\(\int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx\) [277]

Optimal result

Integrand size = 26, antiderivative size = 107 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.86 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx=-\frac {i \left (-1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\text {arcsinh}\left (e^{i (c+d x)}\right )-\sqrt {2} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a-a \cos (c+d x)}} \] 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))] - 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)) 
]]))/(Sqrt[2]*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a - a*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3256, 3042, 3254, 220, 3261, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}\)

Input:

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

Output:

(2*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d 
*x]])])/(Sqrt[a]*d) - (Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqr 
t[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/(Sqrt[a]*d)
 

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 221

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

rule 3042

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

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]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(321\) vs. \(2(88)=176\).

Time = 6.82 (sec) , antiderivative size = 322, normalized size of antiderivative = 3.01

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )+\ln \left (\frac {2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )-\operatorname {arctanh}\left (\frac {-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\right )\right ) \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )-\csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {4}}{4 d \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}{\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}}\)	\(322\)

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)*(2*2^(1/2)*arctanh(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))+ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d 
*x+1/2*c)+1)^2)^(1/2)*cos(1/2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos( 
1/2*d*x+1/2*c)+1)^2)^(1/2)-2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)) 
-arctanh((-1+2*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)))/(a*sin(1/2*d*x+1/2*c)^2)^(1/ 
2)/((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)-csc(1/2*d*x+1/2*c))*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {a} \log \left (-\frac {2 \, \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )} + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{2 \, a d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (88) = 176\).

Time = 0.79 (sec) , antiderivative size = 269, normalized size of antiderivative = 2.51 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (2 \, \sqrt {2} \log \left (\frac {2 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 2 \, \sqrt {2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}}{{\left | -2 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 4 \, \sqrt {2} + 2 \, \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} - 2 \right |}}\right ) - \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right ) + \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right ) + \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )\right )}}{2 \, \sqrt {a} d \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

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

Output:

-1/2*sqrt(2)*(2*sqrt(2)*log(2*(tan(1/4*d*x + 1/4*c)^2 + 2*sqrt(2) - sqrt(t 
an(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1)/abs(-2*tan(1/4* 
d*x + 1/4*c)^2 + 4*sqrt(2) + 2*sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x 
 + 1/4*c)^2 + 1) - 2)) - log(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1 
/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1) + log(abs(-tan(1/4*d*x + 1/4* 
c)^2 + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 3)) + 
 log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4 
*d*x + 1/4*c)^2 + 1) + 1)))/(sqrt(a)*d*sgn(sin(1/2*d*x + 1/2*c)))
 

Mupad [F(-1)]

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

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