Integrand size = 31, antiderivative size = 176 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {5 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{16 b}-\frac {5 c \sin (2 a+2 b x)}{16 b \sqrt {-c+c \sec (2 a+2 b x)}}+\frac {5 c \cos (2 a+2 b x) \sin (2 a+2 b x)}{24 b \sqrt {-c+c \sec (2 a+2 b x)}}-\frac {c \cos ^2(2 a+2 b x) \sin (2 a+2 b x)}{6 b \sqrt {-c+c \sec (2 a+2 b x)}} \]
5/16*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))*c^(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)
Time = 1.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.66 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\left (-26 \cot (a+b x)+15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x)+30 \sin (2 (a+b x))-2 \sin (4 (a+b x))+4 \sin (6 (a+b x))\right ) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{96 b} \]
((-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)
Time = 0.81 (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 |
\(\displaystyle -\frac {5}{6} \left (-\frac {3}{4} \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos (2 a+2 b x)}{4 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c \sin (2 a+2 b x) \cos ^2(2 a+2 b x)}{6 b \sqrt {c \sec (2 a+2 b x)-c}}\) |
-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
3.7.10.3.1 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. \(810\) vs. \(2(156)=312\).
Time = 7.42 (sec) , antiderivative size = 811, normalized size of antiderivative = 4.61
-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)-3/8*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)+3/32*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+c os(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)-1/192*2^(1/2) /b*csc(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)*(128*cos(b*x+a)^7+ 16*cos(b*x+a)^5+15*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)+15*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)+20*cos(b*x+a)^3-30*cos(b*x+a))*4^(1/2)
Time = 0.28 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.73 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\left [\frac {15 \, {\left (\tan \left (b x + a\right )^{7} + 3 \, \tan \left (b x + a\right )^{5} + 3 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} + 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) + 4 \, \sqrt {2} {\left (33 \, \tan \left (b x + a\right )^{6} - 19 \, \tan \left (b x + a\right )^{4} - \tan \left (b x + a\right )^{2} - 13\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{192 \, {\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, -\frac {15 \, {\left (\tan \left (b x + a\right )^{7} + 3 \, \tan \left (b x + a\right )^{5} + 3 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (33 \, \tan \left (b x + a\right )^{6} - 19 \, \tan \left (b x + a\right )^{4} - \tan \left (b x + a\right )^{2} - 13\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{96 \, {\left (b \tan \left (b x + a\right )^{7} + 3 \, b \tan \left (b x + a\right )^{5} + 3 \, b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]
[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))]
Timed out. \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2333 vs. \(2 (156) = 312\).
Time = 0.76 (sec) , antiderivative size = 2333, normalized size of antiderivative = 13.26 \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Too large to display} \]
-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(...
\[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int { \sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )} \cos \left (2 \, b x + 2 \, a\right )^{3} \,d x } \]
Timed out. \[ \int \cos ^3(2 (a+b x)) \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int {\cos \left (2\,a+2\,b\,x\right )}^3\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \]