3.6.77 \(\int \frac {\sqrt {x}}{\sqrt [3]{x}+x} \, dx\) [577]

Optimal. Leaf size=108 \[ 2 \sqrt {x}+\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}-\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}} \]

[Out]

-3/2*arctan(-1+x^(1/6)*2^(1/2))*2^(1/2)-3/2*arctan(1+x^(1/6)*2^(1/2))*2^(1/2)-3/4*ln(1+x^(1/3)-x^(1/6)*2^(1/2)
)*2^(1/2)+3/4*ln(1+x^(1/3)+x^(1/6)*2^(1/2))*2^(1/2)+2*x^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {1598, 348, 327, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {3 \text {ArcTan}\left (1-\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}-\frac {3 \text {ArcTan}\left (\sqrt {2} \sqrt [6]{x}+1\right )}{\sqrt {2}}+2 \sqrt {x}-\frac {3 \log \left (\sqrt [3]{x}-\sqrt {2} \sqrt [6]{x}+1\right )}{2 \sqrt {2}}+\frac {3 \log \left (\sqrt [3]{x}+\sqrt {2} \sqrt [6]{x}+1\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(x^(1/3) + x),x]

[Out]

2*Sqrt[x] + (3*ArcTan[1 - Sqrt[2]*x^(1/6)])/Sqrt[2] - (3*ArcTan[1 + Sqrt[2]*x^(1/6)])/Sqrt[2] - (3*Log[1 - Sqr
t[2]*x^(1/6) + x^(1/3)])/(2*Sqrt[2]) + (3*Log[1 + Sqrt[2]*x^(1/6) + x^(1/3)])/(2*Sqrt[2])

Rule 210

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 303

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 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

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 348

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

Rule 631

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 642

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 1176

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 1179

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 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {\sqrt {x}}{\sqrt [3]{x}+x} \, dx &=\int \frac {\sqrt [6]{x}}{1+x^{2/3}} \, dx\\ &=3 \text {Subst}\left (\int \frac {x^{5/2}}{1+x^2} \, dx,x,\sqrt [3]{x}\right )\\ &=2 \sqrt {x}-3 \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\sqrt [3]{x}\right )\\ &=2 \sqrt {x}-6 \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt [6]{x}\right )\\ &=2 \sqrt {x}+3 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt [6]{x}\right )-3 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt [6]{x}\right )\\ &=2 \sqrt {x}-\frac {3}{2} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt [6]{x}\right )-\frac {3}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt [6]{x}\right )-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt [6]{x}\right )}{2 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt [6]{x}\right )}{2 \sqrt {2}}\\ &=2 \sqrt {x}-\frac {3 \log \left (1-\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}}-\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}+\frac {3 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}\\ &=2 \sqrt {x}+\frac {3 \tan ^{-1}\left (1-\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}-\frac {3 \tan ^{-1}\left (1+\sqrt {2} \sqrt [6]{x}\right )}{\sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} \sqrt [6]{x}+\sqrt [3]{x}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 62, normalized size = 0.57 \begin {gather*} 2 \sqrt {x}-\frac {3 \tan ^{-1}\left (\frac {-1+\sqrt [3]{x}}{\sqrt {2} \sqrt [6]{x}}\right )}{\sqrt {2}}+\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [6]{x}}{1+\sqrt [3]{x}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(x^(1/3) + x),x]

[Out]

2*Sqrt[x] - (3*ArcTan[(-1 + x^(1/3))/(Sqrt[2]*x^(1/6))])/Sqrt[2] + (3*ArcTanh[(Sqrt[2]*x^(1/6))/(1 + x^(1/3))]
)/Sqrt[2]

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Maple [A]
time = 0.30, size = 66, normalized size = 0.61

method	result	size
derivativedivides	\(2 \sqrt {x}-\frac {3 \sqrt {2}\, \left (\ln \left (\frac {1+x^{\frac {1}{3}}-x^{\frac {1}{6}} \sqrt {2}}{1+x^{\frac {1}{3}}+x^{\frac {1}{6}} \sqrt {2}}\right )+2 \arctan \left (1+x^{\frac {1}{6}} \sqrt {2}\right )+2 \arctan \left (-1+x^{\frac {1}{6}} \sqrt {2}\right )\right )}{4}\)	\(66\)
default	\(2 \sqrt {x}-\frac {3 \sqrt {2}\, \left (\ln \left (\frac {1+x^{\frac {1}{3}}-x^{\frac {1}{6}} \sqrt {2}}{1+x^{\frac {1}{3}}+x^{\frac {1}{6}} \sqrt {2}}\right )+2 \arctan \left (1+x^{\frac {1}{6}} \sqrt {2}\right )+2 \arctan \left (-1+x^{\frac {1}{6}} \sqrt {2}\right )\right )}{4}\)	\(66\)
meijerg	\(2 \sqrt {x}-\frac {3 \sqrt {x}\, \left (\frac {\sqrt {2}\, \ln \left (1+x^{\frac {1}{3}}-x^{\frac {1}{6}} \sqrt {2}\right )}{2 \sqrt {x}}+\frac {\sqrt {2}\, \arctan \left (\frac {x^{\frac {1}{6}} \sqrt {2}}{2-x^{\frac {1}{6}} \sqrt {2}}\right )}{\sqrt {x}}-\frac {\sqrt {2}\, \ln \left (1+x^{\frac {1}{3}}+x^{\frac {1}{6}} \sqrt {2}\right )}{2 \sqrt {x}}+\frac {\sqrt {2}\, \arctan \left (\frac {x^{\frac {1}{6}} \sqrt {2}}{2+x^{\frac {1}{6}} \sqrt {2}}\right )}{\sqrt {x}}\right )}{2}\)	\(112\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x^(1/3)+x),x,method=_RETURNVERBOSE)

