3.8.94 \(\int \frac {\sqrt [3]{1+2 x+x^2}}{3+x^2} \, dx\)
Optimal. Leaf size=61 \[ \frac {\sqrt [3]{(x+1)^2} \text {RootSum}\left [\text {$\#$1}^6-2 \text {$\#$1}^3+4\& ,\frac {\text {$\#$1}^2 \log \left (\sqrt [3]{x+1}-\text {$\#$1}\right )}{\text {$\#$1}^3-1}\& \right ]}{2 (x+1)^{2/3}} \]
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Rubi [C] time = 0.48, 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
= {970, 712, 50, 55, 617, 204, 31} \begin {gather*} \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}}+\frac {i \left (1-i \sqrt {3}\right )^{2/3} \sqrt [3]{x^2+2 x+1} \tan ^{-1}\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} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{x+1}}{\sqrt [3]{1+i \sqrt {3}}}}{\sqrt {3}}\right )}{2 (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 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 55
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 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 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 712
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 970
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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 [C] time = 0.44, size = 241, normalized size = 3.95 \begin {gather*} \frac {\sqrt [3]{(x+1)^2} \left (\frac {\left (3+i \sqrt {3}\right ) \left (-\log \left (x+i \sqrt {3}\right )+3 \log \left (-\sqrt [3]{x+1}+\sqrt [3]{1-i \sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{x+1}}{\sqrt [3]{1-i \sqrt {3}}}}{\sqrt {3}}\right )\right )}{\sqrt [3]{1-i \sqrt {3}}}+\frac {\left (3-i \sqrt {3}\right ) \left (-\log \left (\sqrt {3}+i x\right )+3 \log \left (-\sqrt [3]{x+1}+\sqrt [3]{1+i \sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{x+1}}{\sqrt [3]{1+i \sqrt {3}}}}{\sqrt {3}}\right )\right )}{\sqrt [3]{1+i \sqrt {3}}}\right )}{12 (x+1)^{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)*(((3 + I*Sqrt[3])*(2*Sqrt[3]*ArcTan[(1 + (2*(1 + x)^(1/3))/(1 - I*Sqrt[3])^(1/3))/Sqrt[3]]
- Log[I*Sqrt[3] + x] + 3*Log[(1 - I*Sqrt[3])^(1/3) - (1 + x)^(1/3)]))/(1 - I*Sqrt[3])^(1/3) + ((3 - I*Sqrt[3])
*(2*Sqrt[3]*ArcTan[(1 + (2*(1 + x)^(1/3))/(1 + I*Sqrt[3])^(1/3))/Sqrt[3]] - Log[Sqrt[3] + I*x] + 3*Log[(1 + I*
Sqrt[3])^(1/3) - (1 + x)^(1/3)]))/(1 + I*Sqrt[3])^(1/3)))/(12*(1 + x)^(2/3))
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IntegrateAlgebraic [A] time = 4.26, 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]
IntegrateAlgebraic[(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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fricas [B] time = 0.74, size = 1132, normalized size = 18.56
result too large to display
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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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}}{x^{2} + 3}\,{d x} \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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maple [B] time = 2.49, size = 2779, normalized size = 45.56
| | |
| method |
result |
size |
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| trager |
\(\text {Expression too large to display}\) |
\(2779\) |
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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]
-1/3*ln(-(12*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2+36*(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+48*RootOf(108*_Z^6+18*
_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x+36*(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)+36*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+6*RootOf(_Z^3+216*RootOf(108*_
Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)*x+4*x*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2+12*(x^2+2*x+1)
^(2/3)+6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)+3*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(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^
3+36)-3*ln(-(12*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2+36*(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+48*RootOf(108*_Z^6+
18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x+36*(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)+36*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+6*RootOf(_Z^3+216*RootOf(10
8*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)*x+4*x*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2+12*(x^2+2*x
+1)^(2/3)+6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)+3*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)-18*RootOf(108*_Z^6+18*_Z^3+1)^4*ln((36*RootOf(108*_Z^6+18*_Z^3+1)^5*x^2+
36*RootOf(108*_Z^6+18*_Z^3+1)^5*x+36*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4*x+36*(x^2+2*x+1)^(1/3)*Roo
tOf(108*_Z^6+18*_Z^3+1)^4+18*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3+3*RootOf(108*_Z^6+18*_Z^3+1)^2*x^2
+6*RootOf(108*_Z^6+18*_Z^3+1)^2*x+3*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)*x+3*RootOf(108*_Z^6+18*_Z^3+1
)^2+3*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)+(x^2+2*x+1)^(2/3))/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z^3+1)^3
+x-1))-2*RootOf(108*_Z^6+18*_Z^3+1)*ln((36*RootOf(108*_Z^6+18*_Z^3+1)^5*x^2+36*RootOf(108*_Z^6+18*_Z^3+1)^5*x+
36*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4*x+36*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4+18*(x^2+
2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3+3*RootOf(108*_Z^6+18*_Z^3+1)^2*x^2+6*RootOf(108*_Z^6+18*_Z^3+1)^2*x+
3*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)*x+3*RootOf(108*_Z^6+18*_Z^3+1)^2+3*(x^2+2*x+1)^(1/3)*RootOf(108
*_Z^6+18*_Z^3+1)+(x^2+2*x+1)^(2/3))/(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((36*RootOf(108*_Z^6+18*_Z^3+1)^5*x^2+144*RootOf(108*_Z^6+18*_Z^3+1)^5*x+18*(x^2+2*x+1)^(1/3)*RootOf(10
8*_Z^6+18*_Z^3+1)^4*x+108*RootOf(108*_Z^6+18*_Z^3+1)^5+18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4+3*Roo
tOf(108*_Z^6+18*_Z^3+1)^2*x^2+12*RootOf(108*_Z^6+18*_Z^3+1)^2*x+9*RootOf(108*_Z^6+18*_Z^3+1)^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((36*RootOf(108*_Z^6+18*_Z^3+1
)^5*x^2+144*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*x+108*RootOf(108*
_Z^6+18*_Z^3+1)^5+18*(x^2+2*x+1)^(1/3)*RootOf(108*_Z^6+18*_Z^3+1)^4+3*RootOf(108*_Z^6+18*_Z^3+1)^2*x^2+12*Root
Of(108*_Z^6+18*_Z^3+1)^2*x+9*RootOf(108*_Z^6+18*_Z^3+1)^2-(x^2+2*x+1)^(2/3))/(1+x)/(12*x*RootOf(108*_Z^6+18*_Z
^3+1)^3+x-1))+3*ln(-(12*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2+72
*(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*RootOf(1
08*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x+216*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^
6+18*_Z^3+1)^3+72*(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)+RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2+6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*
(x^2+2*x+1)^(1/3)*x+24*(x^2+2*x+1)^(2/3)+6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)-
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)+1/6*ln(-(12*RootOf(108*_Z^6+18*_Z^3+1)^3*Ro
otOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*x^2+72*(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*RootOf(108*_Z^6+18*_Z^3+1)^3*RootOf(_Z^3+216*RootOf(108*_Z^6+18
*_Z^3+1)^3+36)^2*x+216*(x^2+2*x+1)^(2/3)*RootOf(108*_Z^6+18*_Z^3+1)^3+72*(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)+RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)^2*
x^2+6*RootOf(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)*x+24*(x^2+2*x+1)^(2/3)+6*RootOf(_Z^3+
216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)*(x^2+2*x+1)^(1/3)-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(_Z^3+216*RootOf(108*_Z^6+18*_Z^3+1)^3+36)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}}{x^{2} + 3}\,{d x} \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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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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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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