3.19 \(\int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{1+x^2+x^4} \, dx\)

Optimal. Leaf size=151 \[ -\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g-i)}{2 \sqrt {3}}+\frac {1}{4} (g-i) \log \left (x^4+x^2+1\right )+h x+\frac {i x^2}{2} \]

[Out]

h*x+1/2*i*x^2-1/4*(d-f)*ln(x^2-x+1)+1/4*(d-f)*ln(x^2+x+1)+1/4*(g-i)*ln(x^4+x^2+1)-1/6*(d+f-2*h)*arctan(1/3*(1-
2*x)*3^(1/2))*3^(1/2)+1/6*(d+f-2*h)*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/6*(2*e-g-i)*arctan(1/3*(2*x^2+1)*3^(
1/2))*3^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1673, 1676, 1169, 634, 618, 204, 628, 1663, 1657} \[ -\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g-i)}{2 \sqrt {3}}+\frac {1}{4} (g-i) \log \left (x^4+x^2+1\right )+h x+\frac {i x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(1 + x^2 + x^4),x]

[Out]

h*x + (i*x^2)/2 - ((d + f - 2*h)*ArcTan[(1 - 2*x)/Sqrt[3]])/(2*Sqrt[3]) + ((d + f - 2*h)*ArcTan[(1 + 2*x)/Sqrt
[3]])/(2*Sqrt[3]) + ((2*e - g - i)*ArcTan[(1 + 2*x^2)/Sqrt[3]])/(2*Sqrt[3]) - ((d - f)*Log[1 - x + x^2])/4 + (
(d - f)*Log[1 + x + x^2])/4 + ((g - i)*Log[1 + x^2 + x^4])/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 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 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2+g x^3+h x^4+19 x^5}{1+x^2+x^4} \, dx &=\int \frac {x \left (e+g x^2+19 x^4\right )}{1+x^2+x^4} \, dx+\int \frac {d+f x^2+h x^4}{1+x^2+x^4} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {e+g x+19 x^2}{1+x+x^2} \, dx,x,x^2\right )+\int \left (h+\frac {d-h+(f-h) x^2}{1+x^2+x^4}\right ) \, dx\\ &=h x+\frac {1}{2} \operatorname {Subst}\left (\int \left (19-\frac {19-e+(19-g) x}{1+x+x^2}\right ) \, dx,x,x^2\right )+\int \frac {d-h+(f-h) x^2}{1+x^2+x^4} \, dx\\ &=h x+\frac {19 x^2}{2}+\frac {1}{2} \int \frac {d-h-(d-f) x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {d-h+(d-f) x}{1+x+x^2} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {19-e+(19-g) x}{1+x+x^2} \, dx,x,x^2\right )\\ &=h x+\frac {19 x^2}{2}+\frac {1}{4} (d-f) \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{4} (-d+f) \int \frac {-1+2 x}{1-x+x^2} \, dx-\frac {1}{4} (19-g) \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,x^2\right )-\frac {1}{4} (19-2 e+g) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )+\frac {1}{4} (d+f-2 h) \int \frac {1}{1-x+x^2} \, dx+\frac {1}{4} (d+f-2 h) \int \frac {1}{1+x+x^2} \, dx\\ &=h x+\frac {19 x^2}{2}-\frac {1}{4} (d-f) \log \left (1-x+x^2\right )+\frac {1}{4} (d-f) \log \left (1+x+x^2\right )-\frac {1}{4} (19-g) \log \left (1+x^2+x^4\right )-\frac {1}{2} (-19+2 e-g) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right )+\frac {1}{2} (-d-f+2 h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{2} (-d-f+2 h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=h x+\frac {19 x^2}{2}-\frac {(d+f-2 h) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {(d+f-2 h) \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {(19-2 e+g) \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (1-x+x^2\right )+\frac {1}{4} (d-f) \log \left (1+x+x^2\right )-\frac {1}{4} (19-g) \log \left (1+x^2+x^4\right )\\ \end {align*}

