\(\int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx\) [264]

Optimal result

Integrand size = 26, antiderivative size = 85 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {a-a \cos (c+d x)}} \] Output:

a^(1/2)*arctanh(a^(1/2)*sin(d*x+c)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2) 
)/d-a*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a-a*cos(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.89 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.79 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=-\frac {i \left (\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}{d \left (-1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3042, 3249, 3042, 3254, 220}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3249

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

\(\displaystyle \frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}\)

Input:

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

Output:

(Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[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 220

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

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 3254

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(213\) vs. \(2(73)=146\).

Time = 8.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.52

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}+\csc \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )\right ) \sqrt {4}}{4 d \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(214\)

Input:

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

Output:

1/4*2^(1/2)/d*(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1 
/2)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c) 
+1)^2)^(1/2)*(cot(1/2*d*x+1/2*c)*(-2*cos(1/2*d*x+1/2*c)-2)*((2*cos(1/2*d*x 
+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)+csc(1/2*d*x+1/2*c)*2^(1/2)*ar 
ctanh(cos(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1 
)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*2^(1/2)))*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.67 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\frac {\sqrt {a} \log \left (-\frac {4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \sqrt {\cos \left (d x + c\right )} + {\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{4 \, d \sin \left (d x + c\right )} \] Input:

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

Output:

1/4*(sqrt(a)*log(-(4*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d 
*x + c) + 1)*sqrt(a)*sqrt(cos(d*x + c)) + (8*a*cos(d*x + c)^2 + 8*a*cos(d* 
x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*sqrt(-a*cos(d*x + 
 c) + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/(d*sin(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (73) = 146\).

Time = 0.29 (sec) , antiderivative size = 795, normalized size of antiderivative = 9.35 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(1/2)*(a-a*cos(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), c 
os(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(s 
in(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)^...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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