Integrand size = 25, antiderivative size = 79 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\frac {\operatorname {EllipticF}\left (e+f x+\tan ^{-1}(b,c),-\frac {b^2+c^2}{a}\right ) \sqrt {1+\frac {(c \cos (e+f x)+b \sin (e+f x))^2}{a}}}{f \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \]
(cos(e+f*x+arctan(b,c))^2)^(1/2)/cos(e+f*x+arctan(b,c))*EllipticF(sin(e+f* x+arctan(b,c)),((-b^2-c^2)/a)^(1/2))*(1+(c*cos(f*x+e)+b*sin(f*x+e))^2/a)^( 1/2)/f/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 1.86 (sec) , antiderivative size = 529, normalized size of antiderivative = 6.70 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {3}{2},\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}},\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}\right ) \sec \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right ) \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (-1+\sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (1+\sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}}{b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} f} \]
(Sqrt[2]*AppellF1[1/2, 1/2, 1/2, 3/2, (2*a + b^2 + c^2 + b*c*Sqrt[(b^2 + c ^2)^2/(b^2*c^2)]*Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]])/(2*a + b ^2 + c^2 - b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]), (2*a + b^2 + c^2 + b*c*Sqrt [(b^2 + c^2)^2/(b^2*c^2)]*Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]) /(2*a + b^2 + c^2 + b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)])]*Sec[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]*Sqrt[-((b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*(- 1 + Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]))/(2*a + b^2 + c^2 + b *c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]))]*Sqrt[-((b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^ 2)]*(1 + Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]))/(2*a + b^2 + c^ 2 - b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]))]*Sqrt[2*a + b^2 + c^2 + b*c*Sqrt[( b^2 + c^2)^2/(b^2*c^2)]*Sin[2*(e + f*x) + ArcTan[(-b^2 + c^2)/(2*b*c)]]])/ (b*c*Sqrt[(b^2 + c^2)^2/(b^2*c^2)]*f)
Time = 0.46 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3720, 3042, 3719, 3042, 3661}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}dx\) |
\(\Big \downarrow \) 3720 |
\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \int \frac {1}{\sqrt {\frac {(c \cos (e+f x)+b \sin (e+f x))^2}{a}+1}}dx}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \int \frac {1}{\sqrt {\frac {(c \cos (e+f x)+b \sin (e+f x))^2}{a}+1}}dx}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\) |
\(\Big \downarrow \) 3719 |
\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \int \frac {1}{\sqrt {\frac {\left (b^2+c^2\right ) \sin ^2\left (e+f x+\tan ^{-1}(b,c)\right )}{a}+1}}dx}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \int \frac {1}{\sqrt {\frac {\left (b^2+c^2\right ) \sin \left (e+f x+\tan ^{-1}(b,c)\right )^2}{a}+1}}dx}{\sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\) |
\(\Big \downarrow \) 3661 |
\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \operatorname {EllipticF}\left (e+f x+\tan ^{-1}(b,c),-\frac {b^2+c^2}{a}\right )}{f \sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\) |
(EllipticF[e + f*x + ArcTan[b, c], -((b^2 + c^2)/a)]*Sqrt[1 + (c*Cos[e + f *x] + b*Sin[e + f*x])^2/a])/(f*Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^ 2])
3.6.94.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
Int[((a_) + (b_.)*(cos[(e_.) + (f_.)*(x_)]*(d_.) + (c_.)*sin[(e_.) + (f_.)* (x_)])^2)^(p_), x_Symbol] :> Int[(a + b*(Sqrt[c^2 + d^2]*Sin[ArcTan[c, d] + e + f*x])^2)^p, x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[p^2, 1/4] && GtQ [a, 0]
Int[((a_) + (b_.)*(cos[(e_.) + (f_.)*(x_)]*(d_.) + (c_.)*sin[(e_.) + (f_.)* (x_)])^2)^(p_), x_Symbol] :> Simp[(a + b*(c*Sin[e + f*x] + d*Cos[e + f*x])^ 2)^p/(1 + (b*(c*Sin[e + f*x] + d*Cos[e + f*x])^2)/a)^p Int[(1 + (b*(c*Sin [e + f*x] + d*Cos[e + f*x])^2)/a)^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[p^2, 1/4] && !GtQ[a, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 6.64 (sec) , antiderivative size = 2174, normalized size of antiderivative = 27.52
-4/f*((RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+ a,index=1)*sin(f*x+e)+cos(f*x+e)-1)/(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2 -2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=2)*sin(f*x+e)+cos(f*x+e)-1)*(RootOf( (c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=2)-Roo tOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=4) )/(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,in dex=1)-RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+ a,index=4)))^(1/2)*(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+ 4*_Z*b*c+c^2+a,index=2)*sin(f*x+e)+cos(f*x+e)-1)^2*(-(RootOf((c^2+a)*_Z^4- 4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=2)-RootOf((c^2+a)*_ Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=1))*(RootOf((c^ 2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=3)*sin(f* x+e)+cos(f*x+e)-1)/(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+ 4*_Z*b*c+c^2+a,index=1)-RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_ Z^2+4*_Z*b*c+c^2+a,index=3))/(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+ 2*a)*_Z^2+4*_Z*b*c+c^2+a,index=2)*sin(f*x+e)+cos(f*x+e)-1))^(1/2)*(-(RootO f((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index=2)-R ootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a,index= 1))*(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^2-2*c^2+2*a)*_Z^2+4*_Z*b*c+c^2+a, index=4)*sin(f*x+e)+cos(f*x+e)-1)/(RootOf((c^2+a)*_Z^4-4*b*c*_Z^3+(4*b^...