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Mathematica [C]  time = 0.58, size = 187, normalized size = 1.24 \[ \frac {1}{12} \left (\left (1+i \sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right ) \left (2 \sqrt {3} d-\left (\sqrt {3}+3 i\right ) f-\left (\sqrt {3}-3 i\right ) h\right )+\left (\sqrt {3}+i\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right ) \left (-2 i \sqrt {3} d+\left (3+i \sqrt {3}\right ) f+i \left (\sqrt {3}+3 i\right ) h\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3}}{2 x^2+1}\right ) (2 e-g-i)+3 (g-i) \log \left (x^4+x^2+1\right )+6 x (2 h+i x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(1 + x^2 + x^4),x]

[Out]

(6*x*(2*h + i*x) + (1 + I*Sqrt[3])*(2*Sqrt[3]*d - (3*I + Sqrt[3])*f - (-3*I + Sqrt[3])*h)*ArcTan[((-I + Sqrt[3
])*x)/2] + (I + Sqrt[3])*((-2*I)*Sqrt[3]*d + (3 + I*Sqrt[3])*f + I*(3*I + Sqrt[3])*h)*ArcTan[((I + Sqrt[3])*x)
/2] - 2*Sqrt[3]*(2*e - g - i)*ArcTan[Sqrt[3]/(1 + 2*x^2)] + 3*(g - i)*Log[1 + x^2 + x^4])/12

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fricas [A]  time = 17.27, size = 106, normalized size = 0.70 \[ \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f + g - 2 \, h + i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f - g - 2 \, h - i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4+x^2+1),x, algorithm="fricas")

[Out]

1/2*i*x^2 + 1/6*sqrt(3)*(d - 2*e + f + g - 2*h + i)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(d + 2*e + f -
g - 2*h - i)*arctan(1/3*sqrt(3)*(2*x - 1)) + h*x + 1/4*(d - f + g - i)*log(x^2 + x + 1) - 1/4*(d - f - g + i)
*log(x^2 - x + 1)

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giac [A]  time = 0.31, size = 108, normalized size = 0.72 \[ \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d + f + g - 2 \, h + i - 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + f - g - 2 \, h - i + 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4+x^2+1),x, algorithm="giac")

[Out]

1/2*i*x^2 + 1/6*sqrt(3)*(d + f + g - 2*h + i - 2*e)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(d + f - g - 2
*h - i + 2*e)*arctan(1/3*sqrt(3)*(2*x - 1)) + h*x + 1/4*(d - f + g - i)*log(x^2 + x + 1) - 1/4*(d - f - g + i)
*log(x^2 - x + 1)

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maple [B]  time = 0.01, size = 303, normalized size = 2.01 \[ \frac {i \,x^{2}}{2}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {d \ln \left (x^{2}-x +1\right )}{4}+\frac {d \ln \left (x^{2}+x +1\right )}{4}-\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {f \ln \left (x^{2}-x +1\right )}{4}-\frac {f \ln \left (x^{2}+x +1\right )}{4}+\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {g \ln \left (x^{2}-x +1\right )}{4}+\frac {g \ln \left (x^{2}+x +1\right )}{4}+h x -\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, i \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, i \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {i \ln \left (x^{2}-x +1\right )}{4}-\frac {i \ln \left (x^{2}+x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4+x^2+1),x)

[Out]

1/2*i*x^2+h*x+1/4*d*ln(x^2+x+1)-1/4*f*ln(x^2+x+1)+1/4*g*ln(x^2+x+1)-1/4*ln(x^2+x+1)*i+1/6*3^(1/2)*d*arctan(1/3
*(2*x+1)*3^(1/2))-1/3*3^(1/2)*e*arctan(1/3*(2*x+1)*3^(1/2))+1/6*3^(1/2)*f*arctan(1/3*(2*x+1)*3^(1/2))+1/6*3^(1
/2)*g*arctan(1/3*(2*x+1)*3^(1/2))-1/3*3^(1/2)*h*arctan(1/3*(2*x+1)*3^(1/2))+1/6*3^(1/2)*arctan(1/3*(2*x+1)*3^(
1/2))*i+1/4*g*ln(x^2-x+1)-1/4*ln(x^2-x+1)*i+1/4*f*ln(x^2-x+1)-1/4*d*ln(x^2-x+1)+1/6*3^(1/2)*d*arctan(1/3*(2*x-
1)*3^(1/2))+1/3*3^(1/2)*e*arctan(1/3*(2*x-1)*3^(1/2))+1/6*3^(1/2)*f*arctan(1/3*(2*x-1)*3^(1/2))-1/6*3^(1/2)*g*
arctan(1/3*(2*x-1)*3^(1/2))-1/3*3^(1/2)*h*arctan(1/3*(2*x-1)*3^(1/2))-1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*
i

