Integrand size = 20, antiderivative size = 45 \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b} \] Output:
-c^(1/2)*arctanh(c^(1/2)*tan(2*b*x+2*a)/(-c+c*sec(2*b*x+2*a))^(1/2))/b
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {2} \cos (a+b x)}{\sqrt {\cos (2 (a+b x))}}\right ) \sqrt {\cos (2 (a+b x))} \csc (a+b x) \sqrt {c \tan (a+b x) \tan (2 (a+b x))}}{\sqrt {2} b} \] Input:
Integrate[Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
Output:
-((ArcTanh[(Sqrt[2]*Cos[a + b*x])/Sqrt[Cos[2*(a + b*x)]]]*Sqrt[Cos[2*(a + b*x)]]*Csc[a + b*x]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(Sqrt[2]*b))
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4897, 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.
\(\displaystyle \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {c \tan (a+b x) \tan (2 a+2 b x)}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \sqrt {c \sec (2 a+2 b x)-c}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {c \csc \left (2 a+2 b x+\frac {\pi }{2}\right )-c}dx\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle -\frac {c \int \frac {1}{\frac {c^2 \tan ^2(2 a+2 b x)}{c \sec (2 a+2 b x)-c}-c}d\left (-\frac {c \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {c \sec (2 a+2 b x)-c}}\right )}{b}\) |
Input:
Int[Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]
Output:
-((Sqrt[c]*ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/Sqrt[-c + c*Sec[2*a + 2*b*x] ]])/b)
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]
Leaf count of result is larger than twice the leaf count of optimal. \(121\) vs. \(2(39)=78\).
Time = 1.13 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.71
method | result | size |
default | \(\frac {\sqrt {4}\, \sqrt {\frac {c \sin \left (b x +a \right )^{2}}{2 \cos \left (b x +a \right )^{2}-1}}\, \sin \left (b x +a \right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \cos \left (b x +a \right )}{\left (\cos \left (b x +a \right )+1\right ) \sqrt {\frac {2 \cos \left (b x +a \right )^{2}-1}{\left (\cos \left (b x +a \right )+1\right )^{2}}}}\right )}{2 b \left (\cos \left (b x +a \right )-1\right )}\) | \(122\) |
Input:
int((c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2/b*4^(1/2)*(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)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(2^(1/2)*cos(b*x+a)/(cos(b*x+a )+1)/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))/(cos(b*x+a)-1)
Time = 0.11 (sec) , antiderivative size = 201, normalized size of antiderivative = 4.47 \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\left [\frac {\sqrt {c} \log \left (-\frac {c \tan \left (b x + a\right )^{5} - 14 \, c \tan \left (b x + a\right )^{3} - 4 \, \sqrt {2} {\left (\tan \left (b x + a\right )^{4} - 4 \, \tan \left (b x + a\right )^{2} + 3\right )} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} \sqrt {c} + 17 \, c \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{5} + 2 \, \tan \left (b x + a\right )^{3} + \tan \left (b x + a\right )}\right )}{4 \, b}, \frac {\sqrt {-c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-c}}{c \tan \left (b x + a\right )^{3} - 3 \, c \tan \left (b x + a\right )}\right )}{2 \, b}\right ] \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="fricas")
Output:
[1/4*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 + t an(b*x + a)))/b, 1/2*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*ta n(b*x + a)))/b]
Timed out. \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Timed out} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (39) = 78\).
Time = 0.16 (sec) , antiderivative size = 430, normalized size of antiderivative = 9.56 \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\frac {\sqrt {c} {\left (\log \left (4 \, \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + 4 \, \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + 8 \, {\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right ) + 4\right ) - \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + \sqrt {\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )^{2}\right )} + 2 \, {\left (\cos \left (4 \, b x + 4 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 2 \, \cos \left (4 \, b x + 4 \, a\right ) + 1\right )}^{\frac {1}{4}} {\left (\cos \left (2 \, b x + 2 \, a\right ) \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right ) + \sin \left (2 \, b x + 2 \, a\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, b x + 4 \, a\right ), \cos \left (4 \, b x + 4 \, a\right ) + 1\right )\right )\right )}\right )\right )}}{4 \, b} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="maxima")
Output:
1/4*sqrt(c)*(log(4*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 + 4*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 + 8*(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*ar ctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a) + 1)) + 4) - log(cos(2*b*x + 2*a) ^2 + sin(2*b*x + 2*a)^2 + 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 + 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)* (cos(2*b*x + 2*a)*cos(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a) + 1)) + sin(2*b*x + 2*a)*sin(1/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a) + 1 )))))/b
Exception generated. \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\int \sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )} \,d x \] Input:
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2),x)
Output:
int((c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2), x)
\[ \int \sqrt {c \tan (a+b x) \tan (2 (a+b x))} \, dx=\sqrt {c}\, \left (\int \sqrt {\tan \left (b x +a \right )}\, \sqrt {\tan \left (2 b x +2 a \right )}d x \right ) \] Input:
int((c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)
Output:
sqrt(c)*int(sqrt(tan(a + b*x))*sqrt(tan(2*a + 2*b*x)),x)