3.112 \(\int \frac{x^2 (4+x^2+3 x^4+5 x^6)}{(3+2 x^2+x^4)^2} \, dx\)

Optimal. Leaf size=225 \[ \frac{25 \left (x^2+1\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+5 x+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]

[Out]

5*x + (25*x*(1 + x^2))/(8*(3 + 2*x^2 + x^4)) + (Sqrt[(19291 + 12899*Sqrt[3])/6]*ArcTan[(Sqrt[2*(-1 + Sqrt[3])]
- 2*x)/Sqrt[2*(1 + Sqrt[3])]])/16 - (Sqrt[(19291 + 12899*Sqrt[3])/6]*ArcTan[(Sqrt[2*(-1 + Sqrt[3])] + 2*x)/Sq
rt[2*(1 + Sqrt[3])]])/16 - (Sqrt[(-19291 + 12899*Sqrt[3])/6]*Log[Sqrt[3] - Sqrt[2*(-1 + Sqrt[3])]*x + x^2])/32
+ (Sqrt[(-19291 + 12899*Sqrt[3])/6]*Log[Sqrt[3] + Sqrt[2*(-1 + Sqrt[3])]*x + x^2])/32

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Rubi [A]  time = 0.29714, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {1668, 1676, 1169, 634, 618, 204, 628} \[ \frac{25 \left (x^2+1\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+5 x+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

5*x + (25*x*(1 + x^2))/(8*(3 + 2*x^2 + x^4)) + (Sqrt[(19291 + 12899*Sqrt[3])/6]*ArcTan[(Sqrt[2*(-1 + Sqrt[3])]
- 2*x)/Sqrt[2*(1 + Sqrt[3])]])/16 - (Sqrt[(19291 + 12899*Sqrt[3])/6]*ArcTan[(Sqrt[2*(-1 + Sqrt[3])] + 2*x)/Sq
rt[2*(1 + Sqrt[3])]])/16 - (Sqrt[(-19291 + 12899*Sqrt[3])/6]*Log[Sqrt[3] - Sqrt[2*(-1 + Sqrt[3])]*x + x^2])/32
+ (Sqrt[(-19291 + 12899*Sqrt[3])/6]*Log[Sqrt[3] + Sqrt[2*(-1 + Sqrt[3])]*x + x^2])/32

Rule 1668

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{d = Coeff[PolynomialRemainde
r[x^m*Pq, a + b*x^2 + c*x^4, x], x, 0], e = Coeff[PolynomialRemainder[x^m*Pq, a + b*x^2 + c*x^4, x], x, 2]}, S
imp[(x*(a + b*x^2 + c*x^4)^(p + 1)*(a*b*e - d*(b^2 - 2*a*c) - c*(b*d - 2*a*e)*x^2))/(2*a*(p + 1)*(b^2 - 4*a*c)
), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a
*c)*PolynomialQuotient[x^m*Pq, a + b*x^2 + c*x^4, x] + b^2*d*(2*p + 3) - 2*a*c*d*(4*p + 5) - a*b*e + c*(4*p +
7)*(b*d - 2*a*e)*x^2, x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && GtQ[Expon[Pq, x^2], 1] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && IGtQ[m/2, 0]

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

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{x^2 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx &=\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \frac{-150-186 x^2+240 x^4}{3+2 x^2+x^4} \, dx\\ &=\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \left (240-\frac{6 \left (145+111 x^2\right )}{3+2 x^2+x^4}\right ) \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{8} \int \frac{145+111 x^2}{3+2 x^2+x^4} \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{\int \frac{145 \sqrt{2 \left (-1+\sqrt{3}\right )}-\left (145-111 \sqrt{3}\right ) x}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx}{16 \sqrt{6 \left (-1+\sqrt{3}\right )}}-\frac{\int \frac{145 \sqrt{2 \left (-1+\sqrt{3}\right )}+\left (145-111 \sqrt{3}\right ) x}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx}{16 \sqrt{6 \left (-1+\sqrt{3}\right )}}\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{96} \left (333+145 \sqrt{3}\right ) \int \frac{1}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx-\frac{1}{96} \left (333+145 \sqrt{3}\right ) \int \frac{1}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \int \frac{-\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \int \frac{\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{48} \left (333+145 \sqrt{3}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{3}\right )-x^2} \, dx,x,-\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x\right )+\frac{1}{48} \left (333+145 \sqrt{3}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{3}\right )-x^2} \, dx,x,\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x\right )\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{3}\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )\\ \end{align*}

