Integrand size = 11, antiderivative size = 32 \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=-\frac {\arcsin (\tan (x))}{\sqrt {2}}+\arctan \left (\frac {\sqrt {2} \tan (x)}{\sqrt {1-\tan ^2(x)}}\right ) \] Output:
arctan(2^(1/2)*tan(x)/(1-tan(x)^2)^(1/2))-1/2*arcsin(tan(x))*2^(1/2)
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.62 \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=\frac {\left (\sqrt {2} \arcsin \left (\sqrt {2} \sin (x)\right )-\arctan \left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )\right ) \cos (x) \sqrt {\cot (2 x) \tan (x)}}{\sqrt {\cos (2 x)}} \] Input:
Integrate[Sqrt[Cot[2*x]*Tan[x]],x]
Output:
((Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] - ArcTan[Sin[x]/Sqrt[Cos[2*x]]])*Cos[x]*S qrt[Cot[2*x]*Tan[x]])/Sqrt[Cos[2*x]]
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4889, 27, 301, 223, 291, 216}
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.
\(\displaystyle \int \sqrt {\tan (x) \cot (2 x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\tan (x) \cot (2 x)}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {\sqrt {1-\tan ^2(x)}}{\sqrt {2} \left (\tan ^2(x)+1\right )}d\tan (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sqrt {1-\tan ^2(x)}}{\tan ^2(x)+1}d\tan (x)}{\sqrt {2}}\) |
\(\Big \downarrow \) 301 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {1-\tan ^2(x)} \left (\tan ^2(x)+1\right )}d\tan (x)-\int \frac {1}{\sqrt {1-\tan ^2(x)}}d\tan (x)}{\sqrt {2}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {1-\tan ^2(x)} \left (\tan ^2(x)+1\right )}d\tan (x)-\arcsin (\tan (x))}{\sqrt {2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {2 \int \frac {1}{\frac {2 \tan ^2(x)}{1-\tan ^2(x)}+1}d\frac {\tan (x)}{\sqrt {1-\tan ^2(x)}}-\arcsin (\tan (x))}{\sqrt {2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \tan (x)}{\sqrt {1-\tan ^2(x)}}\right )-\arcsin (\tan (x))}{\sqrt {2}}\) |
Input:
Int[Sqrt[Cot[2*x]*Tan[x]],x]
Output:
(-ArcSin[Tan[x]] + Sqrt[2]*ArcTan[(Sqrt[2]*Tan[x])/Sqrt[1 - Tan[x]^2]])/Sq rt[2]
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[b/ d Int[(a + b*x^2)^(p - 1), x], x] - Simp[(b*c - a*d)/d Int[(a + b*x^2)^ (p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4] || (EqQ[p, 2/3] && E qQ[b*c + 3*a*d, 0]))
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Leaf count of result is larger than twice the leaf count of optimal. \(141\) vs. \(2(26)=52\).
Time = 6.14 (sec) , antiderivative size = 142, normalized size of antiderivative = 4.44
method | result | size |
default | \(\frac {\left (2 \sqrt {2}\, \arctan \left (\frac {\left (\csc \left (x \right )-\cot \left (x \right )\right ) \sqrt {2}}{\sqrt {\frac {2 \cos \left (x \right )^{2}-1}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )-\arctan \left (\frac {2 \sin \left (x \right )-1}{\left (1+\cos \left (x \right )\right ) \sqrt {\frac {2 \cos \left (x \right )^{2}-1}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )-\arctan \left (\frac {1+2 \sin \left (x \right )}{\left (1+\cos \left (x \right )\right ) \sqrt {\frac {2 \cos \left (x \right )^{2}-1}{\left (1+\cos \left (x \right )\right )^{2}}}}\right )\right ) \sqrt {2-\sec \left (x \right )^{2}}\, \cos \left (x \right ) \sqrt {2}}{4 \left (1+\cos \left (x \right )\right ) \sqrt {\frac {2 \cos \left (x \right )^{2}-1}{\left (1+\cos \left (x \right )\right )^{2}}}}\) | \(142\) |
Input:
int((cot(2*x)/cot(x))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4*(2*2^(1/2)*arctan((csc(x)-cot(x))*2^(1/2)/((2*cos(x)^2-1)/(1+cos(x))^2 )^(1/2))-arctan((2*sin(x)-1)/(1+cos(x))/((2*cos(x)^2-1)/(1+cos(x))^2)^(1/2 ))-arctan((1+2*sin(x))/(1+cos(x))/((2*cos(x)^2-1)/(1+cos(x))^2)^(1/2)))*(2 -sec(x)^2)^(1/2)*cos(x)/(1+cos(x))/((2*cos(x)^2-1)/(1+cos(x))^2)^(1/2)*2^( 1/2)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}} \sin \left (2 \, x\right )}{2 \, \cos \left (2 \, x\right )}\right ) + \arctan \left (\frac {\sqrt {\frac {\cos \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 1}} \sin \left (2 \, x\right )}{\cos \left (2 \, x\right )}\right ) \] Input:
