Integrand size = 31, antiderivative size = 129 \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {3 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{8 b}+\frac {3 c \sin (2 a+2 b x)}{8 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos (2 a+2 b x) \sin (2 a+2 b x)}{4 b \sqrt {-c+c \sec (2 a+2 b x)}} \]
-3/8*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))*c^(1/2)/b +3/8*c*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)-1/4*c*cos(2*b*x+2*a)*s in(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
Time = 0.80 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\left (-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x)+2 (\cot (a+b x)-\sin (2 (a+b x))+\sin (4 (a+b x)))\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{16 b} \]
((-3*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[C os[2*(a + b*x)]]*Csc[a + b*x] + 2*(Cot[a + b*x] - Sin[2*(a + b*x)] + Sin[4 *(a + b*x)]))*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(16*b)
Time = 0.61 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4897, 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.
| \(\displaystyle \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (2 a+2 b x)^2 \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \cos ^2(2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle -\frac {3}{4} \int \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3}{4} \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}dx-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle -\frac {3}{4} \left (-\frac {1}{2} \int \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3}{4} \left (-\frac {1}{2} \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle -\frac {3}{4} \left (\frac {c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -\frac {3}{4} \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
-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
3.7.9.3.1 Defintions of rubi rules used
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_)]*(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]
Leaf count of result is larger than twice the leaf count of optimal. \(570\) vs. \(2(113)=226\).
Time = 6.14 (sec) , antiderivative size = 571, normalized size of antiderivative = 4.43
| method | result | size |
| default | \(\frac {\sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \sin \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 {4}}{2 b \left (\cos \left (x b +a \right )-1\right )}+\frac {\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}}\, \sqrt {4}}{4 b}-\frac {\sqrt {2}\, \csc \left (x b +a \right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (16 \cos \left (x b +a \right )^{5}+3 \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 )+3 \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}-6 \cos \left (x b +a \right )\right ) \sqrt {4}}{32 b}\) | \(571\) |
1/2/b*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*sin(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))/(cos(b*x+a)-1)*4^(1/2)+1/4*2^(1/ 2)/b*csc(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*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2 )^(1/2)*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*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)*4^(1/2)-1/32*2^(1/2)/b*csc(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1 ))^(1/2)*(16*cos(b*x+a)^5+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)^2-1)/(1+co s(b*x+a))^2)^(1/2)*cos(b*x+a)+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)^2-1)/( 1+cos(b*x+a))^2)^(1/2)+4*cos(b*x+a)^3-6*cos(b*x+a))*4^(1/2)
Time = 0.30 (sec) , antiderivative size = 419, normalized size of antiderivative = 3.25 \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\left [\frac {3 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 4 \, \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{32 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, \frac {3 \, {\left (\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (5 \, \tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{16 \, {\left (b \tan \left (b x + a\right )^{5} + 2 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]
[1/32*(3*(tan(b*x + a)^5 + 2*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)*(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*s qrt(2)*(5*tan(b*x + a)^4 - 4*tan(b*x + a)^2 - 1)*sqrt(-c*tan(b*x + a)^2/(t an(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*(3*(tan(b*x + a)^5 + 2*tan(b*x + a)^3 + tan(b*x + a))*sqrt(-c)*a rctan(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)*(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*tan(b*x + a)^5 + 2*b*tan(b*x + a)^3 + b*tan(b*x + a))]
Timed out. \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1421 vs. \(2 (113) = 226\).
Time = 0.51 (sec) , antiderivative size = 1421, normalized size of antiderivative = 11.02 \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Too large to display} \]
-1/64*(4*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1 )^(1/4)*((cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(4* b*x + 4*a) - (cos(4*b*x + 4*a) - 2)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos (4*b*x + 4*a) - 1)))*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) - cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin(4*b*x + 4* a) - (cos(4*b*x + 4*a) - 2)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - ((cos(4*b*x + 4*a) - 2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -c os(4*b*x + 4*a) - 1)) + sin(4*b*x + 4*a)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4* a))))*sqrt(c) - 3*sqrt(c)*(log(sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^ 2 + 2*cos(4*b*x + 4*a) + 1)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4 )*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) - log(sqr t(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*cos(1/ 2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + sqrt(cos(4*b*x + 4 *a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)*sin(1/2*arctan2(sin(4 *b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 - 2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*sin(1/2*arctan2(sin(4*b*x + 4...
\[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int { \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} \,d x } \]
Timed out. \[ \int \cos ^2(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int {\cos \left (2\,a+2\,b\,x\right )}^2\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \]