### 3.33 $$\int \frac{A+B x+C x^2}{(a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=368 $\frac{x \left (c x^2 (A b-2 a C)-2 a A c-a b C+A b^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{c} \left (\frac{A \left (b^2-12 a c\right )+4 a b C}{\sqrt{b^2-4 a c}}-2 a C+A b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{c} \left (-\frac{-12 a A c+4 a b C+A b^2}{\sqrt{b^2-4 a c}}-2 a C+A b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{B \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{2 B c \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}$

[Out]

-(B*(b + 2*c*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (x*(A*b^2 - 2*a*A*c - a*b*C + c*(A*b - 2*a*C)*x^2))
/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[c]*(A*b - 2*a*C + (A*(b^2 - 12*a*c) + 4*a*b*C)/Sqrt[b^2 - 4*a
*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4
*a*c]]) + (Sqrt[c]*(A*b - 2*a*C - (A*b^2 - 12*a*A*c + 4*a*b*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/S
qrt[b + Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*B*c*ArcTanh[(b + 2*c
*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)

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Rubi [A]  time = 0.866742, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.36, Rules used = {1673, 1178, 1166, 205, 12, 1107, 614, 618, 206} $\frac{x \left (c x^2 (A b-2 a C)-2 a A c-a b C+A b^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{c} \left (\frac{A \left (b^2-12 a c\right )+4 a b C}{\sqrt{b^2-4 a c}}-2 a C+A b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{c} \left (-\frac{-12 a A c+4 a b C+A b^2}{\sqrt{b^2-4 a c}}-2 a C+A b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{B \left (b+2 c x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{2 B c \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(B*(b + 2*c*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (x*(A*b^2 - 2*a*A*c - a*b*C + c*(A*b - 2*a*C)*x^2))
/(2*a*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[c]*(A*b - 2*a*C + (A*(b^2 - 12*a*c) + 4*a*b*C)/Sqrt[b^2 - 4*a
*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4
*a*c]]) + (Sqrt[c]*(A*b - 2*a*C - (A*b^2 - 12*a*A*c + 4*a*b*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/S
qrt[b + Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*a*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*B*c*ArcTanh[(b + 2*c
*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)

