\(\int \frac {1}{\sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx\) [365]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {2} \arcsin \left (\frac {\sin (c+d x)}{1+\cos (c+d x)}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 \arcsin \left (\frac {\sin (c+d x)}{\sqrt {1+\cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d} \] Output:

-2^(1/2)*arcsin(sin(d*x+c)/(1+cos(d*x+c)))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/ 
2)/d+2*arcsin(sin(d*x+c)/(1+cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c) 
^(1/2)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.85 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.82 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {i \sqrt {2} e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-\text {arcsinh}\left (e^{i (c+d x)}\right )+\sqrt {2} \text {arctanh}\left (\frac {-1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {1+\cos (c+d x)}} \] Input:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4710, 3042, 3256, 3042, 3253, 223, 3260, 223}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

\(\Big \downarrow \) 3260

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

(-((Sqrt[2]*ArcSin[Sin[c + d*x]/(1 + Cos[c + d*x])])/d) + (2*ArcSin[Sin[c 
+ d*x]/Sqrt[1 + Cos[c + d*x]]])/d)*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]
 

Defintions of rubi rules used

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 3260

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-Sqrt[2]/(Sqrt[a]*f)   Subst[Int[1/Sqrt[1 - 
x^2], x], x, b*(Cos[e + f*x]/(a + b*Sin[e + f*x]))], x] /; FreeQ[{a, b, d, 
e, f}, x] && EqQ[a^2 - b^2, 0] && EqQ[d, a/b] && GtQ[a, 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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 3.43 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {csgn}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{d \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(99\)

Input:

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

Output:

1/d*sec(1/2*d*x+1/2*c)*csgn(cos(1/2*d*x+1/2*c))*(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)))/sec(d*x 
+c)^(1/2)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, \arctan \left (\frac {\sqrt {\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right )}{d} \] Input:

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

Output:

(sqrt(2)*arctan(sqrt(2)*sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x 
+ c)) - 2*arctan(sqrt(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sin(d*x + c)))/ 
d
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 689, normalized size of antiderivative = 7.33 \[ \int \frac {1}{\sqrt {1+\cos (c+d x)} \sqrt {\sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-(sqrt(2)*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) + 1 
6)^(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*co 
s(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*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*cos(d*x + c) - 2)/abs(2*e^(I*d*x + I*c) + 2)) - arctan2((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*arc 
tan2(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*arcta 
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + cos(d*x + c)))/d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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)