3.20.42 \(\int \frac {x \sqrt [3]{x^2+x^4}}{1+2 x^2} \, dx\)

Optimal. Leaf size=136 \[ \frac {3}{8} \sqrt [3]{x^4+x^2}-\frac {\log \left (2^{2/3} \sqrt [3]{x^4+x^2}+1\right )}{8\ 2^{2/3}}+\frac {\log \left (-2 \sqrt [3]{2} \left (x^4+x^2\right )^{2/3}+2^{2/3} \sqrt [3]{x^4+x^2}-1\right )}{16\ 2^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{x^4+x^2}}{\sqrt {3}}\right )}{8\ 2^{2/3}} \]

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Rubi [A]  time = 0.21, antiderivative size = 109, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2034, 694, 266, 50, 58, 617, 204, 31} \begin {gather*} \frac {3 \sqrt [3]{\left (2 x^2+1\right )^2-1}}{8\ 2^{2/3}}+\frac {\log \left (2 x^2+1\right )}{8\ 2^{2/3}}-\frac {3 \log \left (\sqrt [3]{\left (2 x^2+1\right )^2-1}+1\right )}{16\ 2^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [3]{\left (2 x^2+1\right )^2-1}}{\sqrt {3}}\right )}{8\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(x^2 + x^4)^(1/3))/(1 + 2*x^2),x]

[Out]

(3*(-1 + (1 + 2*x^2)^2)^(1/3))/(8*2^(2/3)) + (Sqrt[3]*ArcTan[(1 - 2*(-1 + (1 + 2*x^2)^2)^(1/3))/Sqrt[3]])/(8*2
^(2/3)) + Log[1 + 2*x^2]/(8*2^(2/3)) - (3*Log[1 + (-1 + (1 + 2*x^2)^2)^(1/3)])/(16*2^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, -Sim
p[Log[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (Dist[3/(2*b*q), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d
*x)^(1/3)], x] + Dist[3/(2*b*q^2), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x
] && NegQ[(b*c - a*d)/b]

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 266

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

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 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 2034

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

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.42 \begin {gather*} \frac {3 \sqrt [3]{x^4+x^2} \left (\sqrt [3]{x^2+1}-F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-x^2,-2 x^2\right )\right )}{8 \sqrt [3]{x^2+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(x^2 + x^4)^(1/3))/(1 + 2*x^2),x]

[Out]

(3*(x^2 + x^4)^(1/3)*((1 + x^2)^(1/3) - AppellF1[1/3, 2/3, 1, 4/3, -x^2, -2*x^2]))/(8*(1 + x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.25, size = 136, normalized size = 1.00 \begin {gather*} \frac {3}{8} \sqrt [3]{x^2+x^4}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2\ 2^{2/3} \sqrt [3]{x^2+x^4}}{\sqrt {3}}\right )}{8\ 2^{2/3}}-\frac {\log \left (1+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{8\ 2^{2/3}}+\frac {\log \left (-1+2^{2/3} \sqrt [3]{x^2+x^4}-2 \sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{16\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*(x^2 + x^4)^(1/3))/(1 + 2*x^2),x]

[Out]

(3*(x^2 + x^4)^(1/3))/8 + (Sqrt[3]*ArcTan[1/Sqrt[3] - (2*2^(2/3)*(x^2 + x^4)^(1/3))/Sqrt[3]])/(8*2^(2/3)) - Lo
g[1 + 2^(2/3)*(x^2 + x^4)^(1/3)]/(8*2^(2/3)) + Log[-1 + 2^(2/3)*(x^2 + x^4)^(1/3) - 2*2^(1/3)*(x^2 + x^4)^(2/3
)]/(16*2^(2/3))

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fricas [A]  time = 0.50, size = 128, normalized size = 0.94 \begin {gather*} \frac {1}{16} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} - 4^{\frac {1}{3}}\right )}\right ) - \frac {1}{64} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}\right ) + \frac {1}{32} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}\right ) + \frac {3}{8} \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="fricas")

[Out]

1/16*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*4^(2/3)*(-1)^(2/3)*(x^4 + x^2)^(1/3) - 4^(1/3)))
- 1/64*4^(2/3)*(-1)^(1/3)*log(4^(2/3)*(-1)^(1/3)*(x^4 + x^2)^(1/3) + 4^(1/3)*(-1)^(2/3) + 4*(x^4 + x^2)^(2/3)
) + 1/32*4^(2/3)*(-1)^(1/3)*log(-4^(2/3)*(-1)^(1/3) + 4*(x^4 + x^2)^(1/3)) + 3/8*(x^4 + x^2)^(1/3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="giac")

