Integrand size = 29, antiderivative size = 178 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b c^{3/2}}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{4 \sqrt {2} b c^{3/2}}-\frac {\sin (2 a+2 b x)}{4 b (-c+c \sec (2 a+2 b x))^{3/2}}-\frac {3 \sin (2 a+2 b x)}{4 b c \sqrt {-c+c \sec (2 a+2 b x)}} \]
-3/2*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b/c^(3/2) -1/4*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(3/2)+9/8*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/c^(3/2)*2^(1/2)-3/4* sin(2*b*x+2*a)/b/c/(-c+c*sec(2*b*x+2*a))^(1/2)
Time = 5.86 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\frac {\csc (a+b x) \sec (a+b x) \left (6 \text {arctanh}\left (\sqrt {1-\tan ^2(a+b x)}\right ) \cos (2 (a+b x))-6 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\tan ^2(a+b x)}}{\sqrt {2}}\right ) \cos (2 (a+b x))+(1-3 \cos (2 (a+b x))+\cos (4 (a+b x))) \cot ^2(a+b x) \sqrt {\cos (2 (a+b x)) \sec ^2(a+b x)}-3 \arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \cos ^2(a+b x) \sqrt {-\left (-1+\tan ^2(a+b x)\right )^2}\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{8 b c^2 \sqrt {1-\tan ^2(a+b x)}} \]
(Csc[a + b*x]*Sec[a + b*x]*(6*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]]*Cos[2*(a + b*x)] - 6*Sqrt[2]*ArcTanh[Sqrt[1 - Tan[a + b*x]^2]/Sqrt[2]]*Cos[2*(a + b* x)] + (1 - 3*Cos[2*(a + b*x)] + Cos[4*(a + b*x)])*Cot[a + b*x]^2*Sqrt[Cos[ 2*(a + b*x)]*Sec[a + b*x]^2] - 3*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Cos[a + b*x]^2*Sqrt[-(-1 + Tan[a + b*x]^2)^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x )]])/(8*b*c^2*Sqrt[1 - Tan[a + b*x]^2])
Time = 0.95 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.483, Rules used = {3042, 4897, 3042, 4304, 27, 3042, 4510, 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.
| \(\displaystyle \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (2 a+2 b x)}{(c \tan (a+b x) \tan (2 a+2 b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \frac {\cos (2 a+2 b x)}{(c \sec (2 a+2 b x)-c)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4304 |
| \(\displaystyle -\frac {\int \frac {3 \cos (2 a+2 b x) (\sec (2 a+2 b x) c+2 c)}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx}{2 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 \int \frac {\cos (2 a+2 b x) (\sec (2 a+2 b x) c+2 c)}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c+2 c}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4510 |
| \(\displaystyle -\frac {3 \left (\frac {\int \frac {\sec (2 a+2 b x) c^2+2 c^2}{\sqrt {c \sec (2 a+2 b x)-c}}dx}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {\int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right ) c^2+2 c^2}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4408 |
| \(\displaystyle -\frac {3 \left (\frac {3 c^2 \int \frac {\sec (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}dx-2 c \int \sqrt {c \sec (2 a+2 b x)-c}dx}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 \left (\frac {3 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx-2 c \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle -\frac {3 \left (\frac {3 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {2 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {3 \left (\frac {3 c^2 \int \frac {\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}{\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}dx+\frac {2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 4282 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {3 c^2 \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-2 c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {3 \left (\frac {\frac {2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}-\frac {3 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b}}{c}+\frac {c \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\right )}{4 c^2}-\frac {\sin (2 a+2 b x)}{4 b (c \sec (2 a+2 b x)-c)^{3/2}}\) |
-1/4*Sin[2*a + 2*b*x]/(b*(-c + c*Sec[2*a + 2*b*x])^(3/2)) - (3*(((2*c^(3/2 )*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x]]])/b - ( 3*c^(3/2)*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])])/(Sqrt[2]*b))/c + (c*Sin[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b*x]])))/(4*c^2)
3.7.31.3.1 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_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc [e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ [m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
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. \(1074\) vs. \(2(151)=302\).
Time = 2.96 (sec) , antiderivative size = 1075, normalized size of antiderivative = 6.04
-1/64*2^(1/2)/b*csc(b*x+a)/(csc(b*x+a)^2*(1-cos(b*x+a))^2*c/(csc(b*x+a)^4* (1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1))^(3/2)*(1-cos(b*x+a))* (16*csc(b*x+a)^2*2^(1/2)*arctanh((csc(b*x+a)^2*(1-cos(b*x+a))^2-1)*2^(1/2) /(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))* (1-cos(b*x+a))^2-10*csc(b*x+a)^2*ln(csc(b*x+a)^2*(1-cos(b*x+a))^2-3+(csc(b *x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-cos( b*x+a))^2-10*csc(b*x+a)^2*arctanh((3*csc(b*x+a)^2*(1-cos(b*x+a))^2-1)/(csc (b*x+a)^4*(1-cos(b*x+a))^4-6*csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))*(1-co s(b*x+a))^2+csc(b*x+a)^2*(1-cos(b*x+a))^2*(csc(b*x+a)^4*(1-cos(b*x+a))^4-6 *csc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2)-(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*c sc(b*x+a)^2*(1-cos(b*x+a))^2+1)^(1/2))/(csc(b*x+a)^4*(1-cos(b*x+a))^4-6*cs c(b*x+a)^2*(1-cos(b*x+a))^2+1)^(3/2)*4^(1/2)+1/16*2^(1/2)/b*csc(b*x+a)*(4* ((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)^3+10*2^(1/2)*arctan h(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))*cos(b*x+a)-10*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))-6*cos(b*x+a)*((2*cos(b*x+a)^2-1) /(1+cos(b*x+a))^2)^(1/2)-7*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))*cos(b*x+a)+7*ln(2*(cos(b*x+a)*((2*co s(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)))*cos(b*x+a)+7*arctanh((2*cos(b*x...
Time = 0.29 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.97 \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\left [\frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 ) + 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}, \frac {9 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 ) - 12 \, {\left (\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}\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 (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{8 \, {\left (b c^{2} \tan \left (b x + a\right )^{5} + b c^{2} \tan \left (b x + a\right )^{3}\right )}}\right ] \]
[1/16*(9*sqrt(2)*(tan(b*x + a)^5 + tan(b*x + a)^3)*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) + 12*(tan(b*x + a)^5 + tan( b*x + a)^3)*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)*(5*tan(b*x + a)^4 - 4*tan(b*x + a)^2 - 1)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c^2*tan(b*x + a)^5 + b*c^2*tan(b*x + a)^3), 1/8*(9*sqrt(2)*(tan(b*x + a)^5 + tan(b*x + a)^3)*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))) - 12*(tan(b*x + a)^5 + tan(b*x + a)^3)*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)*(5*tan(b*x + a)^4 - 4*tan(b*x + a)^2 - 1)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1 )))/(b*c^2*tan(b*x + a)^5 + b*c^2*tan(b*x + a)^3)]
Timed out. \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int { \frac {\cos \left (2 \, b x + 2 \, a\right )}{\left (c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos (2 (a+b x))}{(c \tan (a+b x) \tan (2 (a+b x)))^{3/2}} \, dx=\int \frac {\cos \left (2\,a+2\,b\,x\right )}{{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2}} \,d x \]