\(\int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) [553]

Optimal result

Integrand size = 31, antiderivative size = 176 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {5 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{16 b}-\frac {5 c \sin (2 a+2 b x)}{16 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {5 c \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt {-c+c \sec (2 a+2 b x)}} \] Output:

5/16*c^(1/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b 
-5/16*c*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+5/24*c*cos(2*b*x+2*a) 
*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-1/6*c*cos(2*b*x+2*a)^2*sin(2 
*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.66 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\left (-26 \cot (a+b x)+15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x)+30 \sin (2 (a+b x))-2 \sin (4 (a+b x))+4 \sin (6 (a+b x))\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{96 b} \] Input:

Integrate[Cos[2*(a + b*x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

((-26*Cot[a + b*x] + 15*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2* 
(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]]*Csc[a + b*x] + 30*Sin[2*(a + b*x)] - 2 
*Sin[4*(a + b*x)] + 4*Sin[6*(a + b*x)])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x 
)]])/(96*b)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4897, 3042, 4292, 3042, 4292, 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.

\(\displaystyle \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (2 a+2 b x)^3 \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int \cos ^3(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4292

\(\displaystyle -\frac {5}{6} \int \cos ^2(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5}{6} \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2}dx-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4292

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \int \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}dx-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4292

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \left (-\frac {1}{2} \int \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \left (-\frac {1}{2} \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 4261

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \left (\frac {c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\)

Input:

Int[Cos[2*(a + b*x)]^3*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
 

Output:

-1/6*(c*Cos[2*a + 2*b*x]^2*Sin[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b* 
x]]) - (5*(-1/4*(c*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2 
*a + 2*b*x]]) - (3*((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]])))/4))/6
 

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. \(595\) vs. \(2(156)=312\).

Time = 1.76 (sec) , antiderivative size = 596, normalized size of antiderivative = 3.39

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 {3 \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}+\frac {3 \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}\, \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 ) \left (3 \cot \left (b x +a \right )+3 \csc \left (b x +a \right )\right )+\left (16 \cos \left (b x +a \right )^{4}+4 \cos \left (b x +a \right )^{2}-6\right ) \cot \left (b x +a \right )\right )}{32 b}-\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}\, \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 ) \left (15 \cot \left (b x +a \right )+15 \csc \left (b x +a \right )\right )+\left (128 \cos \left (b x +a \right )^{6}+16 \cos \left (b x +a \right )^{4}+20 \cos \left (b x +a \right )^{2}-30\right ) \cot \left (b x +a \right )\right )}{192 b}\)	\(596\)

Input:

int(cos(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x,method=_RETURNV 
ERBOSE)
 

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)-3/8*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))+3/32*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)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*a 
rctanh(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))*(3*cot(b*x+a)+3*csc(b*x+a))+(16*cos(b*x+a)^4+4*cos(b*x+a)^2-6) 
*cot(b*x+a))-1/192*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)*((2*cos(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))*(15 
*cot(b*x+a)+15*csc(b*x+a))+(128*cos(b*x+a)^6+16*cos(b*x+a)^4+20*cos(b*x+a) 
^2-30)*cot(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.73 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\left [\frac {15 \, {\left (\tan \left (b x + a\right )^{7} + 3 \, \tan \left (b x + a\right )^{5} + 3 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} {\left (33 \, \tan \left (b x + a\right )^{6} - 19 \, \tan \left (b x + a\right )^{4} - \tan \left (b x + a\right )^{2} - 13\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{192 \, {\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac {15 \, {\left (\tan \left (b x + a\right )^{7} + 3 \, \tan \left (b x + a\right )^{5} + 3 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (33 \, \tan \left (b x + a\right )^{6} - 19 \, \tan \left (b x + a\right )^{4} - \tan \left (b x + a\right )^{2} - 13\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{96 \, {\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \] Input:

integrate(cos(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="fricas")
 

Output:

[1/192*(15*(tan(b*x + a)^7 + 3*tan(b*x + a)^5 + 3*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)*(t 
an(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)*(33*tan(b*x + a)^6 - 19*tan(b*x + a)^4 - tan(b 
*x + a)^2 - 13)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b*x + 
 a)^7 + 3*b*tan(b*x + a)^5 + 3*b*tan(b*x + a)^3 + b*tan(b*x + a)), -1/96*( 
15*(tan(b*x + a)^7 + 3*tan(b*x + a)^5 + 3*tan(b*x + a)^3 + tan(b*x + a))*s 
qrt(-c)*arctan(2*sqrt(2)*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)^3 - 3*c*tan(b*x + a))) - 2*sqrt( 
2)*(33*tan(b*x + a)^6 - 19*tan(b*x + a)^4 - tan(b*x + a)^2 - 13)*sqrt(-c*t 
an(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b*x + a)^7 + 3*b*tan(b*x + a)^ 
5 + 3*b*tan(b*x + a)^3 + b*tan(b*x + a))]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \] Input:

integrate(cos(2*b*x+2*a)**3*(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. 2333 vs. \(2 (156) = 312\).

Time = 0.59 (sec) , antiderivative size = 2333, normalized size of antiderivative = 13.26 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Too large to display} \] Input:

integrate(cos(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="maxima")
 

Output:

-1/384*(8*(cos(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a)))^2 + sin(2/ 
3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a)))^2 + 2*cos(2/3*arctan2(sin(6 
*b*x + 6*a), cos(6*b*x + 6*a))) + 1)^(3/4)*(cos(3/2*arctan2(sin(2/3*arctan 
2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))), -cos(2/3*arctan2(sin(6*b*x + 6*a), 
 cos(6*b*x + 6*a))) - 1))*sin(6*b*x + 6*a) + (cos(6*b*x + 6*a) + 1)*sin(3/ 
2*arctan2(sin(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))), -cos(2/3*a 
rctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))) - 1)))*sqrt(c) + 12*(cos(2/3*a 
rctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a)))^2 + sin(2/3*arctan2(sin(6*b*x 
+ 6*a), cos(6*b*x + 6*a)))^2 + 2*cos(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b 
*x + 6*a))) + 1)^(1/4)*((sin(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a 
))) - 5*sin(1/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))))*cos(1/2*arct 
an2(sin(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))), -cos(2/3*arctan2 
(sin(6*b*x + 6*a), cos(6*b*x + 6*a))) - 1)) + (cos(2/3*arctan2(sin(6*b*x + 
 6*a), cos(6*b*x + 6*a))) - 3*cos(1/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x 
+ 6*a))) - 4)*sin(1/2*arctan2(sin(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x 
+ 6*a))), -cos(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))) - 1)))*sqr 
t(c) + 15*sqrt(c)*(log(sqrt(cos(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 
6*a)))^2 + sin(2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a)))^2 + 2*cos( 
2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))) + 1)*cos(1/2*arctan2(sin( 
2/3*arctan2(sin(6*b*x + 6*a), cos(6*b*x + 6*a))), -cos(2/3*arctan2(sin(...
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(2*b*x+2*a)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorith 
m="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. \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int {\cos \left (2\,a+2\,b\,x\right )}^3\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \] Input:

int(cos(2*a + 2*b*x)^3*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2),x)
 

Output:

int(cos(2*a + 2*b*x)^3*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2), x)
 

Reduce [F]

\[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}\, \cos \left (2 b x +2 a \right )^{3}d x \right ) \] Input:

int(cos(2*b*x+2*a)^3*(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)**3, 
x)