\(\int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx\) [525]

Optimal result

Integrand size = 25, antiderivative size = 79 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\frac {\operatorname {EllipticF}\left (e+f x+\tan ^{-1}(b,c),-\frac {b^2+c^2}{a}\right ) \sqrt {1+\frac {(c \cos (e+f x)+b \sin (e+f x))^2}{a}}}{f \sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \] Output:

InverseJacobiAM(e+f*x+arctan(c,b),(-(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)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 2.11 (sec) , antiderivative size = 529, normalized size of antiderivative = 6.70 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac {3}{2},\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}},\frac {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}\right ) \sec \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right ) \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (-1+\sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {-\frac {b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \left (1+\sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )\right )}{2 a+b^2+c^2-b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}}}} \sqrt {2 a+b^2+c^2+b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} \sin \left (2 (e+f x)+\arctan \left (\frac {-b^2+c^2}{2 b c}\right )\right )}}{b c \sqrt {\frac {\left (b^2+c^2\right )^2}{b^2 c^2}} f} \] Input:

Integrate[1/Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]
 

Output:

(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)
 

Rubi [A] (verified)

Time = 0.48 (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

\(\displaystyle \frac {\sqrt {\frac {(b \sin (e+f x)+c \cos (e+f x))^2}{a}+1} \operatorname {EllipticF}\left (e+f x+\tan ^{-1}(b,c),-\frac {b^2+c^2}{a}\right )}{f \sqrt {a+(b \sin (e+f x)+c \cos (e+f x))^2}}\)

Input:

Int[1/Sqrt[a + (c*Cos[e + f*x] + b*Sin[e + f*x])^2],x]
 

Output:

(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])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3661

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]
 

rule 3719

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]
 

rule 3720

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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2253\) vs. \(2(77)=154\).

Time = 4.95 (sec) , antiderivative size = 2254, normalized size of antiderivative = 28.53

method	result	size
default	\(\text {Expression too large to display}\)	\(2254\)

Input:

int(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/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=2)^2*cos(f*x+e)+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)^2-2*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)*((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)-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))/(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=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,in 
dex=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=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)-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))/(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,in 
dex=3)))^(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=4)*sin(f*x+e)+cos(f*x+e)-1)/(RootOf((c^2+a)*_Z^4-4*b...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 1623, normalized size of antiderivative = 20.54 \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="fricas")
 

Output:

-((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...
 

Sympy [F]

\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int \frac {1}{\sqrt {a + \left (b \sin {\left (e + f x \right )} + c \cos {\left (e + f x \right )}\right )^{2}}}\, dx \] Input:

integrate(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))**2)**(1/2),x)
 

Output:

Integral(1/sqrt(a + (b*sin(e + f*x) + c*cos(e + f*x))**2), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int { \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}} \,d x } \] Input:

integrate(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int { \frac {1}{\sqrt {{\left (c \cos \left (f x + e\right ) + b \sin \left (f x + e\right )\right )}^{2} + a}} \,d x } \] Input:

integrate(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/sqrt((c*cos(f*x + e) + b*sin(f*x + e))^2 + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int \frac {1}{\sqrt {a+{\left (c\,\cos \left (e+f\,x\right )+b\,\sin \left (e+f\,x\right )\right )}^2}} \,d x \] Input:

int(1/(a + (c*cos(e + f*x) + b*sin(e + f*x))^2)^(1/2),x)
 

Output:

int(1/(a + (c*cos(e + f*x) + b*sin(e + f*x))^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a+(c \cos (e+f x)+b \sin (e+f x))^2}} \, dx=\int \frac {\sqrt {\cos \left (f x +e \right )^{2} c^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b c +\sin \left (f x +e \right )^{2} b^{2}+a}}{\cos \left (f x +e \right )^{2} c^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b c +\sin \left (f x +e \right )^{2} b^{2}+a}d x \] Input:

int(1/(a+(c*cos(f*x+e)+b*sin(f*x+e))^2)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int(sqrt(cos(e + f*x)**2*c**2 + 2*cos(e + f*x)*sin(e + f*x)*b*c + sin(e + 
f*x)**2*b**2 + a)/(cos(e + f*x)**2*c**2 + 2*cos(e + f*x)*sin(e + f*x)*b*c 
+ sin(e + f*x)**2*b**2 + a),x)