3.1.1 \(\int \frac {x (1+2 x^2)}{\sqrt {1+x^2} (1+x^2+x^4)} \, dx\)

Optimal. Leaf size=106 \[ \frac {1}{4} \sqrt {3} \log \left (x^2-\sqrt {3} \sqrt {x^2+1}+2\right )-\frac {1}{4} \sqrt {3} \log \left (x^2+\sqrt {3} \sqrt {x^2+1}+2\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt {3}-2 \sqrt {x^2+1}\right )+\frac {1}{2} \tan ^{-1}\left (2 \sqrt {x^2+1}+\sqrt {3}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1685, 826, 1169, 634, 618, 204, 628} \begin {gather*} \frac {1}{4} \sqrt {3} \log \left (x^2-\sqrt {3} \sqrt {x^2+1}+2\right )-\frac {1}{4} \sqrt {3} \log \left (x^2+\sqrt {3} \sqrt {x^2+1}+2\right )-\frac {1}{2} \tan ^{-1}\left (\sqrt {3}-2 \sqrt {x^2+1}\right )+\frac {1}{2} \tan ^{-1}\left (2 \sqrt {x^2+1}+\sqrt {3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 1685

Int[(Px_)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[(Px /. x -> Sqrt[x])*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}
, x] && PolyQ[Px, x^2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.20, size = 125, normalized size = 1.18 \begin {gather*} \frac {\sqrt {2 \left (1-i \sqrt {3}\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x^2+1}}{\sqrt {1-i \sqrt {3}}}\right )}{-1+i \sqrt {3}}+\frac {\sqrt {2 \left (1+i \sqrt {3}\right )} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {x^2+1}}{\sqrt {1+i \sqrt {3}}}\right )}{-1-i \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 0.13, size = 81, normalized size = 0.76 \begin {gather*} \frac {1}{2} \left (1-i \sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (1-i \sqrt {3}\right ) \sqrt {x^2+1}\right )+\frac {1}{2} \left (1+i \sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (1+i \sqrt {3}\right ) \sqrt {x^2+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.19, size = 279, normalized size = 2.63 \begin {gather*} \frac {1}{4} \, \sqrt {3} \log \left (32 \, x^{4} + 80 \, x^{2} + 32 \, \sqrt {3} {\left (x^{3} + x\right )} - 16 \, {\left (2 \, x^{3} + \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 32\right ) - \frac {1}{4} \, \sqrt {3} \log \left (32 \, x^{4} + 80 \, x^{2} - 32 \, \sqrt {3} {\left (x^{3} + x\right )} - 16 \, {\left (2 \, x^{3} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 32\right ) - \arctan \left (2 \, \sqrt {2 \, x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} + \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2} {\left (x + \sqrt {x^{2} + 1}\right )} + \sqrt {3} - 2 \, \sqrt {x^{2} + 1}\right ) - \arctan \left (2 \, \sqrt {2 \, x^{4} + 5 \, x^{2} - 2 \, \sqrt {3} {\left (x^{3} + x\right )} - {\left (2 \, x^{3} - \sqrt {3} {\left (2 \, x^{2} + 1\right )} + 4 \, x\right )} \sqrt {x^{2} + 1} + 2} {\left (x + \sqrt {x^{2} + 1}\right )} - \sqrt {3} - 2 \, \sqrt {x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*log(32*x^4 + 80*x^2 + 32*sqrt(3)*(x^3 + x) - 16*(2*x^3 + sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1)
+ 32) - 1/4*sqrt(3)*log(32*x^4 + 80*x^2 - 32*sqrt(3)*(x^3 + x) - 16*(2*x^3 - sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x
^2 + 1) + 32) - arctan(2*sqrt(2*x^4 + 5*x^2 + 2*sqrt(3)*(x^3 + x) - (2*x^3 + sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x
^2 + 1) + 2)*(x + sqrt(x^2 + 1)) + sqrt(3) - 2*sqrt(x^2 + 1)) - arctan(2*sqrt(2*x^4 + 5*x^2 - 2*sqrt(3)*(x^3 +
x) - (2*x^3 - sqrt(3)*(2*x^2 + 1) + 4*x)*sqrt(x^2 + 1) + 2)*(x + sqrt(x^2 + 1)) - sqrt(3) - 2*sqrt(x^2 + 1))

