\(\int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx\) [371]

Optimal result

Integrand size = 25, antiderivative size = 105 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {2 \arctan \left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d} \] Output:

2*arctan(a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(1/ 
2)/d-2^(1/2)*arctan(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*c 
os(d*x+c))^(1/2))/a^(1/2)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {i \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {a (1+\cos (c+d x))}} \] Input:

Integrate[1/(Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]
 

Output:

(I*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2* 
I)*(c + d*x))]*(-ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*ArcTanh[(-1 + E^(I*(c 
+ d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2 
*I)*(c + d*x))]])*Cos[(c + d*x)/2])/(d*E^((I/2)*(c + d*x))*Sqrt[a*(1 + Cos 
[c + d*x])])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4710, 3042, 3256, 3042, 3253, 223, 3261, 218}

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.

\(\displaystyle \int \frac {1}{\sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {\cos (c+d x) a+a}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\)

\(\Big \downarrow \) 3256

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{a}-\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{a d}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\right )\)

Input:

Int[1/(Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]),x]
 

Output:

((2*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d) - 
 (Sqrt[2]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a 
 + a*Cos[c + d*x]])])/(Sqrt[a]*d))*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 223

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

rule 3042

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

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3256

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

rule 3261

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

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a}\)	\(108\)

Input:

int(1/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*(2^(1/2)*arctan(tan(d*x+c)*(cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2))+arcsin(cot(d*x+c)-csc(d*x+c)))/(cos(d*x+c)+1)/sec 
(d*x+c)^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{a d} \] Input:

integrate(1/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

(sqrt(2)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c) 
)/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt 
(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a*d)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \sqrt {\sec {\left (c + d x \right )}}}\, dx \] Input:

integrate(1/(a+a*cos(d*x+c))**(1/2)/sec(d*x+c)**(1/2),x)
 

Output:

Integral(1/(sqrt(a*(cos(c + d*x) + 1))*sqrt(sec(c + d*x))), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.60 (sec) , antiderivative size = 698, normalized size of antiderivative = 6.65 \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x, algorithm="maxima")
 

Output:

-(sqrt(2)*sqrt(a)*arctan2(((abs(2*e^(I*d*x + I*c) + 2)^4 + 16*cos(d*x + c) 
^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + 
c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 32*(cos(d*x + c 
)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c)^2 - 64*cos(d*x 
+ c) + 16)^(1/4)*sin(1/2*arctan2(8*(cos(d*x + c) - 1)*sin(d*x + c)/abs(2*e 
^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*cos(d*x + c)^2 - 
4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I*c) + 2)^2)) + 2* 
sin(d*x + c))/abs(2*e^(I*d*x + I*c) + 2), ((abs(2*e^(I*d*x + I*c) + 2)^4 + 
 16*cos(d*x + c)^4 + 16*sin(d*x + c)^4 + 8*(cos(d*x + c)^2 - sin(d*x + c)^ 
2 - 2*cos(d*x + c) + 1)*abs(2*e^(I*d*x + I*c) + 2)^2 - 64*cos(d*x + c)^3 + 
 32*(cos(d*x + c)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 96*cos(d*x + c) 
^2 - 64*cos(d*x + c) + 16)^(1/4)*cos(1/2*arctan2(8*(cos(d*x + c) - 1)*sin( 
d*x + c)/abs(2*e^(I*d*x + I*c) + 2)^2, (abs(2*e^(I*d*x + I*c) + 2)^2 + 4*c 
os(d*x + c)^2 - 4*sin(d*x + c)^2 - 8*cos(d*x + c) + 4)/abs(2*e^(I*d*x + I* 
c) + 2)^2)) + 2*cos(d*x + c) - 2)/abs(2*e^(I*d*x + I*c) + 2)) - sqrt(a)*ar 
ctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^( 
1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + 
c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/ 
4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c) 
))/(a*d)
 

Giac [F]

\[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:

integrate(1/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(a*cos(d*x + c) + a)*sqrt(sec(d*x + c))), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(1/2)),x)
 

Output:

int(1/((1/cos(c + d*x))^(1/2)*(a + a*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )+\sec \left (d x +c \right )}d x \right )}{a} \] Input:

int(1/(a+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x)*sec 
(c + d*x) + sec(c + d*x)),x))/a