3.1.47 \(\int \frac {d+e x}{(1+x^2+x^4)^3} \, dx\)

Optimal. Leaf size=185 \[ -\frac {9}{32} d \log \left (x^2-x+1\right )+\frac {9}{32} d \log \left (x^2+x+1\right )+\frac {d x \left (2-7 x^2\right )}{24 \left (x^4+x^2+1\right )}+\frac {d x \left (1-x^2\right )}{12 \left (x^4+x^2+1\right )^2}-\frac {13 d \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {13 d \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {2 e \tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {e \left (2 x^2+1\right )}{6 \left (x^4+x^2+1\right )}+\frac {e \left (2 x^2+1\right )}{12 \left (x^4+x^2+1\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {1673, 12, 1092, 1178, 1169, 634, 618, 204, 628, 1107, 614} \begin {gather*} \frac {d x \left (2-7 x^2\right )}{24 \left (x^4+x^2+1\right )}+\frac {d x \left (1-x^2\right )}{12 \left (x^4+x^2+1\right )^2}-\frac {9}{32} d \log \left (x^2-x+1\right )+\frac {9}{32} d \log \left (x^2+x+1\right )-\frac {13 d \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {13 d \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {e \left (2 x^2+1\right )}{6 \left (x^4+x^2+1\right )}+\frac {e \left (2 x^2+1\right )}{12 \left (x^4+x^2+1\right )^2}+\frac {2 e \tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)/(1 + x^2 + x^4)^3,x]

[Out]

(d*x*(1 - x^2))/(12*(1 + x^2 + x^4)^2) + (e*(1 + 2*x^2))/(12*(1 + x^2 + x^4)^2) + (d*x*(2 - 7*x^2))/(24*(1 + x
^2 + x^4)) + (e*(1 + 2*x^2))/(6*(1 + x^2 + x^4)) - (13*d*ArcTan[(1 - 2*x)/Sqrt[3]])/(48*Sqrt[3]) + (13*d*ArcTa
n[(1 + 2*x)/Sqrt[3]])/(48*Sqrt[3]) + (2*e*ArcTan[(1 + 2*x^2)/Sqrt[3]])/(3*Sqrt[3]) - (9*d*Log[1 - x + x^2])/32
 + (9*d*Log[1 + x + x^2])/32

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 1092

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(b^2 - 2*a*c + b*c*x^2)*(a + b*x^2 + c*x^
4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a*c + 2*(p + 1)
*(b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*
a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1107

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

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 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
 b*x^2 + c*x^4)^(p + 1), 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] && LtQ[p, -1] && IntegerQ[2*p]

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]