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 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 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 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 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 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{A+B x+C x^2}{\left (a+b x^2+c x^4\right )^2} \, dx &=\int \frac{B x}{\left (a+b x^2+c x^4\right )^2} \, dx+\int \frac{A+C x^2}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{x \left (A b^2-2 a A c-a b C+c (A b-2 a C) x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+B \int \frac{x}{\left (a+b x^2+c x^4\right )^2} \, dx-\frac{\int \frac{-A b^2+6 a A c-a b C-c (A b-2 a C) x^2}{a+b x^2+c x^4} \, dx}{2 a \left (b^2-4 a c\right )}\\ &=\frac{x \left (A b^2-2 a A c-a b C+c (A b-2 a C) x^2\right )}{2 a \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{1}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )+\frac{\left (c \left (A \left (b^2-12 a c+b \sqrt{b^2-4 a c}\right )+2 a \left (2 b-\sqrt{b^2-4 a c}\right ) C\right )\right ) \int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{4 a \left (b^2-4 a c\right )^{3/2}}+\frac{\left (c \left (A b-2 a C-\frac{A b^2-12 a A c+4 a b C}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx}{4 a \left (b^2-4 a c\right )}\\ &=-\frac{B \left (b+2 c 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^2-2 a A c-a b C+c (A b-2 a C) x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{c} \left (A \left (b^2-12 a c+b \sqrt{b^2-4 a c}\right )+2 a \left (2 b-\sqrt{b^2-4 a c}\right ) C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{c} \left (A b-2 a C-\frac{A b^2-12 a A c+4 a b 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} a \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(B c) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{b^2-4 a c}\\ &=-\frac{B \left (b+2 c 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^2-2 a A c-a b C+c (A b-2 a C) x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{c} \left (A \left (b^2-12 a c+b \sqrt{b^2-4 a c}\right )+2 a \left (2 b-\sqrt{b^2-4 a c}\right ) C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{c} \left (A b-2 a C-\frac{A b^2-12 a A c+4 a b 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} a \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{(2 B c) \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 (b+2 c 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^2-2 a A c-a b C+c (A b-2 a C) x^2\right )}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{c} \left (A \left (b^2-12 a c+b \sqrt{b^2-4 a c}\right )+2 a \left (2 b-\sqrt{b^2-4 a c}\right ) C\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{c} \left (A b-2 a C-\frac{A b^2-12 a A c+4 a b 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} a \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{2 B c \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 = 1.45573, size = 393, normalized size = 1.07 $\frac{1}{4} \left (\frac{4 a c x (A+x (B+C x))+2 a b (B+C x)-2 A b x \left (b+c x^2\right )}{a \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{2} \sqrt{c} \left (A \left (b \sqrt{b^2-4 a c}-12 a c+b^2\right )-2 a C \left (\sqrt{b^2-4 a c}-2 b\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{a \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{\sqrt{2} \sqrt{c} \left (A \left (-b \sqrt{b^2-4 a c}-12 a c+b^2\right )+2 a C \left (\sqrt{b^2-4 a c}+2 b\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{a \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{4 B c \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{4 B c \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*a*b*(B + C*x) - 2*A*b*x*(b + c*x^2) + 4*a*c*x*(A + x*(B + C*x)))/(a*(-b^2 + 4*a*c)*(a + b*x^2 + c*x^4)) +
(Sqrt[2]*Sqrt[c]*(A*(b^2 - 12*a*c + b*Sqrt[b^2 - 4*a*c]) - 2*a*(-2*b + Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*S
qrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(a*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (Sqrt[2]*Sqrt[c]
*(A*(b^2 - 12*a*c - b*Sqrt[b^2 - 4*a*c]) + 2*a*(2*b + Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b
+ Sqrt[b^2 - 4*a*c]]])/(a*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - (4*B*c*Log[-b + Sqrt[b^2 - 4*a*c]
- 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + (4*B*c*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 = 0.105, size = 1813, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)/a*x*A*b^3+1/4/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/
2)/c+1/2*b/c)/a*x*A*b^3+2*c/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)*a*C*x+c/(4*a*c-b^2)^2/(x^2+1/
2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)*x*A*(-4*a*c+b^2)^(1/2)-c/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)*
A*b*x-c/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)*x*A*(-4*a*c+b^2)^(1/2)-c/(4*a*c-b^2)^2/(x^2+1/2*b
/c-1/2*(-4*a*c+b^2)^(1/2)/c)*A*b*x+2*c/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)*a*C*x-1/4*c/(4*a*c
-b^2)^2/a*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*(-4*
a*c+b^2)^(1/2)*b^2-1/4*c/(4*a*c-b^2)^2/a*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^2-1/2/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)*x*
C*b^2-1/2/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)*x*C*b^2+2*c/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)
^(1/2)/c+1/2*b/c)*B*a+c/(4*a*c-b^2)^2*B*(-4*a*c+b^2)^(1/2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)+2*c/(4*a*c-b^2)^2/
(x^2+1/2*b/c-1/2*(-4*a*c+b^2)^(1/2)/c)*B*a-c/(4*a*c-b^2)^2*B*(-4*a*c+b^2)^(1/2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)
-b)-1/2/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)*B*b^2-1/2/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*c+
b^2)^(1/2)/c)*B*b^2+3*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*(-4*a*c+b^2)^(1/2)-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*b+2*c^2/(4*a*c-b^2)^2*a*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-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))*C*b^2+1/4/(4*a*c-b^2)^2/(x^2+1/2*b/c-1/2*(-4*a*
c+b^2)^(1/2)/c)/a*x*A*(-4*a*c+b^2)^(1/2)*b^2-1/4/(4*a*c-b^2)^2/(x^2+1/2*(-4*a*c+b^2)^(1/2)/c+1/2*b/c)/a*x*A*(-
4*a*c+b^2)^(1/2)*b^2+3*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*(-4*a*c+b^2)^(1/2)+c^2/(4*a*c-b^2)^2*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*arct
anh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*A*b-2*c^2/(4*a*c-b^2)^2*a*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+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))*C*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)*b-1/4*c/(4*a*c-
b^2)^2/a*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^3-
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)*b+1/4*c/(4*a*c-b^2)^2/a*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^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \, B a c x^{2} +{\left (2 \, C a - A b\right )} c x^{3} + B a b +{\left (C a b - A b^{2} + 2 \, A a c\right )} x}{2 \,{\left ({\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )}} + \frac{-\int \frac{4 \, B a c x +{\left (2 \, C a - A b\right )} c x^{2} - C a b - A b^{2} + 6 \, A a c}{c x^{4} + b x^{2} + a}\,{d x}}{2 \,{\left (a b^{2} - 4 \, a^{2} c\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError