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

Optimal. Leaf size=534 \[ -\frac{\left (2 A \left (6 a^2 c^2-6 a b^2 c+b^4\right )-a b C \left (b^2-6 a c\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{-6 a A c-a b C+2 A b^2}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac{(2 A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\log (x) (2 A b-a C)}{a^3}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 x \left (b^2-4 a c\right )}-\frac{B \sqrt{c} \left (\left (3 b^2-10 a c\right ) \sqrt{b^2-4 a c}-16 a b c+3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{B \sqrt{c} \left (-\left (3 b^2-10 a c\right ) \sqrt{b^2-4 a c}-16 a b c+3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{A \left (b^2-2 a c\right )+c x^2 (A b-2 a C)-a b C}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{B \left (-2 a c+b^2+b c x^2\right )}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

-(2*A*b^2 - 6*a*A*c - a*b*C)/(2*a^2*(b^2 - 4*a*c)*x^2) - (B*(3*b^2 - 10*a*c))/(2*a^2*(b^2 - 4*a*c)*x) + (B*(b^
2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*x*(a + b*x^2 + c*x^4)) + (A*(b^2 - 2*a*c) - a*b*C + c*(A*b - 2*a*C)*x
^2)/(2*a*(b^2 - 4*a*c)*x^2*(a + b*x^2 + c*x^4)) - (B*Sqrt[c]*(3*b^3 - 16*a*b*c + (3*b^2 - 10*a*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^2*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sq
rt[b^2 - 4*a*c]]) + (B*Sqrt[c]*(3*b^3 - 16*a*b*c - (3*b^2 - 10*a*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^2*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((2*A*(b^4
- 6*a*b^2*c + 6*a^2*c^2) - a*b*(b^2 - 6*a*c)*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*(b^2 - 4*a*c
)^(3/2)) - ((2*A*b - a*C)*Log[x])/a^3 + ((2*A*b - a*C)*Log[a + b*x^2 + c*x^4])/(4*a^3)

________________________________________________________________________________________

Rubi [A]  time = 1.99236, antiderivative size = 534, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {1662, 1251, 822, 800, 634, 618, 206, 628, 12, 1121, 1281, 1166, 205} \[ -\frac{\left (2 A \left (6 a^2 c^2-6 a b^2 c+b^4\right )-a b C \left (b^2-6 a c\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{-6 a A c-a b C+2 A b^2}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac{(2 A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\log (x) (2 A b-a C)}{a^3}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 x \left (b^2-4 a c\right )}-\frac{B \sqrt{c} \left (\left (3 b^2-10 a c\right ) \sqrt{b^2-4 a c}-16 a b c+3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{B \sqrt{c} \left (-\left (3 b^2-10 a c\right ) \sqrt{b^2-4 a c}-16 a b c+3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{A \left (b^2-2 a c\right )+c x^2 (A b-2 a C)-a b C}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{B \left (-2 a c+b^2+b c x^2\right )}{2 a x \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*A*b^2 - 6*a*A*c - a*b*C)/(2*a^2*(b^2 - 4*a*c)*x^2) - (B*(3*b^2 - 10*a*c))/(2*a^2*(b^2 - 4*a*c)*x) + (B*(b^
2 - 2*a*c + b*c*x^2))/(2*a*(b^2 - 4*a*c)*x*(a + b*x^2 + c*x^4)) + (A*(b^2 - 2*a*c) - a*b*C + c*(A*b - 2*a*C)*x
^2)/(2*a*(b^2 - 4*a*c)*x^2*(a + b*x^2 + c*x^4)) - (B*Sqrt[c]*(3*b^3 - 16*a*b*c + (3*b^2 - 10*a*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^2*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sq
rt[b^2 - 4*a*c]]) + (B*Sqrt[c]*(3*b^3 - 16*a*b*c - (3*b^2 - 10*a*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^2*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((2*A*(b^4
- 6*a*b^2*c + 6*a^2*c^2) - a*b*(b^2 - 6*a*c)*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^3*(b^2 - 4*a*c
)^(3/2)) - ((2*A*b - a*C)*Log[x])/a^3 + ((2*A*b - a*C)*Log[a + b*x^2 + c*x^4])/(4*a^3)

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 822

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

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 1121

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[((d*x)^(m + 1)*(b^2 - 2*a
*c + b*c*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*d*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*
c)), Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m + 4*p + 5) + b*c*(m + 4*p + 7)*
x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (Integer
Q[p] || IntegerQ[m])

Rule 1281

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

Rubi steps

\begin{align*} \int \frac{A+B x+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx &=\int \frac{B}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx+\int \frac{A+C x^2}{x^3 \left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+C x}{x^2 \left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )+B \int \frac{1}{x^2 \left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 A b^2+6 a A c+a b C-2 c (A b-2 a C) x}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}-\frac{B \int \frac{-3 b^2+10 a c-3 b c x^2}{x^2 \left (a+b x^2+c x^4\right )} \, dx}{2 a \left (b^2-4 a c\right )}\\ &=-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{-2 A b^2+6 a A c+a b C}{a x^2}+\frac{\left (-b^2+4 a c\right ) (-2 A b+a C)}{a^2 x}+\frac{-2 A \left (b^4-5 a b^2 c+3 a^2 c^2\right )+a b \left (b^2-5 a c\right ) C-c \left (b^2-4 a c\right ) (2 A b-a C) x}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}+\frac{B \int \frac{-b \left (3 b^2-13 a c\right )-c \left (3 b^2-10 a c\right ) x^2}{a+b x^2+c x^4} \, dx}{2 a^2 \left (b^2-4 a c\right )}\\ &=-\frac{2 A b^2-6 a A c-a b C}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{(2 A b-a C) \log (x)}{a^3}-\frac{\operatorname{Subst}\left (\int \frac{-2 A \left (b^4-5 a b^2 c+3 a^2 c^2\right )+a b \left (b^2-5 a c\right ) C-c \left (b^2-4 a c\right ) (2 A b-a C) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^3 \left (b^2-4 a c\right )}-\frac{\left (B c \left (3 b^2-10 a c+\frac{3 b^3}{\sqrt{b^2-4 a c}}-\frac{16 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^2 \left (b^2-4 a c\right )}-\frac{\left (B c \left (3 b^2-10 a c-\frac{3 b^3}{\sqrt{b^2-4 a c}}+\frac{16 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^2 \left (b^2-4 a c\right )}\\ &=-\frac{2 A b^2-6 a A c-a b C}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{B \sqrt{c} \left (3 b^2-10 a c+\frac{3 b^3}{\sqrt{b^2-4 a c}}-\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \sqrt{c} \left (3 b^2-10 a c-\frac{3 b^3}{\sqrt{b^2-4 a c}}+\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(2 A b-a C) \log (x)}{a^3}+\frac{(2 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^3}+\frac{\left (2 A \left (b^4-6 a b^2 c+6 a^2 c^2\right )-a b \left (b^2-6 a c\right ) C\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 a^3 \left (b^2-4 a c\right )}\\ &=-\frac{2 A b^2-6 a A c-a b C}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{B \sqrt{c} \left (3 b^2-10 a c+\frac{3 b^3}{\sqrt{b^2-4 a c}}-\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \sqrt{c} \left (3 b^2-10 a c-\frac{3 b^3}{\sqrt{b^2-4 a c}}+\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(2 A b-a C) \log (x)}{a^3}+\frac{(2 A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^3}-\frac{\left (2 A \left (b^4-6 a b^2 c+6 a^2 c^2\right )-a b \left (b^2-6 a c\right ) 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^3 \left (b^2-4 a c\right )}\\ &=-\frac{2 A b^2-6 a A c-a b C}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac{B \left (3 b^2-10 a c\right )}{2 a^2 \left (b^2-4 a c\right ) x}+\frac{B \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) x \left (a+b x^2+c x^4\right )}+\frac{A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{B \sqrt{c} \left (3 b^2-10 a c+\frac{3 b^3}{\sqrt{b^2-4 a c}}-\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \sqrt{c} \left (3 b^2-10 a c-\frac{3 b^3}{\sqrt{b^2-4 a c}}+\frac{16 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^2 \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{\left (2 A \left (b^4-6 a b^2 c+6 a^2 c^2\right )-a b \left (b^2-6 a c\right ) C\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{(2 A b-a C) \log (x)}{a^3}+\frac{(2 A b-a C) \log \left (a+b x^2+c x^4\right )}{4 a^3}\\ \end{align*}