Rubi steps

\begin {align*} \int \frac {d+e x}{\left (1+x^2+x^4\right )^3} \, dx &=\int \frac {d}{\left (1+x^2+x^4\right )^3} \, dx+\int \frac {e x}{\left (1+x^2+x^4\right )^3} \, dx\\ &=d \int \frac {1}{\left (1+x^2+x^4\right )^3} \, dx+e \int \frac {x}{\left (1+x^2+x^4\right )^3} \, dx\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {1}{12} d \int \frac {11-5 x^2}{\left (1+x^2+x^4\right )^2} \, dx+\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{\left (1+x+x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {e \left (1+2 x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {d x \left (2-7 x^2\right )}{24 \left (1+x^2+x^4\right )}+\frac {1}{72} d \int \frac {60-21 x^2}{1+x^2+x^4} \, dx+\frac {1}{2} e \operatorname {Subst}\left (\int \frac {1}{\left (1+x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {e \left (1+2 x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {d x \left (2-7 x^2\right )}{24 \left (1+x^2+x^4\right )}+\frac {e \left (1+2 x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{144} d \int \frac {60-81 x}{1-x+x^2} \, dx+\frac {1}{144} d \int \frac {60+81 x}{1+x+x^2} \, dx+\frac {1}{3} e \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {e \left (1+2 x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {d x \left (2-7 x^2\right )}{24 \left (1+x^2+x^4\right )}+\frac {e \left (1+2 x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{96} (13 d) \int \frac {1}{1-x+x^2} \, dx+\frac {1}{96} (13 d) \int \frac {1}{1+x+x^2} \, dx-\frac {1}{32} (9 d) \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{32} (9 d) \int \frac {1+2 x}{1+x+x^2} \, dx-\frac {1}{3} (2 e) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {e \left (1+2 x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {d x \left (2-7 x^2\right )}{24 \left (1+x^2+x^4\right )}+\frac {e \left (1+2 x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {2 e \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {9}{32} d \log \left (1-x+x^2\right )+\frac {9}{32} d \log \left (1+x+x^2\right )-\frac {1}{48} (13 d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac {1}{48} (13 d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac {d x \left (1-x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {e \left (1+2 x^2\right )}{12 \left (1+x^2+x^4\right )^2}+\frac {d x \left (2-7 x^2\right )}{24 \left (1+x^2+x^4\right )}+\frac {e \left (1+2 x^2\right )}{6 \left (1+x^2+x^4\right )}-\frac {13 d \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {13 d \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{48 \sqrt {3}}+\frac {2 e \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {9}{32} d \log \left (1-x+x^2\right )+\frac {9}{32} d \log \left (1+x+x^2\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.75, size = 186, normalized size = 1.01 \begin {gather*} \frac {1}{144} \left (\frac {6 \left (d x \left (2-7 x^2\right )+e \left (8 x^2+4\right )\right )}{x^4+x^2+1}+\frac {12 \left (d \left (x-x^3\right )+2 e x^2+e\right )}{\left (x^4+x^2+1\right )^2}-\frac {\left (7 \sqrt {3}-47 i\right ) d \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right )}{\sqrt {\frac {1}{6} \left (1+i \sqrt {3}\right )}}-\frac {\left (7 \sqrt {3}+47 i\right ) d \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right )}{\sqrt {\frac {1}{6} \left (1-i \sqrt {3}\right )}}-32 \sqrt {3} e \tan ^{-1}\left (\frac {\sqrt {3}}{2 x^2+1}\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((6*(d*x*(2 - 7*x^2) + e*(4 + 8*x^2)))/(1 + x^2 + x^4) + (12*(e + 2*e*x^2 + d*(x - x^3)))/(1 + x^2 + x^4)^2 -
((-47*I + 7*Sqrt[3])*d*ArcTan[((-I + Sqrt[3])*x)/2])/Sqrt[(1 + I*Sqrt[3])/6] - ((47*I + 7*Sqrt[3])*d*ArcTan[((
I + Sqrt[3])*x)/2])/Sqrt[(1 - I*Sqrt[3])/6] - 32*Sqrt[3]*e*ArcTan[Sqrt[3]/(1 + 2*x^2)])/144

