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

Optimal. Leaf size=59 \[ \frac {3}{32} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^2}}\right )-\frac {3}{32} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+x^2}}\right )+\frac {1}{16} \sqrt [4]{x^4+x^2} \left (4 x^3+x\right ) \]

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Rubi [B]  time = 0.11, antiderivative size = 125, normalized size of antiderivative = 2.12, number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {2021, 2024, 2032, 329, 331, 298, 203, 206} \begin {gather*} \frac {1}{16} \sqrt [4]{x^4+x^2} x+\frac {3 \left (x^2+1\right )^{3/4} x^{3/2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{32 \left (x^4+x^2\right )^{3/4}}-\frac {3 \left (x^2+1\right )^{3/4} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{32 \left (x^4+x^2\right )^{3/4}}+\frac {1}{4} \sqrt [4]{x^4+x^2} x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(x^2 + x^4)^(1/4),x]

[Out]

(x*(x^2 + x^4)^(1/4))/16 + (x^3*(x^2 + x^4)^(1/4))/4 + (3*x^(3/2)*(1 + x^2)^(3/4)*ArcTan[Sqrt[x]/(1 + x^2)^(1/
4)])/(32*(x^2 + x^4)^(3/4)) - (3*x^(3/2)*(1 + x^2)^(3/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)])/(32*(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 2021

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

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 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 x^2 \sqrt [4]{x^2+x^4} \, dx &=\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}+\frac {1}{8} \int \frac {x^4}{\left (x^2+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {3}{32} \int \frac {x^2}{\left (x^2+x^4\right )^{3/4}} \, dx\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 x^{3/2} \left (1+x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (1+x^2\right )^{3/4}} \, dx}{32 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 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 )}{16 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 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 )}{16 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}-\frac {\left (3 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 )}{32 \left (x^2+x^4\right )^{3/4}}+\frac {\left (3 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 )}{32 \left (x^2+x^4\right )^{3/4}}\\ &=\frac {1}{16} x \sqrt [4]{x^2+x^4}+\frac {1}{4} x^3 \sqrt [4]{x^2+x^4}+\frac {3 x^{3/2} \left (1+x^2\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}-\frac {3 x^{3/2} \left (1+x^2\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{32 \left (x^2+x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 52, normalized size = 0.88 \begin {gather*} \frac {x \sqrt [4]{x^4+x^2} \left (\left (x^2+1\right )^{5/4}-\, _2F_1\left (-\frac {1}{4},\frac {3}{4};\frac {7}{4};-x^2\right )\right )}{4 \sqrt [4]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(x^2 + x^4)^(1/4),x]

[Out]

(x*(x^2 + x^4)^(1/4)*((1 + x^2)^(5/4) - Hypergeometric2F1[-1/4, 3/4, 7/4, -x^2]))/(4*(1 + x^2)^(1/4))

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IntegrateAlgebraic [A]  time = 0.14, size = 59, normalized size = 1.00 \begin {gather*} \frac {1}{16} \left (x+4 x^3\right ) \sqrt [4]{x^2+x^4}+\frac {3}{32} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3}{32} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((x + 4*x^3)*(x^2 + x^4)^(1/4))/16 + (3*ArcTan[x/(x^2 + x^4)^(1/4)])/32 - (3*ArcTanh[x/(x^2 + x^4)^(1/4)])/32

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fricas [B]  time = 2.56, size = 102, normalized size = 1.73 \begin {gather*} \frac {1}{16} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (4 \, x^{3} + x\right )} + \frac {3}{64} \, \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 {3}{64} \, \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^2*(x^4+x^2)^(1/4),x, algorithm="fricas")

[Out]

1/16*(x^4 + x^2)^(1/4)*(4*x^3 + x) + 3/64*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x^2)^(3/4))/x) + 3/64*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 = 0.29, size = 57, normalized size = 0.97 \begin {gather*} \frac {1}{16} \, {\left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right )} x^{4} - \frac {3}{32} \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{64} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {3}{64} \, \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^2*(x^4+x^2)^(1/4),x, algorithm="giac")

[Out]

1/16*((1/x^2 + 1)^(5/4) + 3*(1/x^2 + 1)^(1/4))*x^4 - 3/32*arctan((1/x^2 + 1)^(1/4)) - 3/64*log((1/x^2 + 1)^(1/
4) + 1) + 3/64*log((1/x^2 + 1)^(1/4) - 1)

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maple [C]  time = 3.08, size = 17, normalized size = 0.29

method	result	size
meijerg	\(\frac {2 x^{\frac {7}{2}} \hypergeom \left (\left [-\frac {1}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x^{2}\right )}{7}\)	\(17\)
trager	\(\frac {x \left (4 x^{2}+1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{4}}}{16}-\frac {3 \ln \left (\frac {2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{2}}\, x +2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}+2 x^{3}+x}{x}\right )}{64}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{2}}\, \RootOf \left (\textit {\_Z}^{2}+1\right ) x -2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{2}\right )^{\frac {3}{4}}-2 x^{2} \left (x^{4}+x^{2}\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+1\right ) x}{x}\right )}{64}\)	\(149\)
risch	\(\frac {x \left (4 x^{2}+1\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{16}+\frac {\left (-\frac {3 \ln \left (\frac {2 x^{6}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+5 x^{4}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}+4 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}+2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+4 x^{2}+2 \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}+1}{\left (x^{2}+1\right )^{2}}\right )}{64}+\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{4}+2 x^{6}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {3}{4}}-4 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}} x^{2}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}\, x^{2}+5 x^{4}-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{8}+3 x^{6}+3 x^{4}+x^{2}\right )^{\frac {1}{4}}-2 \sqrt {x^{8}+3 x^{6}+3 x^{4}+x^{2}}+4 x^{2}+1}{\left (x^{2}+1\right )^{2}}\right )}{64}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} \left (x^{2} \left (x^{2}+1\right )^{3}\right )^{\frac {1}{4}}}{x \left (x^{2}+1\right )}\)	\(416\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*x^(7/2)*hypergeom([-1/4,7/4],[11/4],-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}} x^{2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + x^2)^(1/4)*x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,{\left (x^4+x^2\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(x^2 + x^4)^(1/4),x)

[Out]

int(x^2*(x^2 + x^4)^(1/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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