\(\int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx\) [344]

Optimal result

Integrand size = 25, antiderivative size = 92 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {a \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4710, 3042, 3249, 3042, 3253, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + d* 
x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d* 
Sqrt[a + a*Cos[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 3249

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 
2*n + 1)))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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] && GtQ[n, 0] && IntegerQ[2*n]
 

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 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 [F]

\[\int \frac {\sqrt {a +a \cos \left (d x +c \right )}}{\sqrt {\sec \left (d x +c \right )}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=-\frac {\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) + d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (78) = 156\).

Time = 0.31 (sec) , antiderivative size = 791, normalized size of antiderivative = 8.60 \[ \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(2*(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))*sin(d*x + 
c) - (cos(d*x + c) - 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
 + 1)))*sqrt(a) + sqrt(a)*(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)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 
 2*c), cos(2*d*x + 2*c) + 1))), (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(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2 
*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c) + 1))) + 1) - 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)*(cos(1/2*arctan2(sin(2*d*x + 2*c 
), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin( 
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (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(d*x + c)*cos(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d* 
x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) - 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*arctan2(sin(2*d*x 
+ 2*c), cos(2*d*x + 2*c) + 1)), (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)) + 1) + arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 +...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((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))/sec(c + d*x),x)