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maxima [A]  time = 2.37, size = 106, normalized size = 0.70 \[ \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f + g - 2 \, h + i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f - g - 2 \, h - i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x^5+h*x^4+g*x^3+f*x^2+e*x+d)/(x^4+x^2+1),x, algorithm="maxima")

[Out]

1/2*i*x^2 + 1/6*sqrt(3)*(d - 2*e + f + g - 2*h + i)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(d + 2*e + f -
g - 2*h - i)*arctan(1/3*sqrt(3)*(2*x - 1)) + h*x + 1/4*(d - f + g - i)*log(x^2 + x + 1) - 1/4*(d - f - g + i)
*log(x^2 - x + 1)

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mupad [B]  time = 7.80, size = 1509, normalized size = 9.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2 + g*x^3 + h*x^4 + i*x^5)/(x^2 + x^4 + 1),x)

[Out]

h*x - log(d*g*3i - d*f*9i - d*e*6i + d*h*3i + d*i*3i + e*h*6i - f*h*3i - g*h*3i - h*i*3i - 3*3^(1/2)*d^2 - d^2
*x*6i - f^2*x*3i + d^2*3i + f^2*6i - 2*3^(1/2)*d*e + 3*3^(1/2)*d*f + 3^(1/2)*d*g + 4*3^(1/2)*e*f + 3*3^(1/2)*d
*h + 3^(1/2)*d*i - 2*3^(1/2)*e*h - 2*3^(1/2)*f*g - 3*3^(1/2)*f*h - 2*3^(1/2)*f*i + 3^(1/2)*g*h + 3^(1/2)*h*i +
d*f*x*9i + e*f*x*6i + d*h*x*3i - e*h*x*6i - f*g*x*3i - f*h*x*3i - f*i*x*3i + g*h*x*3i + h*i*x*3i - 3*3^(1/2)*
f^2*x + 3*3^(1/2)*d*f*x - 2*3^(1/2)*d*g*x - 2*3^(1/2)*e*f*x - 3*3^(1/2)*d*h*x - 2*3^(1/2)*d*i*x - 2*3^(1/2)*e*
h*x + 3^(1/2)*f*g*x + 3*3^(1/2)*f*h*x + 3^(1/2)*f*i*x + 3^(1/2)*g*h*x + 3^(1/2)*h*i*x + 4*3^(1/2)*d*e*x)*(d/4
- f/4 - g/4 + i/4 + (3^(1/2)*d*1i)/12 + (3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12 - (3^(1/2)*g*1i)/12 - (3^(1/2)*h*
1i)/6 - (3^(1/2)*i*1i)/12) - log(d*e*6i + d*f*9i - d*g*3i - d*h*3i - d*i*3i - e*h*6i + f*h*3i + g*h*3i + h*i*3
i - 3*3^(1/2)*d^2 + d^2*x*6i + f^2*x*3i - d^2*3i - f^2*6i - 2*3^(1/2)*d*e + 3*3^(1/2)*d*f + 3^(1/2)*d*g + 4*3^
(1/2)*e*f + 3*3^(1/2)*d*h + 3^(1/2)*d*i - 2*3^(1/2)*e*h - 2*3^(1/2)*f*g - 3*3^(1/2)*f*h - 2*3^(1/2)*f*i + 3^(1
/2)*g*h + 3^(1/2)*h*i - d*f*x*9i - e*f*x*6i - d*h*x*3i + e*h*x*6i + f*g*x*3i + f*h*x*3i + f*i*x*3i - g*h*x*3i