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.06, size = 278, normalized size = 1.50 \begin {gather*} -\frac {84 \, d x^{7} - 96 \, e x^{6} + 60 \, d x^{5} - 144 \, e x^{4} + 84 \, d x^{3} - 192 \, e x^{2} - 2 \, \sqrt {3} {\left ({\left (13 \, d - 32 \, e\right )} x^{8} + 2 \, {\left (13 \, d - 32 \, e\right )} x^{6} + 3 \, {\left (13 \, d - 32 \, e\right )} x^{4} + 2 \, {\left (13 \, d - 32 \, e\right )} x^{2} + 13 \, d - 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left ({\left (13 \, d + 32 \, e\right )} x^{8} + 2 \, {\left (13 \, d + 32 \, e\right )} x^{6} + 3 \, {\left (13 \, d + 32 \, e\right )} x^{4} + 2 \, {\left (13 \, d + 32 \, e\right )} x^{2} + 13 \, d + 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 48 \, d x - 81 \, {\left (d x^{8} + 2 \, d x^{6} + 3 \, d x^{4} + 2 \, d x^{2} + d\right )} \log \left (x^{2} + x + 1\right ) + 81 \, {\left (d x^{8} + 2 \, d x^{6} + 3 \, d x^{4} + 2 \, d x^{2} + d\right )} \log \left (x^{2} - x + 1\right ) - 72 \, e}{288 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/288*(84*d*x^7 - 96*e*x^6 + 60*d*x^5 - 144*e*x^4 + 84*d*x^3 - 192*e*x^2 - 2*sqrt(3)*((13*d - 32*e)*x^8 + 2*(
13*d - 32*e)*x^6 + 3*(13*d - 32*e)*x^4 + 2*(13*d - 32*e)*x^2 + 13*d - 32*e)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*
sqrt(3)*((13*d + 32*e)*x^8 + 2*(13*d + 32*e)*x^6 + 3*(13*d + 32*e)*x^4 + 2*(13*d + 32*e)*x^2 + 13*d + 32*e)*ar
ctan(1/3*sqrt(3)*(2*x - 1)) - 48*d*x - 81*(d*x^8 + 2*d*x^6 + 3*d*x^4 + 2*d*x^2 + d)*log(x^2 + x + 1) + 81*(d*x
^8 + 2*d*x^6 + 3*d*x^4 + 2*d*x^2 + d)*log(x^2 - x + 1) - 72*e)/(x^8 + 2*x^6 + 3*x^4 + 2*x^2 + 1)

________________________________________________________________________________________

giac [A]  time = 0.36, size = 131, normalized size = 0.71 \begin {gather*} \frac {1}{144} \, \sqrt {3} {\left (13 \, d - 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{144} \, \sqrt {3} {\left (13 \, d + 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {9}{32} \, d \log \left (x^{2} + x + 1\right ) - \frac {9}{32} \, d \log \left (x^{2} - x + 1\right ) - \frac {7 \, d x^{7} - 8 \, x^{6} e + 5 \, d x^{5} - 12 \, x^{4} e + 7 \, d x^{3} - 16 \, x^{2} e - 4 \, d x - 6 \, e}{24 \, {\left (x^{4} + x^{2} + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*sqrt(3)*(13*d - 32*e)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/144*sqrt(3)*(13*d + 32*e)*arctan(1/3*sqrt(3)*(2*
x - 1)) + 9/32*d*log(x^2 + x + 1) - 9/32*d*log(x^2 - x + 1) - 1/24*(7*d*x^7 - 8*x^6*e + 5*d*x^5 - 12*x^4*e + 7
*d*x^3 - 16*x^2*e - 4*d*x - 6*e)/(x^4 + x^2 + 1)^2

________________________________________________________________________________________

maple [A]  time = 0.02, size = 180, normalized size = 0.97 \begin {gather*} \frac {13 \sqrt {3}\, d \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{144}+\frac {13 \sqrt {3}\, d \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{144}-\frac {9 d \ln \left (x^{2}-x +1\right )}{32}+\frac {9 d \ln \left (x^{2}+x +1\right )}{32}-\frac {2 \sqrt {3}\, e \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {2 \sqrt {3}\, e \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\frac {-6 d \,x^{2}+\left (-\frac {7 d}{3}-\frac {4 e}{3}\right ) x^{3}-4 d +2 e +\left (-\frac {20 d}{3}+\frac {e}{3}\right ) x}{16 \left (x^{2}+x +1\right )^{2}}-\frac {-6 d \,x^{2}+\left (\frac {7 d}{3}-\frac {4 e}{3}\right ) x^{3}-4 d -2 e +\left (\frac {20 d}{3}+\frac {e}{3}\right ) x}{16 \left (x^{2}-x +1\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((-7/3*d-4/3*e)*x^3-6*d*x^2+(-20/3*d+1/3*e)*x-4*d+2*e)/(x^2+x+1)^2+9/32*d*ln(x^2+x+1)+13/144*3^(1/2)*d*ar
ctan(1/3*(2*x+1)*3^(1/2))-2/9*3^(1/2)*e*arctan(1/3*(2*x+1)*3^(1/2))-1/16*((7/3*d-4/3*e)*x^3-6*d*x^2+(20/3*d+1/
3*e)*x-4*d-2*e)/(x^2-x+1)^2-9/32*d*ln(x^2-x+1)+13/144*3^(1/2)*d*arctan(1/3*(2*x-1)*3^(1/2))+2/9*3^(1/2)*e*arct
an(1/3*(2*x-1)*3^(1/2))

