3.28 \(\int \frac{A+B x+C x^2}{x^3 (a+b x^2+c x^4)} \, dx\)

Optimal. Leaf size=288 \[ -\frac{\left (A \left (b^2-2 a c\right )-a b C\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 \sqrt{b^2-4 a c}}+\frac{(A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^2}-\frac{\log (x) (A b-a C)}{a^2}-\frac{A}{2 a x^2}-\frac{B \sqrt{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} a \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \sqrt{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} a \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{B}{a x} \]

[Out]

-A/(2*a*x^2) - B/(a*x) - (B*Sqrt[c]*(1 + b/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4
*a*c]]])/(Sqrt[2]*a*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (B*Sqrt[c]*(1 - b/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c
]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((A*(b^2 - 2*a*c) - a*b*C)*ArcTan
h[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^2*Sqrt[b^2 - 4*a*c]) - ((A*b - a*C)*Log[x])/a^2 + ((A*b - a*C)*Log[a
+ b*x^2 + c*x^4])/(4*a^2)

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Rubi [A]  time = 0.473922, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.393, Rules used = {1662, 1251, 800, 634, 618, 206, 628, 12, 1123, 1166, 205} \[ -\frac{\left (A \left (b^2-2 a c\right )-a b C\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 \sqrt{b^2-4 a c}}+\frac{(A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^2}-\frac{\log (x) (A b-a C)}{a^2}-\frac{A}{2 a x^2}-\frac{B \sqrt{c} \left (\frac{b}{\sqrt{b^2-4 a c}}+1\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} a \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \sqrt{c} \left (1-\frac{b}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} a \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{B}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(x^3*(a + b*x^2 + c*x^4)),x]

[Out]

-A/(2*a*x^2) - B/(a*x) - (B*Sqrt[c]*(1 + b/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4
*a*c]]])/(Sqrt[2]*a*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (B*Sqrt[c]*(1 - b/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c
]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*a*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((A*(b^2 - 2*a*c) - a*b*C)*ArcTan
h[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^2*Sqrt[b^2 - 4*a*c]) - ((A*b - a*C)*Log[x])/a^2 + ((A*b - a*C)*Log[a
+ b*x^2 + c*x^4])/(4*a^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 1251

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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 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])

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 12

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

Rule 1123

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

Rubi steps

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

Mathematica [A]  time = 0.983659, size = 377, normalized size = 1.31 \[ \frac{\frac{\left (A \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right )-a C \left (\sqrt{b^2-4 a c}+b\right )\right ) \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\sqrt{b^2-4 a c}}+\frac{\left (A \left (b \sqrt{b^2-4 a c}+2 a c-b^2\right )+a C \left (b-\sqrt{b^2-4 a c}\right )\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\sqrt{b^2-4 a c}}+4 \log (x) (a C-A b)-\frac{2 a A}{x^2}-\frac{2 \sqrt{2} a B \sqrt{c} \left (\sqrt{b^2-4 a c}+b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{2 \sqrt{2} a B \sqrt{c} \left (\sqrt{b^2-4 a c}-b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{4 a B}{x}}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(x^3*(a + b*x^2 + c*x^4)),x]

[Out]

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

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Maple [B]  time = 0.037, size = 1054, normalized size = 3.7 \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)/x^3/(c*x^4+b*x^2+a),x)

[Out]

-1/2*A/a/x^2-B/a/x-1/a^2*ln(x)*A*b+1/a*ln(x)*C+8/a*c/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*b-2/a^
2/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*b^3+4/a*c/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b
)*(-4*a*c+b^2)^(1/2)*A-2/a^2/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*(-4*a*c+b^2)^(1/2)*A*b^2+2/a/(32
*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*C*b*(-4*a*c+b^2)^(1/2)-8*c/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^
2)^(1/2)-b)*C+2/a/(32*a*c-8*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*C*b^2-4/a*c/(32*a*c-8*b^2)*B*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))*b*(-4*a*c+b^2)^(1/2)+16*c^2/(
32*a*c-8*b^2)*B*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))
-4/a*c/(32*a*c-8*b^2)*B*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))*b^2+8/a*c/(32*a*c-8*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*b-2/a^2/(32*a*c-8*b^2)*ln(2*c*x^2+(-4*a*c
+b^2)^(1/2)+b)*A*b^3-4/a*c/(32*a*c-8*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*(-4*a*c+b^2)^(1/2)*A+2/a^2/(32*a*c-
8*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*(-4*a*c+b^2)^(1/2)*A*b^2-2/a/(32*a*c-8*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1
/2)+b)*C*b*(-4*a*c+b^2)^(1/2)-8*c/(32*a*c-8*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*C+2/a/(32*a*c-8*b^2)*ln(2*c*
x^2+(-4*a*c+b^2)^(1/2)+b)*C*b^2-4/a*c/(32*a*c-8*b^2)*B*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))*b*(-4*a*c+b^2)^(1/2)-16*c^2/(32*a*c-8*b^2)*B*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))+4/a*c/(32*a*c-8*b^2)*B*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))*b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/x^3/(c*x^4+b*x^2+a),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)/x**3/(c*x**4+b*x**2+a),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)/x^3/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError