Integrand size = 31, antiderivative size = 133 \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {7 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{8 b}-\frac {7 c^2 \sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {c^2 \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}} \]
7/8*c^(3/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b- 7/8*c^2*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)+1/4*c^2*cos(2*b*x+2*a )*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
Time = 1.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c \left (-5 \cos (a+b x)+7 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))}-6 \cos (3 (a+b x))+\cos (5 (a+b x))\right ) \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{16 b} \]
(c*(-5*Cos[a + b*x] + 7*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2* (a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]] - 6*Cos[3*(a + b*x)] + Cos[5*(a + b*x) ])*Csc[a + b*x]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(16*b)
Time = 0.68 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.03, 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, 4300, 27, 2011, 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.
| \(\displaystyle \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (2 a+2 b x)^2 (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \cos ^2(2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4300 |
| \(\displaystyle \frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {1}{2} c \int \frac {7 \cos (2 a+2 b x) (c-c \sec (2 a+2 b x))}{2 \sqrt {c \sec (2 a+2 b x)-c}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}-\frac {7}{4} c \int \frac {\cos (2 a+2 b x) (c-c \sec (2 a+2 b x))}{\sqrt {c \sec (2 a+2 b x)-c}}dx\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {7}{4} c \int \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{4} c \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}dx+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {7}{4} c \left (-\frac {1}{2} \int \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{4} c \left (-\frac {1}{2} \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle \frac {7}{4} c \left (\frac {c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {7}{4} c \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )+\frac {c^2 \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
(c^2*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*((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
3.7.17.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
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_)]*(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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[b^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[a/(d*n) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*(b*(m - 2*n - 2) - a*(m + 2*n - 1)*Csc[e + f *x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && (LtQ[n, -1] || (EqQ[m, 3/2] && EqQ[n, -2^(-1)])) && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(701\) vs. \(2(117)=234\).
Time = 5.32 (sec) , antiderivative size = 702, normalized size of antiderivative = 5.28
| method | result | size |
| default | \(-\frac {\sqrt {2}\, c \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (\cot \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+\csc \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+2 \cot \left (x b +a \right )\right ) \sqrt {4}}{2 b \left (2+\sqrt {2}\right ) \left (-2+\sqrt {2}\right )}-\frac {2 \sqrt {2}\, \csc \left (x b +a \right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )^{3}-2 \cos \left (x b +a \right )\right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, c \sqrt {4}}{b \left (-2+\sqrt {2}\right )^{3} \left (2+\sqrt {2}\right )^{3}}+\frac {\sqrt {2}\, \csc \left (x b +a \right ) \left (-16 \cos \left (x b +a \right )^{5}+9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )+9 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+12 \cos \left (x b +a \right )^{3}-18 \cos \left (x b +a \right )\right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, c \sqrt {4}}{b \left (-2+\sqrt {2}\right )^{5} \left (2+\sqrt {2}\right )^{5}}\) | \(702\) |
-1/2*2^(1/2)/b*c*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(cot(b*x+a)*((2 *cos(b*x+a)^2-1)/(1+cos(b*x+a))^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^(1/2)+csc(b*x+a)*( (2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^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^(1/2)+2*cot(b*x+ a))*4^(1/2)/(2+2^(1/2))/(-2+2^(1/2))-2*2^(1/2)/b*csc(b*x+a)*(2^(1/2)*arcta nh(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)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)+2^(1/2)*arc tanh(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)^2-1)/(1+cos(b*x+a))^2)^(1/2)-4*cos(b*x+a)^3-2*cos (b*x+a))*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*c*4^(1/2)/(-2+2^(1/2))^ 3/(2+2^(1/2))^3+2^(1/2)/b*csc(b*x+a)*(-16*cos(b*x+a)^5+9*2^(1/2)*arctanh(c os(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)^2-1)/(1+cos(b*x+a))^2)^(1/2)*cos(b*x+a)+9*2^(1/2)*arcta nh(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)^2-1)/(1+cos(b*x+a))^2)^(1/2)+12*cos(b*x+a)^3-18*cos (b*x+a))*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*c*4^(1/2)/(-2+2^(1/2))^ 5/(2+2^(1/2))^5
Time = 0.29 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.29 \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\left [\frac {7 \, {\left (c \tan \left (b x + a\right )^{5} + 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} {\left (9 \, c \tan \left (b x + a\right )^{4} - 4 \, c \tan \left (b x + a\right )^{2} - 5 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{32 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac {7 \, {\left (c \tan \left (b x + a\right )^{5} + 2 \, c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (9 \, c \tan \left (b x + a\right )^{4} - 4 \, c \tan \left (b x + a\right )^{2} - 5 \, c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]
[1/32*(7*(c*tan(b*x + a)^5 + 2*c*tan(b*x + a)^3 + c*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)*(9*c*tan(b*x + a)^4 - 4*c*tan(b*x + a)^2 - 5*c)*sqrt(-c*tan(b *x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b*x + a)^5 + 2*b*tan(b*x + a)^3 + b*tan(b*x + a)), -1/16*(7*(c*tan(b*x + a)^5 + 2*c*tan(b*x + a)^3 + c*tan(b *x + a))*sqrt(-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)*(9*c*tan(b*x + a)^4 - 4*c*tan(b*x + a)^2 - 5*c)*sqrt(-c*tan(b *x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b*x + a)^5 + 2*b*tan(b*x + a)^3 + b*tan(b*x + a))]
Timed out. \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos ^2(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int {\cos \left (2\,a+2\,b\,x\right )}^2\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2} \,d x \]