________________________________________________________________________________________

maxima [A]  time = 2.55, size = 137, normalized size = 0.74 \begin {gather*} \frac {1}{144} \, \sqrt {3} {\left (13 \, d - 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{144} \, \sqrt {3} {\left (13 \, d + 32 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {9}{32} \, d \log \left (x^{2} + x + 1\right ) - \frac {9}{32} \, d \log \left (x^{2} - x + 1\right ) - \frac {7 \, d x^{7} - 8 \, e x^{6} + 5 \, d x^{5} - 12 \, e x^{4} + 7 \, d x^{3} - 16 \, e x^{2} - 4 \, d x - 6 \, e}{24 \, {\left (x^{8} + 2 \, x^{6} + 3 \, x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*sqrt(3)*(13*d - 32*e)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/144*sqrt(3)*(13*d + 32*e)*arctan(1/3*sqrt(3)*(2*
x - 1)) + 9/32*d*log(x^2 + x + 1) - 9/32*d*log(x^2 - x + 1) - 1/24*(7*d*x^7 - 8*e*x^6 + 5*d*x^5 - 12*e*x^4 + 7
*d*x^3 - 16*e*x^2 - 4*d*x - 6*e)/(x^8 + 2*x^6 + 3*x^4 + 2*x^2 + 1)

________________________________________________________________________________________

mupad [B]  time = 0.26, size = 185, normalized size = 1.00 \begin {gather*} \frac {-\frac {7\,d\,x^7}{24}+\frac {e\,x^6}{3}-\frac {5\,d\,x^5}{24}+\frac {e\,x^4}{2}-\frac {7\,d\,x^3}{24}+\frac {2\,e\,x^2}{3}+\frac {d\,x}{6}+\frac {e}{4}}{x^8+2\,x^6+3\,x^4+2\,x^2+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {9\,d}{32}+\frac {\sqrt {3}\,d\,13{}\mathrm {i}}{288}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {9\,d}{32}-\frac {\sqrt {3}\,d\,13{}\mathrm {i}}{288}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {9\,d}{32}+\frac {\sqrt {3}\,d\,13{}\mathrm {i}}{288}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {9\,d}{32}+\frac {\sqrt {3}\,d\,13{}\mathrm {i}}{288}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)/(x^2 + x^4 + 1)^3,x)

[Out]

(e/4 + (d*x)/6 - (7*d*x^3)/24 - (5*d*x^5)/24 - (7*d*x^7)/24 + (2*e*x^2)/3 + (e*x^4)/2 + (e*x^6)/3)/(2*x^2 + 3*
x^4 + 2*x^6 + x^8 + 1) - log(x - (3^(1/2)*1i)/2 - 1/2)*((9*d)/32 + (3^(1/2)*d*13i)/288 + (3^(1/2)*e*1i)/9) + l
og(x - (3^(1/2)*1i)/2 + 1/2)*((9*d)/32 - (3^(1/2)*d*13i)/288 + (3^(1/2)*e*1i)/9) + log(x + (3^(1/2)*1i)/2 - 1/
2)*((3^(1/2)*d*13i)/288 - (9*d)/32 + (3^(1/2)*e*1i)/9) + log(x + (3^(1/2)*1i)/2 + 1/2)*((9*d)/32 + (3^(1/2)*d*
13i)/288 - (3^(1/2)*e*1i)/9)

