\(\int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx\) [361]

Optimal result

Integrand size = 32, antiderivative size = 88 \[ \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2}+\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )}}\right )}{\sqrt [4]{b^2+c^2} e} \] Output:

2^(1/2)*arctanh(1/2*(b^2+c^2)^(1/4)*sin(d+e*x-arctan(c,b))*2^(1/2)/((b^2+c 
^2)^(1/2)+(b^2+c^2)^(1/2)*cos(d+e*x-arctan(c,b)))^(1/2))/(b^2+c^2)^(1/4)/e
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 54.39 (sec) , antiderivative size = 63264, normalized size of antiderivative = 718.91 \[ \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\text {Result too large to show} \] Input:

Integrate[1/Sqrt[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x]],x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 3594, 3042, 3128, 219}

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.

\(\displaystyle \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}}dx\)

\(\Big \downarrow \) 3594

\(\displaystyle \int \frac {1}{\sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )+\sqrt {b^2+c^2}}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\sqrt {b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x+\frac {\pi }{2}\right )+\sqrt {b^2+c^2}}}dx\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {2 \int \frac {1}{2 \sqrt {b^2+c^2}-\frac {\left (b^2+c^2\right ) \sin ^2\left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )+\sqrt {b^2+c^2}}}d\left (-\frac {\sqrt {b^2+c^2} \sin \left (d+e x-\tan ^{-1}(b,c)\right )}{\sqrt {\sqrt {b^2+c^2} \cos \left (d+e x-\tan ^{-1}(b,c)\right )+\sqrt {b^2+c^2}}}\right )}{e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{b^2+c^2} \sin \left (-\tan ^{-1}(b,c)+d+e x\right )}{\sqrt {2} \sqrt {\sqrt {b^2+c^2} \cos \left (-\tan ^{-1}(b,c)+d+e x\right )+\sqrt {b^2+c^2}}}\right )}{e \sqrt [4]{b^2+c^2}}\)

Input:

Int[1/Sqrt[Sqrt[b^2 + c^2] + b*Cos[d + e*x] + c*Sin[d + e*x]],x]
 

Output:

(Sqrt[2]*ArcTanh[((b^2 + c^2)^(1/4)*Sin[d + e*x - ArcTan[b, c]])/(Sqrt[2]* 
Sqrt[Sqrt[b^2 + c^2] + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, c]]])])/((b 
^2 + c^2)^(1/4)*e)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3594

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( 
x_)]], x_Symbol] :> Int[1/Sqrt[a + Sqrt[b^2 + c^2]*Cos[d + e*x - ArcTan[b, 
c]]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(171\) vs. \(2(75)=150\).

Time = 1.89 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.95

method	result	size
default	\(-\frac {\left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )+1\right ) \sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-\sqrt {b^{2}+c^{2}}\, \left (\sin \left (e x +d -\arctan \left (-b , c\right )\right )-1\right )}\, \sqrt {2}}{2 \left (b^{2}+c^{2}\right )^{\frac {1}{4}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {1}{4}} \cos \left (e x +d -\arctan \left (-b , c\right )\right ) \sqrt {\frac {b^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+c^{2} \sin \left (e x +d -\arctan \left (-b , c\right )\right )+b^{2}+c^{2}}{\sqrt {b^{2}+c^{2}}}}\, e}\)	\(172\)

Input:

int(1/((b^2+c^2)^(1/2)+b*cos(e*x+d)+c*sin(e*x+d))^(1/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-(sin(e*x+d-arctan(-b,c))+1)*(-(b^2+c^2)^(1/2)*(sin(e*x+d-arctan(-b,c))-1) 
)^(1/2)*2^(1/2)/(b^2+c^2)^(1/4)*arctanh(1/2*(-(b^2+c^2)^(1/2)*(sin(e*x+d-a 
rctan(-b,c))-1))^(1/2)*2^(1/2)/(b^2+c^2)^(1/4))/cos(e*x+d-arctan(-b,c))/(( 
b^2*sin(e*x+d-arctan(-b,c))+c^2*sin(e*x+d-arctan(-b,c))+b^2+c^2)/(b^2+c^2) 
^(1/2))^(1/2)/e
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (81) = 162\).

Time = 0.24 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.97 \[ \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\frac {\sqrt {2} \log \left (\frac {{\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} + {\left (b^{2} c + 4 \, c^{3}\right )} \cos \left (e x + d\right ) - {\left (3 \, b^{3} + 4 \, b c^{2} + {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right ) + \frac {2 \, \sqrt {2} {\left (2 \, {\left (b^{3} + b c^{2}\right )} \cos \left (e x + d\right ) + 2 \, {\left (b^{2} c + c^{3}\right )} \sin \left (e x + d\right ) - {\left (2 \, b c \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} + b^{2} + 2 \, c^{2}\right )} \sqrt {b^{2} + c^{2}}\right )} \sqrt {b \cos \left (e x + d\right ) + c \sin \left (e x + d\right ) + \sqrt {b^{2} + c^{2}}}}{{\left (b^{2} + c^{2}\right )}^{\frac {1}{4}}} - 4 \, {\left (2 \, b c \cos \left (e x + d\right )^{2} - {\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) - b c\right )} \sqrt {b^{2} + c^{2}}}{3 \, b^{2} c \cos \left (e x + d\right ) - {\left (3 \, b^{2} c - c^{3}\right )} \cos \left (e x + d\right )^{3} - {\left (b^{3} - {\left (b^{3} - 3 \, b c^{2}\right )} \cos \left (e x + d\right )^{2}\right )} \sin \left (e x + d\right )}\right )}{2 \, {\left (b^{2} + c^{2}\right )}^{\frac {1}{4}} e} \] Input:

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

Output:

1/2*sqrt(2)*log(((3*b^2*c - c^3)*cos(e*x + d)^3 + (b^2*c + 4*c^3)*cos(e*x 
+ d) - (3*b^3 + 4*b*c^2 + (b^3 - 3*b*c^2)*cos(e*x + d)^2)*sin(e*x + d) + 2 
*sqrt(2)*(2*(b^3 + b*c^2)*cos(e*x + d) + 2*(b^2*c + c^3)*sin(e*x + d) - (2 
*b*c*cos(e*x + d)*sin(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + b^2 + 2*c^2) 
*sqrt(b^2 + c^2))*sqrt(b*cos(e*x + d) + c*sin(e*x + d) + sqrt(b^2 + c^2))/ 
(b^2 + c^2)^(1/4) - 4*(2*b*c*cos(e*x + d)^2 - (b^2 - c^2)*cos(e*x + d)*sin 
(e*x + d) - b*c)*sqrt(b^2 + c^2))/(3*b^2*c*cos(e*x + d) - (3*b^2*c - c^3)* 
cos(e*x + d)^3 - (b^3 - (b^3 - 3*b*c^2)*cos(e*x + d)^2)*sin(e*x + d)))/((b 
^2 + c^2)^(1/4)*e)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\sqrt {b^2+c^2}+b \cos (d+e x)+c \sin (d+e x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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