3.930 \(\int \sqrt {\sqrt [4]{x}+x} \, dx\)

Optimal. Leaf size=59 \[ \frac {2}{3} \sqrt {x+\sqrt [4]{x}} x+\frac {1}{3} \sqrt {x+\sqrt [4]{x}} \sqrt [4]{x}-\frac {1}{3} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt [4]{x}}}\right ) \]

[Out]

-1/3*arctanh(x^(1/2)/(x^(1/4)+x)^(1/2))+1/3*x^(1/4)*(x^(1/4)+x)^(1/2)+2/3*x*(x^(1/4)+x)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {2004, 2018, 2024, 2029, 206} \[ \frac {2}{3} \sqrt {x+\sqrt [4]{x}} x+\frac {1}{3} \sqrt {x+\sqrt [4]{x}} \sqrt [4]{x}-\frac {1}{3} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt [4]{x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^(1/4) + x],x]

[Out]

(x^(1/4)*Sqrt[x^(1/4) + x])/3 + (2*x*Sqrt[x^(1/4) + x])/3 - ArcTanh[Sqrt[x]/Sqrt[x^(1/4) + x]]/3

Rule 206

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

Rule 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 2024

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

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps

\begin {align*} \int \sqrt {\sqrt [4]{x}+x} \, dx &=\frac {2}{3} x \sqrt {\sqrt [4]{x}+x}+\frac {1}{4} \int \frac {\sqrt [4]{x}}{\sqrt {\sqrt [4]{x}+x}} \, dx\\ &=\frac {2}{3} x \sqrt {\sqrt [4]{x}+x}+\operatorname {Subst}\left (\int \frac {x^4}{\sqrt {x+x^4}} \, dx,x,\sqrt [4]{x}\right )\\ &=\frac {1}{3} \sqrt [4]{x} \sqrt {\sqrt [4]{x}+x}+\frac {2}{3} x \sqrt {\sqrt [4]{x}+x}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\sqrt {x+x^4}} \, dx,x,\sqrt [4]{x}\right )\\ &=\frac {1}{3} \sqrt [4]{x} \sqrt {\sqrt [4]{x}+x}+\frac {2}{3} x \sqrt {\sqrt [4]{x}+x}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {\sqrt [4]{x}+x}}\right )\\ &=\frac {1}{3} \sqrt [4]{x} \sqrt {\sqrt [4]{x}+x}+\frac {2}{3} x \sqrt {\sqrt [4]{x}+x}-\frac {1}{3} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {\sqrt [4]{x}+x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.97 \[ \frac {3 x^{5/4}-\sqrt {x^{3/4}+1} \sqrt [8]{x} \sinh ^{-1}\left (x^{3/8}\right )+2 x^2+\sqrt {x}}{3 \sqrt {x+\sqrt [4]{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^(1/4) + x],x]

[Out]

(Sqrt[x] + 3*x^(5/4) + 2*x^2 - Sqrt[1 + x^(3/4)]*x^(1/8)*ArcSinh[x^(3/8)])/(3*Sqrt[x^(1/4) + x])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/4)+x)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 4.57, size = 45, normalized size = 0.76 \[ \frac {1}{3} \, \sqrt {x + x^{\frac {1}{4}}} x^{\frac {1}{4}} {\left (2 \, x^{\frac {3}{4}} + 1\right )} - \frac {1}{6} \, \log \left (\sqrt {\frac {1}{x^{\frac {3}{4}}} + 1} + 1\right ) + \frac {1}{6} \, \log \left ({\left | \sqrt {\frac {1}{x^{\frac {3}{4}}} + 1} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/4)+x)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.10, size = 342, normalized size = 5.80 \[ \frac {2 \sqrt {x +x^{\frac {1}{4}}}\, x}{3}+\frac {\sqrt {x +x^{\frac {1}{4}}}\, x^{\frac {1}{4}}}{3}+\frac {\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x^{\frac {1}{4}}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}+1\right )}}\, \left (x^{\frac {1}{4}}+1\right )^{2} \sqrt {-\frac {x^{\frac {1}{4}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}+1\right )}}\, \sqrt {-\frac {x^{\frac {1}{4}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}+1\right )}}\, \left (-\EllipticF \left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x^{\frac {1}{4}}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}+1\right )}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )+\EllipticPi \left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x^{\frac {1}{4}}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}+1\right )}}, \frac {\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {\left (x^{\frac {1}{4}}+1\right ) \left (x^{\frac {1}{4}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x^{\frac {1}{4}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{\frac {1}{4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/4)+x)^(1/2),x)

[Out]

2/3*x*(x^(1/4)+x)^(1/2)+1/3*x^(1/4)*(x^(1/4)+x)^(1/2)+(-1/2-1/2*I*3^(1/2))*((3/2+1/2*I*3^(1/2))*x^(1/4)/(1/2+1
/2*I*3^(1/2))/(x^(1/4)+1))^(1/2)*(x^(1/4)+1)^2*(-(x^(1/4)-1/2+1/2*I*3^(1/2))/(1/2-1/2*I*3^(1/2))/(x^(1/4)+1))^
(1/2)*(-(x^(1/4)-1/2-1/2*I*3^(1/2))/(1/2+1/2*I*3^(1/2))/(x^(1/4)+1))^(1/2)/(3/2+1/2*I*3^(1/2))/(x^(1/4)*(x^(1/
4)+1)*(x^(1/4)-1/2+1/2*I*3^(1/2))*(x^(1/4)-1/2-1/2*I*3^(1/2)))^(1/2)*(-EllipticF(((3/2+1/2*I*3^(1/2))*x^(1/4)/
(1/2+1/2*I*3^(1/2))/(x^(1/4)+1))^(1/2),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1
/2*I*3^(1/2)))^(1/2))+EllipticPi(((3/2+1/2*I*3^(1/2))*x^(1/4)/(1/2+1/2*I*3^(1/2))/(x^(1/4)+1))^(1/2),(1/2+1/2*
I*3^(1/2))/(3/2+1/2*I*3^(1/2)),((-3/2+1/2*I*3^(1/2))*(-1/2-1/2*I*3^(1/2))/(-1/2+1/2*I*3^(1/2))/(-3/2-1/2*I*3^(
1/2)))^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x + x^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^(1/4)+x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x + x^(1/4)), x)

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mupad [B]  time = 3.53, size = 27, normalized size = 0.46 \[ \frac {8\,x\,\sqrt {x+x^{1/4}}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {3}{2};\ \frac {5}{2};\ -x^{3/4}\right )}{9\,\sqrt {x^{3/4}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^(1/4))^(1/2),x)

[Out]

(8*x*(x + x^(1/4))^(1/2)*hypergeom([-1/2, 3/2], 5/2, -x^(3/4)))/(9*(x^(3/4) + 1)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\sqrt [4]{x} + x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**(1/4)+x)**(1/2),x)

[Out]

Integral(sqrt(x**(1/4) + x), x)

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