\sec \left (e x + d\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \sec (d+e x)+c \tan (d+e x)}}{\sqrt {\sec (d+e x)}} \, dx=\int \frac {\sqrt {a+c\,\mathrm {tan}\left (d+e\,x\right )+\frac {b}{\cos \left (d+e\,x\right )}}}{\sqrt {\frac {1}{\cos \left (d+e\,x\right )}}} \,d x \]