Mathematica [C]  time = 0.170382, size = 121, normalized size = 0.54 \[ \frac{25 \left (x^3+x\right )}{8 \left (x^4+2 x^2+3\right )}+5 x-\frac{\left (111 \sqrt{2}-34 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1-i \sqrt{2}}}\right )}{16 \sqrt{2-2 i \sqrt{2}}}-\frac{\left (111 \sqrt{2}+34 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1+i \sqrt{2}}}\right )}{16 \sqrt{2+2 i \sqrt{2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(4 + x^2 + 3*x^4 + 5*x^6))/(3 + 2*x^2 + x^4)^2,x]

[Out]

5*x + (25*(x + x^3))/(8*(3 + 2*x^2 + x^4)) - ((-34*I + 111*Sqrt[2])*ArcTan[x/Sqrt[1 - I*Sqrt[2]]])/(16*Sqrt[2
- (2*I)*Sqrt[2]]) - ((34*I + 111*Sqrt[2])*ArcTan[x/Sqrt[1 + I*Sqrt[2]]])/(16*Sqrt[2 + (2*I)*Sqrt[2]])

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Maple [B]  time = 0.018, size = 412, normalized size = 1.8 \begin{align*} 5\,x-{\frac{1}{{x}^{4}+2\,{x}^{2}+3} \left ( -{\frac{25\,{x}^{3}}{8}}-{\frac{25\,x}{8}} \right ) }-{\frac{47\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{96}}+{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{64}}-{\frac{ \left ( -94+94\,\sqrt{3} \right ) \sqrt{3}}{48\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{-34+34\,\sqrt{3}}{32\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }-{\frac{145\,\sqrt{3}}{24\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{47\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{96}}-{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{64}}-{\frac{ \left ( -94+94\,\sqrt{3} \right ) \sqrt{3}}{48\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{-34+34\,\sqrt{3}}{32\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }-{\frac{145\,\sqrt{3}}{24\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x)

[Out]

5*x-(-25/8*x^3-25/8*x)/(x^4+2*x^2+3)-47/96*ln(x^2+3^(1/2)-x*(-2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))^(1/2)*3^(1/2)
+17/64*ln(x^2+3^(1/2)-x*(-2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))^(1/2)-47/48/(2+2*3^(1/2))^(1/2)*arctan((2*x-(-2+2
*3^(1/2))^(1/2))/(2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))*3^(1/2)+17/32/(2+2*3^(1/2))^(1/2)*arctan((2*x-(-2+2*3^(1/
2))^(1/2))/(2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))-145/24/(2+2*3^(1/2))^(1/2)*arctan((2*x-(-2+2*3^(1/2))^(1/2))/(2
+2*3^(1/2))^(1/2))*3^(1/2)+47/96*ln(x^2+3^(1/2)+x*(-2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))^(1/2)*3^(1/2)-17/64*ln(
x^2+3^(1/2)+x*(-2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))^(1/2)-47/48/(2+2*3^(1/2))^(1/2)*arctan((2*x+(-2+2*3^(1/2))^
(1/2))/(2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))*3^(1/2)+17/32/(2+2*3^(1/2))^(1/2)*arctan((2*x+(-2+2*3^(1/2))^(1/2))
/(2+2*3^(1/2))^(1/2))*(-2+2*3^(1/2))-145/24/(2+2*3^(1/2))^(1/2)*arctan((2*x+(-2+2*3^(1/2))^(1/2))/(2+2*3^(1/2)
)^(1/2))*3^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 5 \, x + \frac{25 \,{\left (x^{3} + x\right )}}{8 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac{1}{8} \, \int \frac{111 \, x^{2} + 145}{x^{4} + 2 \, x^{2} + 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="maxima")

[Out]

5*x + 25/8*(x^3 + x)/(x^4 + 2*x^2 + 3) - 1/8*integrate((111*x^2 + 145)/(x^4 + 2*x^2 + 3), x)

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Fricas [B]  time = 1.75631, size = 2072, normalized size = 9.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="fricas")

[Out]

1/19736089152*(98680445760*x^5 + 31876*499152603^(1/4)*sqrt(2)*(x^4 + 2*x^2 + 3)*sqrt(248834609*sqrt(3) + 4991
52603)*arctan(1/2453286601800494203302*499152603^(3/4)*sqrt(308376393)*sqrt(308376393*x^2 + 499152603^(1/4)*(1
45*sqrt(3)*x - 333*x)*sqrt(248834609*sqrt(3) + 499152603) + 308376393*sqrt(3))*(111*sqrt(3)*sqrt(2) - 145*sqrt
(2))*sqrt(248834609*sqrt(3) + 499152603) - 1/7955494186614*499152603^(3/4)*(111*sqrt(3)*sqrt(2)*x - 145*sqrt(2
)*x)*sqrt(248834609*sqrt(3) + 499152603) + 1/2*sqrt(3)*sqrt(2) - 1/2*sqrt(2)) + 31876*499152603^(1/4)*sqrt(2)*
(x^4 + 2*x^2 + 3)*sqrt(248834609*sqrt(3) + 499152603)*arctan(1/2453286601800494203302*499152603^(3/4)*sqrt(308
376393)*sqrt(308376393*x^2 - 499152603^(1/4)*(145*sqrt(3)*x - 333*x)*sqrt(248834609*sqrt(3) + 499152603) + 308
376393*sqrt(3))*(111*sqrt(3)*sqrt(2) - 145*sqrt(2))*sqrt(248834609*sqrt(3) + 499152603) - 1/7955494186614*4991
52603^(3/4)*(111*sqrt(3)*sqrt(2)*x - 145*sqrt(2)*x)*sqrt(248834609*sqrt(3) + 499152603) - 1/2*sqrt(3)*sqrt(2)
+ 1/2*sqrt(2)) + 259036170120*x^3 + 499152603^(1/4)*(19291*x^4 + 38582*x^2 - 12899*sqrt(3)*(x^4 + 2*x^2 + 3) +
57873)*sqrt(248834609*sqrt(3) + 499152603)*log(308376393*x^2 + 499152603^(1/4)*(145*sqrt(3)*x - 333*x)*sqrt(2
48834609*sqrt(3) + 499152603) + 308376393*sqrt(3)) - 499152603^(1/4)*(19291*x^4 + 38582*x^2 - 12899*sqrt(3)*(x
^4 + 2*x^2 + 3) + 57873)*sqrt(248834609*sqrt(3) + 499152603)*log(308376393*x^2 - 499152603^(1/4)*(145*sqrt(3)*
x - 333*x)*sqrt(248834609*sqrt(3) + 499152603) + 308376393*sqrt(3)) + 357716615880*x)/(x^4 + 2*x^2 + 3)

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Sympy [A]  time = 0.535457, size = 51, normalized size = 0.23 \begin{align*} 5 x + \frac{25 x^{3} + 25 x}{8 x^{4} + 16 x^{2} + 24} + \operatorname{RootSum}{\left (3145728 t^{4} + 39507968 t^{2} + 166384201, \left ( t \mapsto t \log{\left (- \frac{9240576 t^{3}}{102792131} - \frac{95003488 t}{102792131} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(5*x**6+3*x**4+x**2+4)/(x**4+2*x**2+3)**2,x)

[Out]

5*x + (25*x**3 + 25*x)/(8*x**4 + 16*x**2 + 24) + RootSum(3145728*_t**4 + 39507968*_t**2 + 166384201, Lambda(_t
, _t*log(-9240576*_t**3/102792131 - 95003488*_t/102792131 + x)))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (5 \, x^{6} + 3 \, x^{4} + x^{2} + 4\right )} x^{2}}{{\left (x^{4} + 2 \, x^{2} + 3\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(5*x^6+3*x^4+x^2+4)/(x^4+2*x^2+3)^2,x, algorithm="giac")

[Out]

integrate((5*x^6 + 3*x^4 + x^2 + 4)*x^2/(x^4 + 2*x^2 + 3)^2, x)