\(\int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1-\cos (c+d x)}} \, dx\) [284]

Optimal result

Integrand size = 25, antiderivative size = 85 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.88 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1-\cos (c+d x)}} \, dx=-\frac {i \left (-1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\text {arcsinh}\left (e^{i (c+d x)}\right )-\sqrt {2} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)}} \] Input:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3256, 3042, 3254, 220, 3261, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3254

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

\(\Big \downarrow \) 220

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

\(\Big \downarrow \) 3261

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

\(\Big \downarrow \) 219

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

Input:

Int[Sqrt[Cos[c + d*x]]/Sqrt[1 - Cos[c + d*x]],x]
 

Output:

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

Defintions of rubi rules used

rule 219

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

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 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 [C] (warning: unable to verify)

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

Time = 3.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\cos \left (d x +c \right )}\, \left (-\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )\right ) \operatorname {csgn}\left (\sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(103\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (74) = 148\).

Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) + 2 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right )}{2 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (74) = 148\).

Time = 0.46 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.13 \[ \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {1-\cos (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (2 \, \sqrt {2} \log \left (\frac {2 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 2 \, \sqrt {2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}}{{\left | -2 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 4 \, \sqrt {2} + 2 \, \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} - 2 \right |}}\right ) - \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right ) + \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right ) + \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )\right )}}{2 \, d \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \] Input:

integrate(cos(d*x+c)^(1/2)/(1-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)))/(d*sgn(sin(1/2*d*x + 1/2*c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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