- h*i*x*3i - 3*3^(1/2)*f^2*x + 3*3^(1/2)*d*f*x - 2*3^(1/2)*d*g*x - 2*3^(1/2)*e*f*x - 3*3^(1/2)*d*h*x - 2*3^(1/
2)*d*i*x - 2*3^(1/2)*e*h*x + 3^(1/2)*f*g*x + 3*3^(1/2)*f*h*x + 3^(1/2)*f*i*x + 3^(1/2)*g*h*x + 3^(1/2)*h*i*x +
4*3^(1/2)*d*e*x)*(d/4 - f/4 - g/4 + i/4 - (3^(1/2)*d*1i)/12 - (3^(1/2)*e*1i)/6 - (3^(1/2)*f*1i)/12 + (3^(1/2)
*g*1i)/12 + (3^(1/2)*h*1i)/6 + (3^(1/2)*i*1i)/12) - log(d*f*9i - d*e*6i + d*g*3i - d*h*3i + d*i*3i + e*h*6i +
f*h*3i - g*h*3i - h*i*3i - 3*3^(1/2)*d^2 - d^2*x*6i - f^2*x*3i - d^2*3i - f^2*6i + 2*3^(1/2)*d*e + 3*3^(1/2)*d
*f - 3^(1/2)*d*g - 4*3^(1/2)*e*f + 3*3^(1/2)*d*h - 3^(1/2)*d*i + 2*3^(1/2)*e*h + 2*3^(1/2)*f*g - 3*3^(1/2)*f*h
+ 2*3^(1/2)*f*i - 3^(1/2)*g*h - 3^(1/2)*h*i + d*f*x*9i - e*f*x*6i + d*h*x*3i + e*h*x*6i + f*g*x*3i - f*h*x*3i
+ f*i*x*3i - g*h*x*3i - h*i*x*3i + 3*3^(1/2)*f^2*x - 3*3^(1/2)*d*f*x - 2*3^(1/2)*d*g*x - 2*3^(1/2)*e*f*x + 3*
3^(1/2)*d*h*x - 2*3^(1/2)*d*i*x - 2*3^(1/2)*e*h*x + 3^(1/2)*f*g*x - 3*3^(1/2)*f*h*x + 3^(1/2)*f*i*x + 3^(1/2)*
g*h*x + 3^(1/2)*h*i*x + 4*3^(1/2)*d*e*x)*(f/4 - d/4 - g/4 + i/4 + (3^(1/2)*d*1i)/12 - (3^(1/2)*e*1i)/6 + (3^(1
/2)*f*1i)/12 + (3^(1/2)*g*1i)/12 - (3^(1/2)*h*1i)/6 + (3^(1/2)*i*1i)/12) + log(d*f*9i - d*e*6i + d*g*3i - d*h*
3i + d*i*3i + e*h*6i + f*h*3i - g*h*3i - h*i*3i + 3*3^(1/2)*d^2 - d^2*x*6i - f^2*x*3i - d^2*3i - f^2*6i - 2*3^
(1/2)*d*e - 3*3^(1/2)*d*f + 3^(1/2)*d*g + 4*3^(1/2)*e*f - 3*3^(1/2)*d*h + 3^(1/2)*d*i - 2*3^(1/2)*e*h - 2*3^(1
/2)*f*g + 3*3^(1/2)*f*h - 2*3^(1/2)*f*i + 3^(1/2)*g*h + 3^(1/2)*h*i + d*f*x*9i - e*f*x*6i + d*h*x*3i + e*h*x*6
i + f*g*x*3i - f*h*x*3i + f*i*x*3i - g*h*x*3i - h*i*x*3i - 3*3^(1/2)*f^2*x + 3*3^(1/2)*d*f*x + 2*3^(1/2)*d*g*x
+ 2*3^(1/2)*e*f*x - 3*3^(1/2)*d*h*x + 2*3^(1/2)*d*i*x + 2*3^(1/2)*e*h*x - 3^(1/2)*f*g*x + 3*3^(1/2)*f*h*x - 3
^(1/2)*f*i*x - 3^(1/2)*g*h*x - 3^(1/2)*h*i*x - 4*3^(1/2)*d*e*x)*(d/4 - f/4 + g/4 - i/4 + (3^(1/2)*d*1i)/12 - (
3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12 + (3^(1/2)*g*1i)/12 - (3^(1/2)*h*1i)/6 + (3^(1/2)*i*1i)/12) + (i*x^2)/2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((i*x**5+h*x**4+g*x**3+f*x**2+e*x+d)/(x**4+x**2+1),x)

[Out]

Timed out

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