[Out]

integrate((x^4 + x^2)^(1/3)*x/(2*x^2 + 1), x)

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maple [C]  time = 22.48, size = 1334, normalized size = 9.81

method	result	size
risch	\(\text {Expression too large to display}\)	\(1334\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^4+x^2)^(1/3)/(2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

3/8*(x^2*(x^2+1))^(1/3)+(1/16*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*ln(-(1698*RootOf(RootOf(_Z^3+2)^
2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3*x^6-524*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z
^3+2)^2*x^6+3396*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3*x^4-1048*RootOf(RootOf(_Z^3+
2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2*x^4-27168*RootOf(_Z^3+2)*x^6+8384*RootOf(RootOf(_Z^3+2)^2+_Z*R
ootOf(_Z^3+2)+_Z^2)*x^6-1698*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3*x^2+524*RootOf(R
ootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2*x^2+18048*RootOf(_Z^3+2)^2*(x^5+2*x^3+x)^(1/3)*x^3+
30186*RootOf(_Z^3+2)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*(x^5+2*x^3+x)^(1/3)*x^3-9024*(x^5+2*x^3+x
)^(2/3)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^2-54336*RootOf(_Z^3+2)*x^4+16768*RootOf
(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*x^4-3396*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3
+2)^3+1048*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2+18048*RootOf(_Z^3+2)^2*(x^5+2*x^
3+x)^(1/3)*x+30186*(x^5+2*x^3+x)^(1/3)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)*x-21225*
RootOf(_Z^3+2)*x^2+6550*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*x^2-12138*(x^5+2*x^3+x)^(2/3)+5943*Roo
tOf(_Z^3+2)-1834*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2))/(2*x^2+1)^2/(x^2+1))+1/16*RootOf(_Z^3+2)*ln(
-(524*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3*x^6-1698*RootOf(RootOf(_Z^3+2)^2+_Z*Roo
tOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2*x^6+1048*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3
*x^4-3396*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2*x^4+7336*RootOf(_Z^3+2)*x^6-23772
*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*x^6-524*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootO
f(_Z^3+2)^3*x^2+1698*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2*x^2-18048*RootOf(_Z^3+
2)^2*(x^5+2*x^3+x)^(1/3)*x^3+12138*RootOf(_Z^3+2)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*(x^5+2*x^3+x
)^(1/3)*x^3+9024*(x^5+2*x^3+x)^(2/3)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^2+14672*Ro
otOf(_Z^3+2)*x^4-47544*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*x^4-1048*RootOf(RootOf(_Z^3+2)^2+_Z*Roo
tOf(_Z^3+2)+_Z^2)*RootOf(_Z^3+2)^3+3396*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)^2*RootOf(_Z^3+2)^2-180
48*RootOf(_Z^3+2)^2*(x^5+2*x^3+x)^(1/3)*x+12138*(x^5+2*x^3+x)^(1/3)*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+
_Z^2)*RootOf(_Z^3+2)*x+7598*RootOf(_Z^3+2)*x^2-24621*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2)*x^2-30186
*(x^5+2*x^3+x)^(2/3)+262*RootOf(_Z^3+2)-849*RootOf(RootOf(_Z^3+2)^2+_Z*RootOf(_Z^3+2)+_Z^2))/(2*x^2+1)^2/(x^2+
1)))/x*(x^2*(x^2+1))^(1/3)*(x*(x^2+1)^2)^(1/3)/(x^2+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x}{2 \, x^{2} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^4+x^2)^(1/3)/(2*x^2+1),x, algorithm="maxima")

[Out]

integrate((x^4 + x^2)^(1/3)*x/(2*x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (x^4+x^2\right )}^{1/3}}{2\,x^2+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(x^2 + x^4)^(1/3))/(2*x^2 + 1),x)

[Out]

int((x*(x^2 + x^4)^(1/3))/(2*x^2 + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt [3]{x^{2} \left (x^{2} + 1\right )}}{2 x^{2} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**4+x**2)**(1/3)/(2*x**2+1),x)

[Out]

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

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