[Out]

2*x^(1/2)-3/4*2^(1/2)*(ln((1+x^(1/3)-x^(1/6)*2^(1/2))/(1+x^(1/3)+x^(1/6)*2^(1/2)))+2*arctan(1+x^(1/6)*2^(1/2))
+2*arctan(-1+x^(1/6)*2^(1/2)))

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Maxima [A]
time = 0.50, size = 83, normalized size = 0.77 \begin {gather*} -\frac {3}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, x^{\frac {1}{6}}\right )}\right ) - \frac {3}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, x^{\frac {1}{6}}\right )}\right ) + \frac {3}{4} \, \sqrt {2} \log \left (\sqrt {2} x^{\frac {1}{6}} + x^{\frac {1}{3}} + 1\right ) - \frac {3}{4} \, \sqrt {2} \log \left (-\sqrt {2} x^{\frac {1}{6}} + x^{\frac {1}{3}} + 1\right ) + 2 \, \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(x^(1/3)+x),x, algorithm="maxima")

[Out]

-3/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*x^(1/6))) - 3/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*x^(1/6))
) + 3/4*sqrt(2)*log(sqrt(2)*x^(1/6) + x^(1/3) + 1) - 3/4*sqrt(2)*log(-sqrt(2)*x^(1/6) + x^(1/3) + 1) + 2*sqrt(
x)

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Fricas [A]
time = 0.35, size = 120, normalized size = 1.11 \begin {gather*} 3 \, \sqrt {2} \arctan \left (\sqrt {2} \sqrt {\sqrt {2} x^{\frac {1}{6}} + x^{\frac {1}{3}} + 1} - \sqrt {2} x^{\frac {1}{6}} - 1\right ) + 3 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} x^{\frac {1}{6}} + 4 \, x^{\frac {1}{3}} + 4} - \sqrt {2} x^{\frac {1}{6}} + 1\right ) + \frac {3}{4} \, \sqrt {2} \log \left (4 \, \sqrt {2} x^{\frac {1}{6}} + 4 \, x^{\frac {1}{3}} + 4\right ) - \frac {3}{4} \, \sqrt {2} \log \left (-4 \, \sqrt {2} x^{\frac {1}{6}} + 4 \, x^{\frac {1}{3}} + 4\right ) + 2 \, \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(x^(1/3)+x),x, algorithm="fricas")

[Out]

3*sqrt(2)*arctan(sqrt(2)*sqrt(sqrt(2)*x^(1/6) + x^(1/3) + 1) - sqrt(2)*x^(1/6) - 1) + 3*sqrt(2)*arctan(1/2*sqr
t(2)*sqrt(-4*sqrt(2)*x^(1/6) + 4*x^(1/3) + 4) - sqrt(2)*x^(1/6) + 1) + 3/4*sqrt(2)*log(4*sqrt(2)*x^(1/6) + 4*x
^(1/3) + 4) - 3/4*sqrt(2)*log(-4*sqrt(2)*x^(1/6) + 4*x^(1/3) + 4) + 2*sqrt(x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\sqrt [3]{x} + x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)/(x**(1/3)+x),x)

[Out]

Integral(sqrt(x)/(x**(1/3) + x), x)

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Giac [A]
time = 3.43, size = 83, normalized size = 0.77 \begin {gather*} -\frac {3}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, x^{\frac {1}{6}}\right )}\right ) - \frac {3}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, x^{\frac {1}{6}}\right )}\right ) + \frac {3}{4} \, \sqrt {2} \log \left (\sqrt {2} x^{\frac {1}{6}} + x^{\frac {1}{3}} + 1\right ) - \frac {3}{4} \, \sqrt {2} \log \left (-\sqrt {2} x^{\frac {1}{6}} + x^{\frac {1}{3}} + 1\right ) + 2 \, \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)/(x^(1/3)+x),x, algorithm="giac")

[Out]

-3/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*x^(1/6))) - 3/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*x^(1/6))
) + 3/4*sqrt(2)*log(sqrt(2)*x^(1/6) + x^(1/3) + 1) - 3/4*sqrt(2)*log(-sqrt(2)*x^(1/6) + x^(1/3) + 1) + 2*sqrt(
x)

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Mupad [B]
time = 0.08, size = 42, normalized size = 0.39 \begin {gather*} 2\,\sqrt {x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^{1/6}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{2}+\frac {3}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^{1/6}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{2}-\frac {3}{2}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(x + x^(1/3)),x)

[Out]

2*x^(1/2) - 2^(1/2)*atan(2^(1/2)*x^(1/6)*(1/2 + 1i/2))*(3/2 + 3i/2) - 2^(1/2)*atan(2^(1/2)*x^(1/6)*(1/2 - 1i/2
))*(3/2 - 3i/2)

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