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

Optimal. Leaf size=98 \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (x)}+1\right )}{\sqrt{2}}+\frac{\log \left (\tan (x)-\sqrt{2} \sqrt{\tan (x)}+1\right )}{2 \sqrt{2}}-\frac{\log \left (\tan (x)+\sqrt{2} \sqrt{\tan (x)}+1\right )}{2 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0687999, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.333, Rules used = {3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (x)}+1\right )}{\sqrt{2}}+\frac{\log \left (\tan (x)-\sqrt{2} \sqrt{\tan (x)}+1\right )}{2 \sqrt{2}}-\frac{\log \left (\tan (x)+\sqrt{2} \sqrt{\tan (x)}+1\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rubi steps

\begin{align*} \int \sqrt{\tan (x)} \, dx &=\operatorname{Subst}\left (\int \frac{\sqrt{x}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\sqrt{\tan (x)}\right )\\ &=-\operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (x)}\right )+\operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (x)}\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (x)}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (x)}\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (x)}\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (x)}\right )}{2 \sqrt{2}}\\ &=\frac{\log \left (1-\sqrt{2} \sqrt{\tan (x)}+\tan (x)\right )}{2 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{\tan (x)}+\tan (x)\right )}{2 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}\\ &=-\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}+\frac{\tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (x)}\right )}{\sqrt{2}}+\frac{\log \left (1-\sqrt{2} \sqrt{\tan (x)}+\tan (x)\right )}{2 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} \sqrt{\tan (x)}+\tan (x)\right )}{2 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0122845, size = 24, normalized size = 0.24 \[ \frac{2}{3} \tan ^{\frac{3}{2}}(x) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-\tan ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[3/4, 1, 7/4, -Tan[x]^2]*Tan[x]^(3/2))/3

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Maple [A]  time = 0.001, size = 49, normalized size = 0.5 \begin{align*}{\frac{\cos \left ( x \right ) \sqrt{2}\arccos \left ( \cos \left ( x \right ) -\sin \left ( x \right ) \right ) }{2}\sqrt{\tan \left ( x \right ) }{\frac{1}{\sqrt{\cos \left ( x \right ) \sin \left ( x \right ) }}}}-{\frac{\sqrt{2}}{2}\ln \left ( \cos \left ( x \right ) +\sqrt{2}\sqrt{\tan \left ( x \right ) }\cos \left ( x \right ) +\sin \left ( x \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

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Maxima [A]  time = 1.41856, size = 108, normalized size = 1.1 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.88925, size = 572, normalized size = 5.84 \begin{align*} -\sqrt{2} \arctan \left (\sqrt{2} \sqrt{\frac{\sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} \cos \left (x\right ) + \cos \left (x\right ) + \sin \left (x\right )}{\cos \left (x\right )}} - \sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} - 1\right ) - \sqrt{2} \arctan \left (\sqrt{2} \sqrt{-\frac{\sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} \cos \left (x\right ) - \cos \left (x\right ) - \sin \left (x\right )}{\cos \left (x\right )}} - \sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} + 1\right ) - \frac{1}{4} \, \sqrt{2} \log \left (\frac{4 \,{\left (\sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} \cos \left (x\right ) + \cos \left (x\right ) + \sin \left (x\right )\right )}}{\cos \left (x\right )}\right ) + \frac{1}{4} \, \sqrt{2} \log \left (-\frac{4 \,{\left (\sqrt{2} \sqrt{\frac{\sin \left (x\right )}{\cos \left (x\right )}} \cos \left (x\right ) - \cos \left (x\right ) - \sin \left (x\right )\right )}}{\cos \left (x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tan{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(tan(x)), x)

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Giac [A]  time = 1.1425, size = 108, normalized size = 1.1 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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