\(\int \frac {\cos (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) [567]

Optimal result

Integrand size = 29, antiderivative size = 138 \[ \int \frac {\cos (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b \sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.18 \[ \int \frac {\cos (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\tan (a+b x) \left (\text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right )-\sqrt {2} \left (\text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right )+\cos ^2(a+b x) \sqrt {\frac {1}{1+\sec (2 (a+b x))}} \left (2+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sec (2 (a+b x)) \sqrt {-1+\tan ^2(a+b x)}\right )\right )\right )}{2 b \sqrt {1-\tan ^2(a+b x)} \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \] Input:

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

Output:

-1/2*(Tan[a + b*x]*(ArcTanh[Sqrt[1 - Tan[a + b*x]^2]] - Sqrt[2]*(ArcTanh[S 
qrt[1 - Tan[a + b*x]^2]/Sqrt[2]] + Cos[a + b*x]^2*Sqrt[(1 + Sec[2*(a + b*x 
)])^(-1)]*(2 + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sec[2*(a + b*x)]*Sqrt[-1 
+ Tan[a + b*x]^2]))))/(b*Sqrt[1 - Tan[a + b*x]^2]*Sqrt[c*Tan[a + b*x]*Tan[ 
2*(a + b*x)]])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {3042, 4897, 3042, 4310, 25, 3042, 4392, 3042, 4375, 383, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4897

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4310

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4392

\(\displaystyle \frac {1}{2} c \int \frac {\tan ^2(2 a+2 b x)}{(c \sec (2 a+2 b x)-c)^{3/2}}dx+\frac {\sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} c \int \frac {\cot \left (2 a+2 b x+\frac {\pi }{2}\right )^2}{\left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}dx+\frac {\sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 383

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {c \left (\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{c^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{c^{3/2}}\right )}{2 b}\)

Input:

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

Output:

-1/2*(c*(-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x] 
]]/c^(3/2)) + (Sqrt[2]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c 
 + c*Sec[2*a + 2*b*x]])])/c^(3/2)))/b + Sin[2*a + 2*b*x]/(2*b*Sqrt[-c + c* 
Sec[2*a + 2*b*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 383

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_Sym 
bol] :> Simp[(-a)*(e^2/(b*c - a*d))   Int[(e*x)^(m - 2)/(a + b*x^2), x], x] 
 + Simp[c*(e^2/(b*c - a*d))   Int[(e*x)^(m - 2)/(c + d*x^2), x], x] /; Free 
Q[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LeQ[2, m, 3]
 

rule 3042

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

rule 4310

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

rule 4375

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

rule 4392

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m   Int[Cot[e + f*x]^(2*m)*( 
c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E 
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !( 
IntegerQ[n] && GtQ[m - n, 0])
 

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. \(562\) vs. \(2(117)=234\).

Time = 1.07 (sec) , antiderivative size = 563, normalized size of antiderivative = 4.08

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {4}\, \sin \left (b x +a \right ) \left (2 \ln \left (\frac {2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )+2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-4 \cos \left (b x +a \right )-2}{\cos \left (b x +a \right )+1}\right )-2 \,\operatorname {arctanh}\left (\frac {2 \cos \left (b x +a \right )-1}{\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 )+3 \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 )+\cos \left (b x +a \right ) \left (2 \cos \left (b x +a \right )+2\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\right )}{8 b \left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}+\frac {\sqrt {2}\, \sqrt {4}\, \left (2 \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 )+\ln \left (\frac {2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \cos \left (b x +a \right )+2 \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}-4 \cos \left (b x +a \right )-2}{\cos \left (b x +a \right )+1}\right )-\operatorname {arctanh}\left (\frac {2 \cos \left (b x +a \right )-1}{\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 )\right ) \left (\cot \left (b x +a \right )-\csc \left (b x +a \right )\right )}{8 b \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\)	\(563\)

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:

1/8*2^(1/2)/b*4^(1/2)*sin(b*x+a)*(2*ln(2*(((2*cos(b*x+a)^2-1)/(cos(b*x+a)+ 
1)^2)^(1/2)*cos(b*x+a)+((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b 
*x+a)-1)/(cos(b*x+a)+1))-2*arctanh((2*cos(b*x+a)-1)/(cos(b*x+a)+1)/((2*cos 
(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))+3*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)*(2 
*cos(b*x+a)+2)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))/(cos(b*x+a)+1) 
/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1)/(cos(b*x+a) 
+1)^2)^(1/2)+1/8*2^(1/2)/b*4^(1/2)*(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))+ln(2*(((2*cos(b 
*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)+((2*cos(b*x+a)^2-1)/(cos(b*x 
+a)+1)^2)^(1/2)-2*cos(b*x+a)-1)/(cos(b*x+a)+1))-arctanh((2*cos(b*x+a)-1)/( 
cos(b*x+a)+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)))/(c*sin(b*x+a)^ 
2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(c 
ot(b*x+a)-csc(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 481, normalized size of antiderivative = 3.49 \[ \int \frac {\cos (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {\sqrt {2} {\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 )^{3} - 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} - 2 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right ) + {\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 )^{3} + 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} - 3 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3} + \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 c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac {\sqrt {2} {\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {\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 )}\right ) - {\left (\tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {\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}}{2 \, c \tan \left (b x + a\right )}\right ) + \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 )}}{2 \, {\left (b c \tan \left (b x + a\right )^{3} + b c \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:

[1/4*(sqrt(2)*(tan(b*x + a)^3 + tan(b*x + a))*sqrt(c)*log((c*tan(b*x + a)^ 
3 - 2*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sq 
rt(c) - 2*c*tan(b*x + a))/tan(b*x + a)^3) + (tan(b*x + a)^3 + tan(b*x + a) 
)*sqrt(c)*log((c*tan(b*x + a)^3 + 2*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b* 
x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(c) - 3*c*tan(b*x + a))/(tan(b*x + 
 a)^3 + 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*c*tan(b*x + a)^3 + b*c*tan(b*x + a)), -1/2* 
(sqrt(2)*(tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*arctan(sqrt(-c*tan(b*x + 
 a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b*x + a)) 
) - (tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-c*ta 
n(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-c)/(c*tan(b* 
x + a))) + sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + 
 a)^2 - 1))/(b*c*tan(b*x + a)^3 + b*c*tan(b*x + a))]
 

Sympy [F(-1)]

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

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

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

Giac [F(-2)]

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

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