integrate((cot(2*x)/cot(x))^(1/2),x, algorithm="fricas")
Output:
-1/2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(cos(2*x)/(cos(2*x) + 1))*sin(2*x)/cos (2*x)) + arctan(sqrt(cos(2*x)/(cos(2*x) + 1))*sin(2*x)/cos(2*x))
\[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=\int \sqrt {\frac {\cot {\left (2 x \right )}}{\cot {\left (x \right )}}}\, dx \] Input:
integrate((cot(2*x)/cot(x))**(1/2),x)
Output:
Integral(sqrt(cot(2*x)/cot(x)), x)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 507, normalized size of antiderivative = 15.84 \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx =\text {Too large to display} \] Input:
integrate((cot(2*x)/cot(x))^(1/2),x, algorithm="maxima")
Output:
1/4*sqrt(2)*(sqrt(2)*arctan2((cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1 /4)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1)) + sin(2*x), (cos(4*x)^2 + sin (4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) + cos(2*x)) - 2*arctan2(((abs(2*e^(2*I*x) + 2)^4 + 16*cos(2*x)^4 + 16*sin(2 *x)^4 + 8*(cos(2*x)^2 - sin(2*x)^2 - 2*cos(2*x) + 1)*abs(2*e^(2*I*x) + 2)^ 2 - 64*cos(2*x)^3 + 32*(cos(2*x)^2 - 2*cos(2*x) + 1)*sin(2*x)^2 + 96*cos(2 *x)^2 - 64*cos(2*x) + 16)^(1/4)*sin(1/2*arctan2(8*(cos(2*x) - 1)*sin(2*x)/ abs(2*e^(2*I*x) + 2)^2, (abs(2*e^(2*I*x) + 2)^2 + 4*cos(2*x)^2 - 4*sin(2*x )^2 - 8*cos(2*x) + 4)/abs(2*e^(2*I*x) + 2)^2)) + 2*sin(2*x))/abs(2*e^(2*I* x) + 2), ((abs(2*e^(2*I*x) + 2)^4 + 16*cos(2*x)^4 + 16*sin(2*x)^4 + 8*(cos (2*x)^2 - sin(2*x)^2 - 2*cos(2*x) + 1)*abs(2*e^(2*I*x) + 2)^2 - 64*cos(2*x )^3 + 32*(cos(2*x)^2 - 2*cos(2*x) + 1)*sin(2*x)^2 + 96*cos(2*x)^2 - 64*cos (2*x) + 16)^(1/4)*cos(1/2*arctan2(8*(cos(2*x) - 1)*sin(2*x)/abs(2*e^(2*I*x ) + 2)^2, (abs(2*e^(2*I*x) + 2)^2 + 4*cos(2*x)^2 - 4*sin(2*x)^2 - 8*cos(2* x) + 4)/abs(2*e^(2*I*x) + 2)^2)) + 2*cos(2*x) - 2)/abs(2*e^(2*I*x) + 2)))
Result contains complex when optimal does not.
Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 4.31 \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=\frac {1}{2} \, {\left (\pi - \sqrt {2} \arctan \left (-i\right ) - \sqrt {2} \arctan \left (\sqrt {2}\right ) - i \, \log \left (2 \, \sqrt {2} + 3\right )\right )} \mathrm {sgn}\left (\sin \left (2 \, x\right )\right ) - \frac {\sqrt {2} {\left (-i \, \sqrt {2} \log \left (2 i \, \sqrt {2} + 3 i\right ) - 2 \, \arctan \left (-i\right )\right )} \mathrm {sgn}\left (\cos \left (x\right )\right ) + 2 \, {\left (\sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} {\left (\frac {3 \, {\left (2 \, \sqrt {2} \sqrt {-2 \, \cos \left (x\right )^{4} + 3 \, \cos \left (x\right )^{2} - 1} - 1\right )}}{4 \, \cos \left (x\right )^{2} - 3} - 1\right )}\right ) + \arcsin \left (4 \, \cos \left (x\right )^{2} - 3\right )\right )} \mathrm {sgn}\left (\cos \left (x\right )\right )}{4 \, \mathrm {sgn}\left (\cos \left (x\right )\right ) \mathrm {sgn}\left (\sin \left (2 \, x\right )\right )} \] Input:
integrate((cot(2*x)/cot(x))^(1/2),x, algorithm="giac")
Output:
1/2*(pi - sqrt(2)*arctan(-I) - sqrt(2)*arctan(sqrt(2)) - I*log(2*sqrt(2) + 3))*sgn(sin(2*x)) - 1/4*(sqrt(2)*(-I*sqrt(2)*log(2*I*sqrt(2) + 3*I) - 2*a rctan(-I))*sgn(cos(x)) + 2*(sqrt(2)*arctan(1/4*sqrt(2)*(3*(2*sqrt(2)*sqrt( -2*cos(x)^4 + 3*cos(x)^2 - 1) - 1)/(4*cos(x)^2 - 3) - 1)) + arcsin(4*cos(x )^2 - 3))*sgn(cos(x)))/(sgn(cos(x))*sgn(sin(2*x)))
Timed out. \[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=\int \sqrt {\frac {\mathrm {cot}\left (2\,x\right )}{\mathrm {cot}\left (x\right )}} \,d x \] Input:
int((cot(2*x)/cot(x))^(1/2),x)
Output:
int((cot(2*x)/cot(x))^(1/2), x)
\[ \int \sqrt {\cot (2 x) \tan (x)} \, dx=\int \frac {\sqrt {\cot \left (x \right )}\, \sqrt {\cot \left (2 x \right )}}{\cot \left (x \right )}d x \] Input:
int((cot(2*x)/cot(x))^(1/2),x)
Output:
int((sqrt(cot(x))*sqrt(cot(2*x)))/cot(x),x)