Integrand size = 31, antiderivative size = 182 \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{8 b \sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}}+\frac {\sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {\cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}} \]
7/8*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) )/b*2^(1/2)/c^(1/2)+1/8*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/4*c os(2*b*x+2*a)*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
Time = 3.98 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\tan (a+b x) \left (7 \text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right )-\sqrt {2} \left (7 \text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right )+\cos ^2(a+b x) \sec (2 (a+b x)) \sqrt {\frac {1}{1+\sec (2 (a+b x))}} \left (2 (1+\cos (2 (a+b x))+\cos (4 (a+b x)))+\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)}\right )\right )\right )}{8 b \sqrt {1-\tan ^2(a+b x)} \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \]
-1/8*(Tan[a + b*x]*(7*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]] - Sqrt[2]*(7*ArcTa nh[Sqrt[1 - Tan[a + b*x]^2]/Sqrt[2]] + Cos[a + b*x]^2*Sec[2*(a + b*x)]*Sqr t[(1 + Sec[2*(a + b*x)])^(-1)]*(2*(1 + Cos[2*(a + b*x)] + Cos[4*(a + b*x)] ) + ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-1 + Tan[a + b*x]^2]))))/(b*Sqr t[1 - Tan[a + b*x]^2]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])
Time = 1.00 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4897, 3042, 4310, 25, 3042, 4510, 27, 3042, 4408, 3042, 4261, 220, 4282, 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.
| \(\displaystyle \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (2 a+2 b x)^2}{\sqrt {c \tan (a+b x) \tan (2 a+2 b x)}}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cos ^2(2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2 \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx\) |
| \(\Big \downarrow \) 4310 |
| \(\displaystyle \frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {\int -\frac {\cos (2 a+2 b x) (3 \sec (2 a+2 b x) c+c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (2 a+2 b x) (3 \sec (2 a+2 b x) c+c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 \csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+c}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (2 a+2 b x) c^2+7 c^2}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (2 a+2 b x) c^2+7 c^2}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c^2+7 c^2}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle \frac {\frac {8 c^2 \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx-7 c \int \sqrt {c \sec (2 a+2 b x)-c}dx}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {8 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx-7 c \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {\frac {8 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {7 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {8 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {7 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle \frac {\frac {\frac {7 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {8 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {\frac {7 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {4 \sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{2 c}+\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}}{4 c}+\frac {\sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
(Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(4*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (((7*c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b* x]]])/b - (4*Sqrt[2]*c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*S qrt[-c + c*Sec[2*a + 2*b*x]])])/b)/(2*c) + (c*Sin[2*a + 2*b*x])/(2*b*Sqrt[ -c + c*Sec[2*a + 2*b*x]]))/(4*c)
3.7.25.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d *n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(988\) vs. \(2(157)=314\).
Time = 3.57 (sec) , antiderivative size = 989, normalized size of antiderivative = 5.43
1/8*2^(1/2)/b*sin(b*x+a)*(2*2^(1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/((2* cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2))+ln(2*(cos(b*x+a)*((2*cos( b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^ (1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))-arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+ a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)))/(1+cos(b*x+a))/(c*sin(b* x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/ 2)*4^(1/2)-1/4*2^(1/2)/b*sin(b*x+a)*(2*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^ 2)^(1/2)*cos(b*x+a)^2+3*2^(1/2)*arctanh(cos(b*x+a)/(1+cos(b*x+a))/((2*cos( b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2))+2*cos(b*x+a)*((2*cos(b*x+a)^2 -1)/(1+cos(b*x+a))^2)^(1/2)-2*arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2* cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2))+2*ln(2*(cos(b*x+a)*((2*cos(b*x+a) ^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)- 2*cos(b*x+a)-1)/(1+cos(b*x+a))))/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos (b*x+a))^2)^(1/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*4^(1/2)+1/32*2 ^(1/2)/b*sin(b*x+a)*(8*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x +a)^4+8*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^3+14*((2*co s(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^2+23*2^(1/2)*arctanh(cos( b*x+a)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*2^(1/2)) +14*cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)-16*arctanh((2*c os(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)...
Time = 0.28 (sec) , antiderivative size = 569, normalized size of antiderivative = 3.13 \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + 2 \, \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 ) + 7 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \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} {\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}}}{16 \, {\left (b c \tan \left (b x + a\right )^{5} + 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}, -\frac {4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + 2 \, \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 ) - 7 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \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} {\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}}}{8 \, {\left (b c \tan \left (b x + a\right )^{5} + 2 \, b c \tan \left (b x + a\right )^{3} + b c \tan \left (b x + a\right )\right )}}\right ] \]
[1/16*(4*sqrt(2)*(tan(b*x + a)^5 + 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)*sqrt(c) - 2*c*tan(b*x + a))/tan(b*x + a)^3) + 7*(tan(b *x + a)^5 + 2*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)*(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)))/(b*c*tan(b*x + a)^5 + 2*b*c*tan(b*x + a)^3 + b*c*tan(b*x + a)), -1/8*(4*sqrt(2)*(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))*s qrt(-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))) - 7*(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*arctan(1/2*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))) - 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)))/(b*c*tan(b*x + a)^5 + 2*b*c*tan(b*x + a)^3 + b*c*tan(b*x + a))]
Timed out. \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\cos \left (2 \, b x + 2 \, a\right )^{2}}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^2(2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {{\cos \left (2\,a+2\,b\,x\right )}^2}{\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]