Result contains complex when optimal does not.
Time = 0.14 (sec) , antiderivative size = 1623, normalized size of antiderivative = 20.54 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\text {Too large to display} \]
-((b^3 + b*c^2 - I*c^3 + 2*a*b - I*(b^2 + 2*a)*c - 2*(b^3 - 3*I*b^2*c - 3* b*c^2 + I*c^3)*sqrt((a*b^6 - 4*I*a*b*c^5 + a*c^6 + a^2*b^4 - 4*I*a^2*b*c^3 - (5*a*b^2 - a^2)*c^4 - (5*a*b^4 + 6*a^2*b^2)*c^2 + 4*I*(a*b^5 + a^2*b^3) *c)/(b^8 + 4*b^6*c^2 + 6*b^4*c^4 + 4*b^2*c^6 + c^8)))*sqrt((b^4 + 2*I*b*c^ 3 - c^4 + 2*a*b^2 - 2*a*c^2 + 2*I*(b^3 + 2*a*b)*c + 2*(b^4 + 2*b^2*c^2 + c ^4)*sqrt((a*b^6 - 4*I*a*b*c^5 + a*c^6 + a^2*b^4 - 4*I*a^2*b*c^3 - (5*a*b^2 - a^2)*c^4 - (5*a*b^4 + 6*a^2*b^2)*c^2 + 4*I*(a*b^5 + a^2*b^3)*c)/(b^8 + 4*b^6*c^2 + 6*b^4*c^4 + 4*b^2*c^6 + c^8)))/(b^4 + 2*b^2*c^2 + c^4))*ellipt ic_f(arcsin(sqrt((b^4 + 2*I*b*c^3 - c^4 + 2*a*b^2 - 2*a*c^2 + 2*I*(b^3 + 2 *a*b)*c + 2*(b^4 + 2*b^2*c^2 + c^4)*sqrt((a*b^6 - 4*I*a*b*c^5 + a*c^6 + a^ 2*b^4 - 4*I*a^2*b*c^3 - (5*a*b^2 - a^2)*c^4 - (5*a*b^4 + 6*a^2*b^2)*c^2 + 4*I*(a*b^5 + a^2*b^3)*c)/(b^8 + 4*b^6*c^2 + 6*b^4*c^4 + 4*b^2*c^6 + c^8))) /(b^4 + 2*b^2*c^2 + c^4))*(cos(f*x + e) + I*sin(f*x + e))), (b^4 + c^4 + 8 *a*b^2 + 2*(b^2 + 4*a)*c^2 + 8*a^2 - 4*(b^4 - 2*I*b*c^3 - c^4 + 2*a*b^2 - 2*a*c^2 - 2*I*(b^3 + 2*a*b)*c)*sqrt((a*b^6 - 4*I*a*b*c^5 + a*c^6 + a^2*b^4 - 4*I*a^2*b*c^3 - (5*a*b^2 - a^2)*c^4 - (5*a*b^4 + 6*a^2*b^2)*c^2 + 4*I*( a*b^5 + a^2*b^3)*c)/(b^8 + 4*b^6*c^2 + 6*b^4*c^4 + 4*b^2*c^6 + c^8)))/(b^4 + 2*b^2*c^2 + c^4)) + (b^3 + b*c^2 + I*c^3 + 2*a*b + I*(b^2 + 2*a)*c - 2* (b^3 + 3*I*b^2*c - 3*b*c^2 - I*c^3)*sqrt((a*b^6 + 4*I*a*b*c^5 + a*c^6 + a^ 2*b^4 + 4*I*a^2*b*c^3 - (5*a*b^2 - a^2)*c^4 - (5*a*b^4 + 6*a^2*b^2)*c^2...
\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int \frac {1}{\sqrt {a + \left (b \sin {\left (e + f x \right )} + c \cos {\left (e + f x \right )}\right )^{2}}}\, dx \]
\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int { \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}} \,d x } \]
\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int { \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int \frac {1}{\sqrt {a+{\left (c\,\cos \left (e+f\,x\right )+b\,\sin \left (e+f\,x\right )\right )}^2}} \,d x \]