3.7.70 \(\int \sqrt [4]{x^2+x^4} \, dx\)

Optimal. Leaf size=53 \[ \frac {1}{2} \sqrt [4]{x^4+x^2} x-\frac {1}{4} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^2}}\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^2}}\right ) \]

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Rubi [B]  time = 0.06, antiderivative size = 107, normalized size of antiderivative = 2.02, number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {2004, 2032, 329, 331, 298, 203, 206} \begin {gather*} \frac {1}{2} \sqrt [4]{x^4+x^2} x-\frac {\left (x^2+1\right )^{3/4} x^{3/2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{4 \left (x^4+x^2\right )^{3/4}}+\frac {\left (x^2+1\right )^{3/4} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{4 \left (x^4+x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + x^4)^(1/4),x]

[Out]

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

Rule 203

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

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 298

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

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.75 \begin {gather*} \frac {2 x \sqrt [4]{x^4+x^2} \, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x^2\right )}{3 \sqrt [4]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + x^4)^(1/4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 53, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \sqrt [4]{x^2+x^4}-\frac {1}{4} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2 + x^4)^(1/4),x]

[Out]

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

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fricas [B]  time = 1.06, size = 95, normalized size = 1.79 \begin {gather*} \frac {1}{2} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x - \frac {1}{8} \, \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) + \frac {1}{8} \, \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^4 + x^2)^(1/4)*x - 1/8*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x^2)^(3/4))/x) + 1/8*log((2*x^3 + 2*(x^
4 + x^2)^(1/4)*x^2 + 2*sqrt(x^4 + x^2)*x + x + 2*(x^4 + x^2)^(3/4))/x)

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giac [A]  time = 0.15, size = 47, normalized size = 0.89 \begin {gather*} \frac {1}{2} \, x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + \frac {1}{4} \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{8} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*(1/x^2 + 1)^(1/4) + 1/4*arctan((1/x^2 + 1)^(1/4)) + 1/8*log((1/x^2 + 1)^(1/4) + 1) - 1/8*log((1/x^2 +
1)^(1/4) - 1)

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maple [C]  time = 3.61, size = 17, normalized size = 0.32

method	result	size
meijerg	\(\frac {2 x^{\frac {3}{2}} \hypergeom \left (\left [-\frac {1}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x^{2}\right )}{3}\)	\(17\)
trager	\(\frac {x \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-2 \sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{8}-\frac {\ln \left (\frac {2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-2 \sqrt {x^{4}+x^{2}}\, x +2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}-2 x^{3}-x}{x}\right )}{8}\)	\(143\)
risch	\(\frac {x \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2}+\frac {\left (\frac {\ln \left (\frac {2 x^{6}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+5 x^{4}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+4 x^{2}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}+1}{\left (x^{2}+1\right )^{2}}\right )}{8}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+2 x^{6}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+5 x^{4}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+4 x^{2}+1}{\left (x^{2}+1\right )^{2}}\right )}{8}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}+1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{2}+1\right )}\)	\(409\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+x^2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + x^2)^(1/4), x)

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mupad [B]  time = 0.57, size = 29, normalized size = 0.55 \begin {gather*} \frac {2\,x\,{\left (x^4+x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ -x^2\right )}{3\,{\left (x^2+1\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^4)^(1/4),x)

[Out]

(2*x*(x^2 + x^4)^(1/4)*hypergeom([-1/4, 3/4], 7/4, -x^2))/(3*(x^2 + 1)^(1/4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [4]{x^{4} + x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**4 + x**2)**(1/4), x)

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