3.42 \(\int \frac{x^2 (A+B x+C x^2)}{(a+b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=356 \[ -\frac{x \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (-\frac{4 A b c-C \left (4 a c+b^2\right )}{\sqrt{b^2-4 a c}}+2 A c-b C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\left (\frac{4 A b c-C \left (4 a c+b^2\right )}{\sqrt{b^2-4 a c}}+2 A c-b C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b B \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

[Out]

(B*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (x*(A*b - 2*a*C + (2*A*c - b*C)*x^2))/(2*(b^2 - 4*a*
c)*(a + b*x^2 + c*x^4)) - ((2*A*c - b*C - (4*A*b*c - (b^2 + 4*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[
c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - ((2*A*c -
b*C + (4*A*b*c - (b^2 + 4*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/
(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - (b*B*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])
/(b^2 - 4*a*c)^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.923822, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {1662, 1275, 1166, 205, 12, 1114, 638, 618, 206} \[ -\frac{x \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (-\frac{4 A b c-C \left (4 a c+b^2\right )}{\sqrt{b^2-4 a c}}+2 A c-b C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\left (\frac{4 A b c-C \left (4 a c+b^2\right )}{\sqrt{b^2-4 a c}}+2 A c-b C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{b B \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2,x]

[Out]

(B*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (x*(A*b - 2*a*C + (2*A*c - b*C)*x^2))/(2*(b^2 - 4*a*
c)*(a + b*x^2 + c*x^4)) - ((2*A*c - b*C - (4*A*b*c - (b^2 + 4*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[
c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - ((2*A*c -
b*C + (4*A*b*c - (b^2 + 4*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/
(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - (b*B*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])
/(b^2 - 4*a*c)^(3/2)

Rule 1662

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

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(f*
(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1)*(b*d - 2*a*e - (b*e - 2*c*d)*x^2))/(2*(p + 1)*(b^2 - 4*a*c)), x] - D
ist[f^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1)*Simp[(m - 1)*(b*d - 2*a*e) -
(4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[
p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1166

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 12

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

Rule 1114

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

Rule 638

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^2 \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx &=\int \frac{B x^3}{\left (a+b x^2+c x^4\right )^2} \, dx+\int \frac{x^2 \left (A+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=-\frac{x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+B \int \frac{x^3}{\left (a+b x^2+c x^4\right )^2} \, dx+\frac{\int \frac{A b-2 a C+(-2 A c+b C) x^2}{a+b x^2+c x^4} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac{x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{1}{2} B \operatorname{Subst}\left (\int \frac{x}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )-\frac{\left (2 A c-b C-\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{4 \left (b^2-4 a c\right )}-\frac{\left (2 A c-b C+\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{4 \left (b^2-4 a c\right )}\\ &=\frac{B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (2 A c-b C-\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\left (2 A c-b C+\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=\frac{B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (2 A c-b C-\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\left (2 A c-b C+\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(b B) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{b^2-4 a c}\\ &=\frac{B \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{x \left (A b-2 a C+(2 A c-b C) x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (2 A c-b C-\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\left (2 A c-b C+\frac{4 A b c-\left (b^2+4 a c\right ) C}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{b B \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.18665, size = 378, normalized size = 1.06 \[ \frac{1}{4} \left (\frac{4 a (B+C x)+2 x \left (b x (B+C x)-A \left (b+2 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{2} \left (C \left (b \sqrt{b^2-4 a c}-4 a c-b^2\right )-2 A c \left (\sqrt{b^2-4 a c}-2 b\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} \left (C \left (b \sqrt{b^2-4 a c}+4 a c+b^2\right )-2 A c \left (\sqrt{b^2-4 a c}+2 b\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{2 b B \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{2 b B \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2,x]

[Out]

((4*a*(B + C*x) + 2*x*(b*x*(B + C*x) - A*(b + 2*c*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[2]*(-2*A*
c*(-2*b + Sqrt[b^2 - 4*a*c]) + (-b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqr
t[b^2 - 4*a*c]]])/(Sqrt[c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*(-2*A*c*(2*b + Sqrt[b^2
- 4*a*c]) + (b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(
Sqrt[c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*b*B*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/(b^2
- 4*a*c)^(3/2) - (2*b*B*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/4

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Maple [B]  time = 0., size = 1119, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x)

[Out]

(1/2*(2*A*c-C*b)/(4*a*c-b^2)*x^3-1/2*B*b/(4*a*c-b^2)*x^2+1/2*(A*b-2*C*a)/(4*a*c-b^2)*x-B*a/(4*a*c-b^2))/(c*x^4
+b*x^2+a)+1/2/(4*a*c-b^2)^2*B*(-4*a*c+b^2)^(1/2)*b*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)-c/(4*a*c-b^2)^2*2^(1/2)/(
((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*A*(-4*a*c+b^2)^(1/2)*b-2
*c^2/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/
2))*A*a+1/2*c/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-
b)*c)^(1/2))*A*b^2+c/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)
^(1/2)-b)*c)^(1/2))*C*(-4*a*c+b^2)^(1/2)*a+1/4/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(
c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*C*(-4*a*c+b^2)^(1/2)*b^2+c/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^
(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*C*a*b-1/4/(4*a*c-b^2)^2*2^(1/2)/(((-4*
a*c+b^2)^(1/2)-b)*c)^(1/2)*arctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*C*b^3-1/2/(4*a*c-b^2)^2*B*(-4
*a*c+b^2)^(1/2)*b*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)-c/(4*a*c-b^2)^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*ar
ctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*A*(-4*a*c+b^2)^(1/2)*b+2*c^2/(4*a*c-b^2)^2*2^(1/2)/((b+(-4*
a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*A*a-1/2*c/(4*a*c-b^2)^2*2^(1/2)/
((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*A*b^2+c/(4*a*c-b^2)^2*2^
(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*C*(-4*a*c+b^2)^(1/
2)*a+1/4/(4*a*c-b^2)^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^
(1/2))*C*(-4*a*c+b^2)^(1/2)*b^2-c/(4*a*c-b^2)^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^(1/2)/((
b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*C*a*b+1/4/(4*a*c-b^2)^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x*2^
(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*C*b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{B b x^{2} +{\left (C b - 2 \, A c\right )} x^{3} + 2 \, B a +{\left (2 \, C a - A b\right )} x}{2 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} x^{2}\right )}} - \frac{-\int \frac{2 \, B b x +{\left (C b - 2 \, A c\right )} x^{2} - 2 \, C a + A b}{c x^{4} + b x^{2} + a}\,{d x}}{2 \,{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

1/2*(B*b*x^2 + (C*b - 2*A*c)*x^3 + 2*B*a + (2*C*a - A*b)*x)/((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 -
4*a*b*c)*x^2) - 1/2*integrate(-(2*B*b*x + (C*b - 2*A*c)*x^2 - 2*C*a + A*b)/(c*x^4 + b*x^2 + a), x)/(b^2 - 4*a*
c)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(C*x**2+B*x+A)/(c*x**4+b*x**2+a)**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError