3.8.94 \(\int \frac {\sqrt [3]{1+2 x+x^2}}{3+x^2} \, dx\) [794]

Optimal. Leaf size=61 \[ \frac {\sqrt [3]{(1+x)^2} \text {RootSum}\left [4-2 \text {$\#$1}^3+\text {$\#$1}^6\& ,\frac {\log \left (\sqrt [3]{1+x}-\text {$\#$1}\right ) \text {$\#$1}^2}{-1+\text {$\#$1}^3}\& \right ]}{2 (1+x)^{2/3}} \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 0.32, antiderivative size = 417, normalized size of antiderivative = 6.84, number of steps used = 13, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {984, 726, 52, 57, 631, 210, 31} \begin {gather*} \frac {i \left (1-i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{x+1}}{\sqrt [3]{1-i \sqrt {3}}}}{\sqrt {3}}\right )}{2 (x+1)^{2/3}}-\frac {i \left (1+i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{x+1}}{\sqrt [3]{1+i \sqrt {3}}}}{\sqrt {3}}\right )}{2 (x+1)^{2/3}}+\frac {i \left (1+i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \log \left (\sqrt {3}+i x\right )}{4 \sqrt {3} (x+1)^{2/3}}-\frac {i \left (1-i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \log \left (x+i \sqrt {3}\right )}{4 \sqrt {3} (x+1)^{2/3}}+\frac {i \sqrt {3} \left (1-i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \log \left (-\sqrt [3]{2} \sqrt [3]{x+1}+\sqrt [3]{2 \left (1-i \sqrt {3}\right )}\right )}{4 (x+1)^{2/3}}-\frac {i \sqrt {3} \left (1+i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \log \left (-\sqrt [3]{2} \sqrt [3]{x+1}+\sqrt [3]{2 \left (1+i \sqrt {3}\right )}\right )}{4 (x+1)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(1 - I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*ArcTan[(1 + (2*(1 + x)^(1/3))/(1 - I*Sqrt[3])^(1/3))/Sqrt[3
]])/(1 + x)^(2/3) - ((I/2)*(1 + I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*ArcTan[(1 + (2*(1 + x)^(1/3))/(1 + I*Sq
rt[3])^(1/3))/Sqrt[3]])/(1 + x)^(2/3) + ((I/4)*(1 + I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*Log[Sqrt[3] + I*x])
/(Sqrt[3]*(1 + x)^(2/3)) - ((I/4)*(1 - I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*Log[I*Sqrt[3] + x])/(Sqrt[3]*(1
+ x)^(2/3)) + ((I/4)*Sqrt[3]*(1 - I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*Log[(2*(1 - I*Sqrt[3]))^(1/3) - 2^(1/
3)*(1 + x)^(1/3)])/(1 + x)^(2/3) - ((I/4)*Sqrt[3]*(1 + I*Sqrt[3])^(2/3)*(1 + 2*x + x^2)^(1/3)*Log[(2*(1 + I*Sq
rt[3]))^(1/3) - 2^(1/3)*(1 + x)^(1/3)])/(1 + x)^(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 52

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 57

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

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 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 726

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

Rule 984

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_.), x_Symbol] :> Dist[(a + b*x + c*x^2)^F
racPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; Free
Q[{a, b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 61, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{(1+x)^2} \text {RootSum}\left [4-2 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {\log \left (\sqrt [3]{1+x}-\text {$\#$1}\right ) \text {$\#$1}^2}{-1+\text {$\#$1}^3}\&\right ]}{2 (1+x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((1 + x)^2)^(1/3)*RootSum[4 - 2*#1^3 + #1^6 & , (Log[(1 + x)^(1/3) - #1]*#1^2)/(-1 + #1^3) & ])/(2*(1 + x)^(2
/3))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 1.
time = 3.50, size = 2244, normalized size = 36.79

method	result	size
trager	\(\text {Expression too large to display}\)	\(2244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*ln(-(-108*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*RootOf(108*_Z^6+18*_Z^3+1)^6*x^2-108*RootOf(_Z
^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*RootOf(108*_Z^6+18*_Z^3+1)^6*x-21*x^2*RootOf(108*_Z^6+18*_Z^3+1)^3*R
ootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^
3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*x-12*x*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*
_Z^3+1)^3+36)^2+108*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^
3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)+9*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(
108*_Z^6+18*_Z^3+1)^3+36)^2-RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2-3*(x^2+2*x+1)^(1/3)*RootOf(
_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*x+12*(x^2+2*x+1)^(2/3)-3*(x^2+2*x+1)^(1/3)*RootOf(_Z^3+216*RootOf(10
8*_Z^6+18*_Z^3+1)^3+36)+RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2)/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^
3+1)^3+x+1))*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)-1/3*ln(-(-108*RootO
f(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*RootOf(108*_Z^6+18*_Z^3+1)^6*x^2-108*RootOf(_Z^3+216*RootOf(108*
_Z^6+18*_Z^3+1)^3+36)^2*RootOf(108*_Z^6+18*_Z^3+1)^6*x-21*x^2*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*Roo
tOf(108*_Z^6+18*_Z^3+1)^3+36)^2-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_
Z^6+18*_Z^3+1)^3+36)*x-12*x*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2+10
8*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3
+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)+9*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1
)^3+36)^2-RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2-3*(x^2+2*x+1)^(1/3)*RootOf(_Z^3+216*RootOf(10
8*_Z^6+18*_Z^3+1)^3+36)*x+12*(x^2+2*x+1)^(2/3)-3*(x^2+2*x+1)^(1/3)*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^
3+36)+RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2)/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^3+1)^3+x+1))*RootO
f(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)+1/6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*ln(-(-x^2*Roo
tOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2+6*(x^2+2*x+1)^(1/3)*RootOf(108*_Z
^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*x-4*x*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^
3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2+6*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*Root
Of(108*_Z^6+18*_Z^3+1)^3+36)-3*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2
+2*(x^2+2*x+1)^(2/3))/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^3+1)^3+x+1))+RootOf(108*_Z^6+18*_Z^3+1)*ln(-(18*RootOf
(108*_Z^6+18*_Z^3+1)^5*x^2-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4*x+72*RootOf(108*_Z^6+18*_Z^3+1)^5
*x-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4+54*RootOf(108*_Z^6+18*_Z^3+1)^5+3*RootOf(108*_Z^6+18*_Z^3
+1)^2*x^2-3*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)*x+12*x*RootOf(108*_Z^6+18*_Z^3+1)^2+(x^2+2*x+1)^(2/3)
-3*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)+9*RootOf(108*_Z^6+18*_Z^3+1)^2)/(1+x)/(12*x*RootOf(108*_Z^6+18
*_Z^3+1)^3+x-1))+18*RootOf(108*_Z^6+18*_Z^3+1)^4*ln((648*RootOf(108*_Z^6+18*_Z^3+1)^8*x^2+648*RootOf(108*_Z^6+
18*_Z^3+1)^8*x+90*RootOf(108*_Z^6+18*_Z^3+1)^5*x^2-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4*x+144*Roo
tOf(108*_Z^6+18*_Z^3+1)^5*x+18*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3-18*(x^2+2*x+1)^(1/3)*RootOf(108*
_Z^6+18*_Z^3+1)^4+54*RootOf(108*_Z^6+18*_Z^3+1)^5+3*RootOf(108*_Z^6+18*_Z^3+1)^2*x^2+6*x*RootOf(108*_Z^6+18*_Z
^3+1)^2+(x^2+2*x+1)^(2/3)+3*RootOf(108*_Z^6+18*_Z^3+1)^2)/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^3+1)^3+x-1))+RootO
f(108*_Z^6+18*_Z^3+1)*ln((648*RootOf(108*_Z^6+18*_Z^3+1)^8*x^2+648*RootOf(108*_Z^6+18*_Z^3+1)^8*x+90*RootOf(10
8*_Z^6+18*_Z^3+1)^5*x^2-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4*x+144*RootOf(108*_Z^6+18*_Z^3+1)^5*x
+18*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3-18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4+54*RootOf
(108*_Z^6+18*_Z^3+1)^5+3*RootOf(108*_Z^6+18*_Z^3+1)^2*x^2+6*x*RootOf(108*_Z^6+18*_Z^3+1)^2+(x^2+2*x+1)^(2/3)+3
*RootOf(108*_Z^6+18*_Z^3+1)^2)/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^3+1)^3+x-1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 3 vs. order 1.
time = 0.58, size = 1132, normalized size = 18.56 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/36*36^(2/3)*12^(1/6)*cos(2/3*arctan(sqrt(3) + 2))*log(-144*(36^(2/3)*12^(1/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3)*
(x + 1)*sin(2/3*arctan(sqrt(3) + 2)) + 3*36^(2/3)*12^(1/6)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*cos(2/3*arctan(sqrt(3
) + 2)) - 3*36^(1/3)*12^(1/3)*(x^2 + 2*x + 1) - 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1)) - 1/9*36^(2/3)*12^(
1/6)*arctan(-1/108*(6*36^(1/3)*12^(5/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3)*cos(2/3*arctan(sqrt(3) + 2)) - 18*(24*(x
+ 1)*cos(2/3*arctan(sqrt(3) + 2)) - 36^(1/3)*12^(5/6)*(x^2 + 2*x + 1)^(1/3))*sin(2/3*arctan(sqrt(3) + 2)) - 1
08*sqrt(3)*(x + 1) - (36^(1/3)*12^(5/6)*sqrt(3)*(x + 1)*cos(2/3*arctan(sqrt(3) + 2)) + 3*36^(1/3)*12^(5/6)*(x
+ 1)*sin(2/3*arctan(sqrt(3) + 2)))*sqrt(-(36^(2/3)*12^(1/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*sin(2/3*arct
an(sqrt(3) + 2)) + 3*36^(2/3)*12^(1/6)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*cos(2/3*arctan(sqrt(3) + 2)) - 3*36^(1/3)
*12^(1/3)*(x^2 + 2*x + 1) - 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1)))/(4*(x + 1)*cos(2/3*arctan(sqrt(3) + 2)
)^2 - 3*x - 3))*sin(2/3*arctan(sqrt(3) + 2)) - 1/18*(36^(2/3)*12^(1/6)*sqrt(3)*cos(2/3*arctan(sqrt(3) + 2)) +
36^(2/3)*12^(1/6)*sin(2/3*arctan(sqrt(3) + 2)))*arctan(1/108*(6*36^(1/3)*12^(5/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3
)*cos(2/3*arctan(sqrt(3) + 2)) - 18*(24*(x + 1)*cos(2/3*arctan(sqrt(3) + 2)) + 36^(1/3)*12^(5/6)*(x^2 + 2*x +
1)^(1/3))*sin(2/3*arctan(sqrt(3) + 2)) + 108*sqrt(3)*(x + 1) - (36^(1/3)*12^(5/6)*sqrt(3)*(x + 1)*cos(2/3*arct
an(sqrt(3) + 2)) - 3*36^(1/3)*12^(5/6)*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)))*sqrt(-(36^(2/3)*12^(1/6)*sqrt(3)*
(x^2 + 2*x + 1)^(1/3)*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)) - 3*36^(2/3)*12^(1/6)*(x^2 + 2*x + 1)^(1/3)*(x + 1)
*cos(2/3*arctan(sqrt(3) + 2)) - 3*36^(1/3)*12^(1/3)*(x^2 + 2*x + 1) - 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1
)))/(4*(x + 1)*cos(2/3*arctan(sqrt(3) + 2))^2 - 3*x - 3)) + 1/18*(36^(2/3)*12^(1/6)*sqrt(3)*cos(2/3*arctan(sqr
t(3) + 2)) - 36^(2/3)*12^(1/6)*sin(2/3*arctan(sqrt(3) + 2)))*arctan(1/216*(36^(1/3)*12^(5/6)*sqrt(3)*(x + 1)*s
qrt((2*36^(2/3)*12^(1/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)) + 3*36^(1/3)*12^(1
/3)*(x^2 + 2*x + 1) + 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1)) - 6*36^(1/3)*12^(5/6)*sqrt(3)*(x^2 + 2*x + 1)
^(1/3) - 216*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)))/((x + 1)*cos(2/3*arctan(sqrt(3) + 2)))) - 1/72*(36^(2/3)*12
^(1/6)*sqrt(3)*sin(2/3*arctan(sqrt(3) + 2)) + 36^(2/3)*12^(1/6)*cos(2/3*arctan(sqrt(3) + 2)))*log(576*(2*36^(2
/3)*12^(1/6)*sqrt(3)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)) + 3*36^(1/3)*12^(1/3)*(x^2 + 2
*x + 1) + 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1)) + 1/72*(36^(2/3)*12^(1/6)*sqrt(3)*sin(2/3*arctan(sqrt(3)
+ 2)) - 36^(2/3)*12^(1/6)*cos(2/3*arctan(sqrt(3) + 2)))*log(-576*(36^(2/3)*12^(1/6)*sqrt(3)*(x^2 + 2*x + 1)^(1
/3)*(x + 1)*sin(2/3*arctan(sqrt(3) + 2)) - 3*36^(2/3)*12^(1/6)*(x^2 + 2*x + 1)^(1/3)*(x + 1)*cos(2/3*arctan(sq
rt(3) + 2)) - 3*36^(1/3)*12^(1/3)*(x^2 + 2*x + 1) - 36*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (x^2+2\,x+1\right )}^{1/3}}{x^2+3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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