Integrand size = 29, antiderivative size = 86 \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=-\frac {3 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{2 b}+\frac {c^2 \sin (2 a+2 b x)}{2 b \sqrt {-c+c \sec (2 a+2 b x)}} \]
-3/2*c^(3/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b +1/2*c^2*sin(2*b*x+2*a)/b/(-c+c*sec(2*b*x+2*a))^(1/2)
Time = 0.60 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08 \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\frac {c \left (\cos (a+b x)-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))}+\cos (3 (a+b x))\right ) \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{4 b} \]
(c*(Cos[a + b*x] - 3*Sqrt[2]*ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]] + Cos[3*(a + b*x)])*Csc[a + b*x]*Sqrt[c*T an[a + b*x]*Tan[2*(a + b*x)]])/(4*b)
Time = 0.60 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.44, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4897, 3042, 4301, 27, 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.
| \(\displaystyle \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (2 a+2 b x) (c \tan (a+b x) \tan (2 a+2 b x))^{3/2}dx\) |
| \(\Big \downarrow \) 4897 |
| \(\displaystyle \int \cos (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c\right )^{3/2}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4301 |
| \(\displaystyle -2 c \int \frac {3}{2} \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -3 c \int \cos (2 a+2 b x) \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -3 c \int \frac {\sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}}{\csc \left (2 a+2 b x+\frac {\pi }{2}\right )}dx-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle -3 c \left (-\frac {1}{2} \int \sqrt {c \sec (2 a+2 b x)-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -3 c \left (-\frac {1}{2} \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 4261 |
| \(\displaystyle -3 c \left (\frac {c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle -3 c \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{2 b}-\frac {c \sin (2 a+2 b x)}{2 b \sqrt {c \sec (2 a+2 b x)-c}}\right )-\frac {c^2 \sin (2 a+2 b x)}{b \sqrt {c \sec (2 a+2 b x)-c}}\) |
-((c^2*Sin[2*a + 2*b*x])/(b*Sqrt[-c + c*Sec[2*a + 2*b*x]])) - 3*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]]))
3.7.16.3.1 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_)]*(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*(m + n - 1))), x] + Simp[b/(m + n - 1) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(456\) vs. \(2(74)=148\).
Time = 4.95 (sec) , antiderivative size = 457, normalized size of antiderivative = 5.31
| method | result | size |
| default | \(\frac {\sqrt {2}\, c \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \left (\cot \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+\csc \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {2}+2 \cot \left (x b +a \right )\right ) \sqrt {4}}{2 b \left (2+\sqrt {2}\right ) \left (-2+\sqrt {2}\right )}+\frac {\sqrt {2}\, \csc \left (x b +a \right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}\, \cos \left (x b +a \right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\cos \left (x b +a \right ) \sqrt {2}}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )^{3}-2 \cos \left (x b +a \right )\right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, c \sqrt {4}}{b \left (-2+\sqrt {2}\right )^{3} \left (2+\sqrt {2}\right )^{3}}\) | \(457\) |
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^(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)*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))*((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
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (74) = 148\).
Time = 0.29 (sec) , antiderivative size = 369, normalized size of antiderivative = 4.29 \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\left [\frac {3 \, {\left (c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right ) - 4 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{2} - c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{8 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}, \frac {3 \, {\left (c \tan \left (b x + a\right )^{3} + c \tan \left (b x + a\right )\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right ) - 2 \, \sqrt {2} {\left (c \tan \left (b x + a\right )^{2} - c\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{4 \, {\left (b \tan \left (b x + a\right )^{3} + b \tan \left (b x + a\right )\right )}}\right ] \]
[1/8*(3*(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)*(c*tan(b* x + a)^2 - c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b*x + a )^3 + b*tan(b*x + a)), 1/4*(3*(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)*(c*t an(b*x + a)^2 - c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*tan(b* x + a)^3 + b*tan(b*x + a))]
Timed out. \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 1058 vs. \(2 (74) = 148\).
Time = 0.46 (sec) , antiderivative size = 1058, normalized size of antiderivative = 12.30 \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Too large to display} \]
-1/16*(4*(c*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))*sin( 2*b*x + 2*a) + (c*cos(2*b*x + 2*a) + c)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos( 4*b*x + 4*a) + 1)^(1/4)*sqrt(c) - 3*(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) - 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(si n(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + 1) + c*log(((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2)*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + (cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2)*sin(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) + 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)*(cos(1/2*arctan2(sin(4*b*x + 4*a...
Timed out. \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \cos (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx=\int \cos \left (2\,a+2\,b\,x\right )\,{\left (c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )\right )}^{3/2} \,d x \]