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

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

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Rubi [A]  time = 0.07, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.01, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan (x)} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Sqrt[Tan[x]],x]

[Out]

Could not integrate

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fricas [B]  time = 0.92, size = 180, normalized size = 1.84 \[ -\sqrt {2} \arctan \left (\sqrt {2} \sqrt {\frac {\sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} \cos \relax (x) + \cos \relax (x) + \sin \relax (x)}{\cos \relax (x)}} - \sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} - 1\right ) - \sqrt {2} \arctan \left (\sqrt {2} \sqrt {-\frac {\sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} \cos \relax (x) - \cos \relax (x) - \sin \relax (x)}{\cos \relax (x)}} - \sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} \cos \relax (x) + \cos \relax (x) + \sin \relax (x)\right )}}{\cos \relax (x)}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {\sin \relax (x)}{\cos \relax (x)}} \cos \relax (x) - \cos \relax (x) - \sin \relax (x)\right )}}{\cos \relax (x)}\right ) \]

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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giac [A]  time = 0.60, size = 80, normalized size = 0.82 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \relax (x)}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \relax (x)}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) \]

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)

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maple [A]  time = 0.12, size = 49, normalized size = 0.50

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.96, size = 80, normalized size = 0.82 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \relax (x)}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \relax (x)}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) \]

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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mupad [B]  time = 0.13, size = 65, normalized size = 0.66 \[ \frac {\sqrt {2}\,\left (\ln \left (\sqrt {2}\,\sqrt {\mathrm {tan}\relax (x)}-\mathrm {tan}\relax (x)-1\right )-\ln \left (\mathrm {tan}\relax (x)+\sqrt {2}\,\sqrt {\mathrm {tan}\relax (x)}+1\right )\right )}{4}+\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\relax (x)}-1\right )+\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\relax (x)}+1\right )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/2)*(log(2^(1/2)*tan(x)^(1/2) - tan(x) - 1) - log(tan(x) + 2^(1/2)*tan(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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan {\relax (x )}}\, dx \]

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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