\(\int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) [551]

Optimal result

Integrand size = 29, antiderivative size = 84 \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}} \] Output:

1/2*c^(1/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b- 
1/2*c*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.10 \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\left (\cos (a+b x)-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))}+\cos (3 (a+b x))\right ) \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{4 b} \] Input:

Integrate[Cos[2*(a + b*x)]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

-1/4*((Cos[a + b*x] - Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a 
 + b*x)]]]*Sqrt[Cos[2*(a + b*x)]] + Cos[3*(a + b*x)])*Csc[a + b*x]*Sqrt[c* 
Tan[a + b*x]*Tan[2*(a + b*x)]])/b
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 4897, 3042, 4292, 3042, 4261, 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.

\(\displaystyle \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (2 a+2 b x) \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4292

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

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

\(\Big \downarrow \) 220

\(\displaystyle \frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\)

Input:

Int[Cos[2*(a + b*x)]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

(Sqrt[c]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]] 
)/(2*b) - (c*Sin[2*a + 2*b*x])/(2*b*Sqrt[-c + c*Sec[2*a + 2*b*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 4261

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4292

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a 
+ b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n))   Int[Sqrt[a + b*Csc 
[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && 
 EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
 

rule 4897

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(72)=144\).

Time = 1.20 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.19

method	result	size
default	\(-\frac {\sqrt {4}\, \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sin \left (b x +a \right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )}{2 b \left (\cos \left (b x +a \right )-1\right )}-\frac {\sqrt {2}\, \sqrt {4}\, \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \left (\cot \left (b x +a \right )+\csc \left (b x +a \right )\right )+\left (4 \cos \left (b x +a \right )^{2}-2\right ) \cot \left (b x +a \right )\right )}{8 b}\)	\(268\)

Input:

int(cos(2*b*x+2*a)*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/2/b*4^(1/2)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*sin(b*x+a)*((2*co 
s(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(2^(1/2)*cos(b*x+a)/(cos(b*x+ 
a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))/(cos(b*x+a)-1)-1/8*2^(1 
/2)/b*4^(1/2)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(2^(1/2)*arctanh(2 
^(1/2)*cos(b*x+a)/(cos(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/ 
2))*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(cot(b*x+a)+csc(b*x+a))+(4 
*cos(b*x+a)^2-2)*cot(b*x+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (72) = 144\).

Time = 0.12 (sec) , antiderivative size = 351, normalized size of antiderivative = 4.18 \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\left [\frac {{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{8 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac {{\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{4 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \] Input:

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

Output:

[1/8*((tan(b*x + a)^3 + tan(b*x + a))*sqrt(c)*log(-(c*tan(b*x + a)^5 - 14* 
c*tan(b*x + a)^3 + 4*sqrt(2)*(tan(b*x + a)^4 - 4*tan(b*x + a)^2 + 3)*sqrt( 
-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*sqrt(c) + 17*c*tan(b*x + a))/(tan( 
b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))) + 4*sqrt(2)*sqrt(-c*tan(b*x 
 + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^3 + b* 
tan(b*x + a)), -1/4*((tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*arctan(2*sqr 
t(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqr 
t(-c)/(c*tan(b*x + a)^3 - 3*c*tan(b*x + a))) - 2*sqrt(2)*sqrt(-c*tan(b*x + 
 a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^3 + b*ta 
n(b*x + a))]
 

Sympy [F(-1)]

Timed out. \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \] Input:

integrate(cos(2*b*x+2*a)*(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1049 vs. \(2 (72) = 144\).

Time = 0.24 (sec) , antiderivative size = 1049, normalized size of antiderivative = 12.49 \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*(4*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1) 
^(1/4)*(cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(2*b* 
x + 2*a) + (cos(2*b*x + 2*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4 
*b*x + 4*a) - 1)))*sqrt(c) - sqrt(c)*(log(sqrt(cos(4*b*x + 4*a)^2 + sin(4* 
b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(1/2*arctan2(sin(4*b*x + 4*a), - 
cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 
2*cos(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a 
) - 1))^2 + 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a 
) + 1)^(1/4)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1 
) - log(sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) 
+ 1)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sqrt(co 
s(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*ar 
ctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 - 2*(cos(4*b*x + 4*a)^2 
+ sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*sin(1/2*arctan2(sin(4 
*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) + log(((cos(2*b*x + 2*a)^2 + sin 
(2*b*x + 2*a)^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) 
^2 + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2)*sin(1/2*arctan2(sin(4*b*x + 
 4*a), -cos(4*b*x + 4*a) - 1))^2)*sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4* 
a)^2 + 2*cos(4*b*x + 4*a) + 1) + 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^ 
2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*(cos(1/2*arctan2(sin(4*b*x + 4*a), -c...
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int \cos \left (2\,a+2\,b\,x\right )\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos (2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \cos \left (2 b x +2 a \right )d x \right ) \] Input:

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

Output:

sqrt(c)*int(sqrt(tan(a + b*x))*sqrt(tan(2*a + 2*b*x))*cos(2*a + 2*b*x),x)