\(\int \sqrt {\tan (x)} \, dx\) [10]

Optimal result

Integrand size = 6, antiderivative size = 71 \[ \int \sqrt {\tan (x)} \, dx=-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (x)}}{1+\tan (x)}\right )}{\sqrt {2}} \] Output:

1/2*arctan(-1+tan(x)^(1/2)*2^(1/2))*2^(1/2)+1/2*arctan(1+tan(x)^(1/2)*2^(1 
/2))*2^(1/2)-1/2*arctanh(2^(1/2)*tan(x)^(1/2)/(1+tan(x)))*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \sqrt {\tan (x)} \, dx=\frac {\left (\arctan \left (\sqrt [4]{-\tan ^2(x)}\right )-\text {arctanh}\left (\sqrt [4]{-\tan ^2(x)}\right )\right ) \sqrt [4]{-\tan (x)}}{\sqrt [4]{\tan (x)}} \] Input:

Integrate[Sqrt[Tan[x]],x]
 

Output:

((ArcTan[(-Tan[x]^2)^(1/4)] - ArcTanh[(-Tan[x]^2)^(1/4)])*(-Tan[x])^(1/4)) 
/Tan[x]^(1/4)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.55, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.833, Rules used = {3042, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \sqrt {\tan (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\tan (x)}dx\)

\(\Big \downarrow \) 3957

\(\displaystyle \int \frac {\sqrt {\tan (x)}}{\tan ^2(x)+1}d\tan (x)\)

\(\Big \downarrow \) 266

\(\displaystyle 2 \int \frac {\tan (x)}{\tan ^2(x)+1}d\sqrt {\tan (x)}\)

\(\Big \downarrow \) 826

\(\displaystyle 2 \left (\frac {1}{2} \int \frac {\tan (x)+1}{\tan ^2(x)+1}d\sqrt {\tan (x)}-\frac {1}{2} \int \frac {1-\tan (x)}{\tan ^2(x)+1}d\sqrt {\tan (x)}\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}+\frac {1}{2} \int \frac {1}{\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}\right )-\frac {1}{2} \int \frac {1-\tan (x)}{\tan ^2(x)+1}d\sqrt {\tan (x)}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\tan (x)-1}d\left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (x)-1}d\left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (x)}{\tan ^2(x)+1}d\sqrt {\tan (x)}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-\tan (x)}{\tan ^2(x)+1}d\sqrt {\tan (x)}\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (x)}}{\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (x)}}{\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (x)}}{\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (x)}+1}{\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1}d\sqrt {\tan (x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\right )\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1\right )}{2 \sqrt {2}}\right )\right )\)

Input:

Int[Sqrt[Tan[x]],x]
 

Output:

2*((-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[ 
Tan[x]]]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*Sqrt[Tan[x]] + Tan[x]]/(2*Sqrt[2]) 
- Log[1 + Sqrt[2]*Sqrt[Tan[x]] + Tan[x]]/(2*Sqrt[2]))/2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.69

method	result	size
lookup	\(\frac {\sqrt {\tan \left (x \right )}\, \cos \left (x \right ) \sqrt {2}\, \arccos \left (\cos \left (x \right )-\sin \left (x \right )\right )}{2 \sqrt {\cos \left (x \right ) \sin \left (x \right )}}-\frac {\sqrt {2}\, \ln \left (\cos \left (x \right )+\sqrt {2}\, \sqrt {\tan \left (x \right )}\, \cos \left (x \right )+\sin \left (x \right )\right )}{2}\)	\(49\)
default	\(\frac {\sqrt {\tan \left (x \right )}\, \cos \left (x \right ) \sqrt {2}\, \arccos \left (\cos \left (x \right )-\sin \left (x \right )\right )}{2 \sqrt {\cos \left (x \right ) \sin \left (x \right )}}-\frac {\sqrt {2}\, \ln \left (\cos \left (x \right )+\sqrt {2}\, \sqrt {\tan \left (x \right )}\, \cos \left (x \right )+\sin \left (x \right )\right )}{2}\)	\(49\)
derivativedivides	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {\tan \left (x \right )-\sqrt {2}\, \sqrt {\tan \left (x \right )}+1}{\tan \left (x \right )+\sqrt {2}\, \sqrt {\tan \left (x \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (x \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (x \right )}\right )\right )}{4}\)	\(62\)

Input:

int(tan(x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*tan(x)^(1/2)/(cos(x)*sin(x))^(1/2)*cos(x)*2^(1/2)*arccos(cos(x)-sin(x) 
)-1/2*2^(1/2)*ln(cos(x)+2^(1/2)*tan(x)^(1/2)*cos(x)+sin(x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \sqrt {\tan (x)} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\tan \left (x\right )} + 1\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\tan \left (x\right )} - 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) \] Input:

integrate(tan(x)^(1/2),x, algorithm="fricas")
 

Output:

1/2*sqrt(2)*arctan(sqrt(2)*sqrt(tan(x)) + 1) + 1/2*sqrt(2)*arctan(sqrt(2)* 
sqrt(tan(x)) - 1) - 1/4*sqrt(2)*log(sqrt(2)*sqrt(tan(x)) + tan(x) + 1) + 1 
/4*sqrt(2)*log(-sqrt(2)*sqrt(tan(x)) + tan(x) + 1)
 

Sympy [F]

\[ \int \sqrt {\tan (x)} \, dx=\int \sqrt {\tan {\left (x \right )}}\, dx \] Input:

integrate(tan(x)**(1/2),x)
 

Output:

Integral(sqrt(tan(x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \sqrt {\tan (x)} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (x\right )}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (x\right )}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) \] Input:

integrate(tan(x)^(1/2),x, algorithm="maxima")
 

Output:

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(x)))) + 1/2*sqrt(2)*a 
rctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(x)))) - 1/4*sqrt(2)*log(sqrt(2)*s 
qrt(tan(x)) + tan(x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*sqrt(tan(x)) + tan(x) 
 + 1)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \sqrt {\tan (x)} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (x\right )}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (x\right )}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (x\right )} + \tan \left (x\right ) + 1\right ) \] Input:

integrate(tan(x)^(1/2),x, algorithm="giac")
 

Output:

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(x)))) + 1/2*sqrt(2)*a 
rctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(x)))) - 1/4*sqrt(2)*log(sqrt(2)*s 
qrt(tan(x)) + tan(x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*sqrt(tan(x)) + tan(x) 
 + 1)
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \sqrt {\tan (x)} \, dx=\frac {\sqrt {2}\,\left (\ln \left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (x\right )}-\mathrm {tan}\left (x\right )-1\right )-\ln \left (\mathrm {tan}\left (x\right )+\sqrt {2}\,\sqrt {\mathrm {tan}\left (x\right )}+1\right )\right )}{4}+\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (x\right )}-1\right )+\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (x\right )}+1\right )\right )}{2} \] Input:

int(tan(x)^(1/2),x)
 

Output:

(2^(1/2)*(log(2^(1/2)*tan(x)^(1/2) - tan(x) - 1) - log(tan(x) + 2^(1/2)*ta 
n(x)^(1/2) + 1)))/4 + (2^(1/2)*(atan(2^(1/2)*tan(x)^(1/2) - 1) + atan(2^(1 
/2)*tan(x)^(1/2) + 1)))/2
 

Reduce [F]

\[ \int \sqrt {\tan (x)} \, dx=\int \sqrt {\tan \left (x \right )}d x \] Input:

int(tan(x)^(1/2),x)
 

Output:

int(sqrt(tan(x)),x)