________________________________________________________________________________________

giac [A]  time = 0.65, size = 80, normalized size = 0.75 \begin {gather*} -\frac {1}{4} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) + \frac {1}{4} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} \sqrt {x^{2} + 1} + 2\right ) + \frac {1}{2} \, \arctan \left (\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) + \frac {1}{2} \, \arctan \left (-\sqrt {3} + 2 \, \sqrt {x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.03, size = 296, normalized size = 2.79 \begin {gather*} -\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \sqrt {3}}{2}\right )+\arctan \left (\frac {\sqrt {\frac {2 \left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+2}\, \left (x +1\right )}{\left (\frac {\left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+1\right ) \left (-x +1\right )}\right )\right )}{4 \sqrt {\frac {\frac {\left (x +1\right )^{2}}{\left (-x +1\right )^{2}}+1}{\left (\frac {x +1}{-x +1}+1\right )^{2}}}\, \left (\frac {x +1}{-x +1}+1\right )}-\frac {\sqrt {2}\, \sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \sqrt {3}}{2}\right )+\arctan \left (\frac {\sqrt {\frac {2 \left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+2}\, \left (x -1\right )}{\left (\frac {\left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+1\right ) \left (-x -1\right )}\right )\right )}{4 \sqrt {\frac {\frac {\left (x -1\right )^{2}}{\left (-x -1\right )^{2}}+1}{\left (\frac {x -1}{-x -1}+1\right )^{2}}}\, \left (\frac {x -1}{-x -1}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*2^(1/2)*(2*(x+1)^2/(1-x)^2+2)^(1/2)*(3^(1/2)*arctanh(1/2*(2*(x+1)^2/(1-x)^2+2)^(1/2)*3^(1/2))+arctan(1/((
x+1)^2/(1-x)^2+1)*(2*(x+1)^2/(1-x)^2+2)^(1/2)*(x+1)/(1-x)))/(((x+1)^2/(1-x)^2+1)/((x+1)/(1-x)+1)^2)^(1/2)/((x+
1)/(1-x)+1)-1/4*2^(1/2)*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*(3^(1/2)*arctanh(1/2*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*3^(1/2)
)+arctan(1/((x-1)^2/(-1-x)^2+1)*(2*(x-1)^2/(-1-x)^2+2)^(1/2)*(x-1)/(-1-x)))/(((x-1)^2/(-1-x)^2+1)/((x-1)/(-1-x
)+1)^2)^(1/2)/((x-1)/(-1-x)+1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.85, size = 397, normalized size = 3.75 \begin {gather*} \frac {\left (\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (\frac {x}{2}+\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}+1+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+1+\sqrt {3}\,1{}\mathrm {i}\right )}+\frac {\left (\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (1+\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}-\frac {x}{2}+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-1+\sqrt {3}\,1{}\mathrm {i}\right )}+\frac {\left (\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (\frac {x}{2}+\left (\frac {\sqrt {3}}{2}-\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}+1-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3-1+\sqrt {3}\,1{}\mathrm {i}\right )}+\frac {\left (\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-\ln \left (1+\left (\frac {\sqrt {3}}{2}+\frac {1}{2}{}\mathrm {i}\right )\,\sqrt {x^2+1}-\frac {x}{2}-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\right )\,\left (2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{\sqrt {{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2+1}\,\left (4\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^3+1+\sqrt {3}\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((log(x - (3^(1/2)*1i)/2 - 1/2) - log(x/2 + (3^(1/2)/2 + 1i/2)*(x^2 + 1)^(1/2) + (3^(1/2)*x*1i)/2 + 1))*((3^(1
/2)*1i)/2 + 2*((3^(1/2)*1i)/2 + 1/2)^3 + 1/2))/((((3^(1/2)*1i)/2 + 1/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)
*1i)/2 + 1/2)^3 + 1)) + ((log(x - (3^(1/2)*1i)/2 + 1/2) - log((3^(1/2)/2 - 1i/2)*(x^2 + 1)^(1/2) - x/2 + (3^(1
/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 - 1/2)^3 - 1/2))/((((3^(1/2)*1i)/2 - 1/2)^2 + 1)^(1/2)*(
3^(1/2)*1i + 4*((3^(1/2)*1i)/2 - 1/2)^3 - 1)) + ((log(x + (3^(1/2)*1i)/2 - 1/2) - log(x/2 + (3^(1/2)/2 - 1i/2)
*(x^2 + 1)^(1/2) - (3^(1/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 - 1/2)^3 - 1/2))/((((3^(1/2)*1i)
/2 - 1/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)*1i)/2 - 1/2)^3 - 1)) + ((log(x + (3^(1/2)*1i)/2 + 1/2) - log(
(3^(1/2)/2 + 1i/2)*(x^2 + 1)^(1/2) - x/2 - (3^(1/2)*x*1i)/2 + 1))*((3^(1/2)*1i)/2 + 2*((3^(1/2)*1i)/2 + 1/2)^3
+ 1/2))/((((3^(1/2)*1i)/2 + 1/2)^2 + 1)^(1/2)*(3^(1/2)*1i + 4*((3^(1/2)*1i)/2 + 1/2)^3 + 1))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(2*x**2 + 1)/(sqrt(x**2 + 1)*(x**2 - x + 1)*(x**2 + x + 1)), x)

________________________________________________________________________________________