________________________________________________________________________________________

sympy [C]  time = 3.62, size = 1103, normalized size = 5.96

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288)*log(x + (-1025428432*d**4*e - 334752912*d**4*(-9*d/32 - sqrt(3)*I*(13*
d + 32*e)/288) - 431308800*d**2*e**3 - 3143688192*d**2*e**2*(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288) + 99170058
24*d**2*e*(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288)**2 + 11878244352*d**2*(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288
)**3 + 142606336*e**5 + 754974720*e**4*(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288) + 3850371072*e**3*(-9*d/32 - sq
rt(3)*I*(13*d + 32*e)/288)**2 + 20384317440*e**2*(-9*d/32 - sqrt(3)*I*(13*d + 32*e)/288)**3)/(217696167*d**5 -
 1217128448*d**3*e**2 - 617611264*d*e**4)) + (-9*d/32 + sqrt(3)*I*(13*d + 32*e)/288)*log(x + (-1025428432*d**4
*e - 334752912*d**4*(-9*d/32 + sqrt(3)*I*(13*d + 32*e)/288) - 431308800*d**2*e**3 - 3143688192*d**2*e**2*(-9*d
/32 + sqrt(3)*I*(13*d + 32*e)/288) + 9917005824*d**2*e*(-9*d/32 + sqrt(3)*I*(13*d + 32*e)/288)**2 + 1187824435
2*d**2*(-9*d/32 + sqrt(3)*I*(13*d + 32*e)/288)**3 + 142606336*e**5 + 754974720*e**4*(-9*d/32 + sqrt(3)*I*(13*d
 + 32*e)/288) + 3850371072*e**3*(-9*d/32 + sqrt(3)*I*(13*d + 32*e)/288)**2 + 20384317440*e**2*(-9*d/32 + sqrt(
3)*I*(13*d + 32*e)/288)**3)/(217696167*d**5 - 1217128448*d**3*e**2 - 617611264*d*e**4)) + (9*d/32 - sqrt(3)*I*
(13*d - 32*e)/288)*log(x + (-1025428432*d**4*e - 334752912*d**4*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288) - 43130
8800*d**2*e**3 - 3143688192*d**2*e**2*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288) + 9917005824*d**2*e*(9*d/32 - sqr
t(3)*I*(13*d - 32*e)/288)**2 + 11878244352*d**2*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288)**3 + 142606336*e**5 + 7
54974720*e**4*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288) + 3850371072*e**3*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288)*
*2 + 20384317440*e**2*(9*d/32 - sqrt(3)*I*(13*d - 32*e)/288)**3)/(217696167*d**5 - 1217128448*d**3*e**2 - 6176
11264*d*e**4)) + (9*d/32 + sqrt(3)*I*(13*d - 32*e)/288)*log(x + (-1025428432*d**4*e - 334752912*d**4*(9*d/32 +
 sqrt(3)*I*(13*d - 32*e)/288) - 431308800*d**2*e**3 - 3143688192*d**2*e**2*(9*d/32 + sqrt(3)*I*(13*d - 32*e)/2
88) + 9917005824*d**2*e*(9*d/32 + sqrt(3)*I*(13*d - 32*e)/288)**2 + 11878244352*d**2*(9*d/32 + sqrt(3)*I*(13*d
 - 32*e)/288)**3 + 142606336*e**5 + 754974720*e**4*(9*d/32 + sqrt(3)*I*(13*d - 32*e)/288) + 3850371072*e**3*(9
*d/32 + sqrt(3)*I*(13*d - 32*e)/288)**2 + 20384317440*e**2*(9*d/32 + sqrt(3)*I*(13*d - 32*e)/288)**3)/(2176961
67*d**5 - 1217128448*d**3*e**2 - 617611264*d*e**4)) + (-7*d*x**7 - 5*d*x**5 - 7*d*x**3 + 4*d*x + 8*e*x**6 + 12
*e*x**4 + 16*e*x**2 + 6*e)/(24*x**8 + 48*x**6 + 72*x**4 + 48*x**2 + 24)

________________________________________________________________________________________