Mathematica [A]  time = 2.76088, size = 655, normalized size = 1.23 \[ \frac{-\frac{2 a \left (2 a^2 c C+A \left (-3 a b c-2 a c^2 x^2+b^2 c x^2+b^3\right )-a \left (b^2 C+b c x (3 B+C x)+2 B c^2 x^3\right )+b^2 B x \left (b+c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{\left (2 A \left (6 a^2 c^2+b^3 \sqrt{b^2-4 a c}-6 a b^2 c-4 a b c \sqrt{b^2-4 a c}+b^4\right )+a C \left (-b^2 \sqrt{b^2-4 a c}+4 a c \sqrt{b^2-4 a c}+6 a b c-b^3\right )\right ) \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\left (2 A \left (-6 a^2 c^2+b^3 \sqrt{b^2-4 a c}+6 a b^2 c-4 a b c \sqrt{b^2-4 a c}-b^4\right )+a C \left (-b^2 \sqrt{b^2-4 a c}+4 a c \sqrt{b^2-4 a c}-6 a b c+b^3\right )\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+4 \log (x) (a C-2 A b)-\frac{2 a A}{x^2}+\frac{\sqrt{2} a B \sqrt{c} \left (-3 b^2 \sqrt{b^2-4 a c}+10 a c \sqrt{b^2-4 a c}+16 a b c-3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} a B \sqrt{c} \left (-3 b^2 \sqrt{b^2-4 a c}+10 a c \sqrt{b^2-4 a c}-16 a b c+3 b^3\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{4 a B}{x}}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*A)/x^2 - (4*a*B)/x - (2*a*(2*a^2*c*C + b^2*B*x*(b + c*x^2) + A*(b^3 - 3*a*b*c + b^2*c*x^2 - 2*a*c^2*x^2
) - a*(b^2*C + 2*B*c^2*x^3 + b*c*x*(3*B + C*x))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (Sqrt[2]*a*B*Sqrt[c]*(
-3*b^3 + 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sq
rt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*a*B*Sqrt[c]*(3*b^3 - 16*a*b*c
- 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])
/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + 4*(-2*A*b + a*C)*Log[x] + ((2*A*(b^4 - 6*a*b^2*c + 6*a^2*
c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*a*c]) + a*(-b^3 + 6*a*b*c - b^2*Sqrt[b^2 - 4*a*c] + 4*a*c*S
qrt[b^2 - 4*a*c])*C)*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((2*A*(-b^4 + 6*a*b^2*c - 6*
a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*a*c]) + a*(b^3 - 6*a*b*c - b^2*Sqrt[b^2 - 4*a*c] + 4*a*
c*Sqrt[b^2 - 4*a*c])*C)*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/(4*a^3)

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Maple [B]  time = 0.062, size = 2512, normalized size = 4.7 \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)^2,x)

[Out]

-22/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^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))*B*b^2+22/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arcta
n(c*x*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*B*b^2-3/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^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))*B*b^4+3/a^2*c/(4*a*c-b^2)/(16*a*c-4*b
^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))*B*b^4-16/a*
c^2/(4*a*c-b^2)/(16*a*c-4*b^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))*B*(-4*a*c+b^2)^(1/2)*b-16/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^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))*B*(-4*a*c+b^2)^(1/2)*b+3/a^2*c/(4*a*c-b^2)/(16*a*c-
4*b^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))*B*(-4*a*
c+b^2)^(1/2)*b^3+3/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^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))*B*(-4*a*c+b^2)^(1/2)*b^3-1/a/(c*x^4+b*x^2+a)*B*c^2/(4*a*c-b^2)*x^3-1/a/(c*
x^4+b*x^2+a)*c^2/(4*a*c-b^2)*x^2*A+1/2/a^2/(c*x^4+b*x^2+a)*B*b^3/(4*a*c-b^2)*x-3/2/a/(c*x^4+b*x^2+a)/(4*a*c-b^
2)*A*b*c+2/a^3/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*b^5+2/a^3/(4*a*c-b^2)/(16*a*c-4*
b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*b^5-1/a^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)
*C*b^4-1/a^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*C*b^4-2/a^3*ln(x)*A*b+1/(c*x^4+b*x^2+
a)/(4*a*c-b^2)*c*C+1/2/a^2/(c*x^4+b*x^2+a)/(4*a*c-b^2)*A*b^3-1/2/a/(c*x^4+b*x^2+a)/(4*a*c-b^2)*C*b^2-16*c^2/(4
*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*C-16*c^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*
a*c+b^2)^(1/2)+b)*C-B/a^2/x+1/a^2*ln(x)*C-1/2*A/a^2/x^2-40*c^3/(4*a*c-b^2)/(16*a*c-4*b^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))*B+12/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^2
)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*(-4*a*c+b^2)^(1/2)-3/2/a/(c*x^4+b*x^2+a)*B*b/(4*a*c-b^2)*x*c+1/2/a^2/(c*
x^4+b*x^2+a)*B*c/(4*a*c-b^2)*x^3*b^2+1/2/a^2/(c*x^4+b*x^2+a)*c/(4*a*c-b^2)*x^2*A*b^2-1/2/a/(c*x^4+b*x^2+a)*c/(
4*a*c-b^2)*x^2*b*C-1/a^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*C*(-4*a*c+b^2)^(1/2)*b^3
+1/a^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*C*(-4*a*c+b^2)^(1/2)*b^3+8/a*c/(4*a*c-b^2)/
(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*C*b^2+2/a^3/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2
)^(1/2)-b)*A*(-4*a*c+b^2)^(1/2)*b^4+32/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*b-
12/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*(-4*a*c+b^2)^(1/2)-2/a^3/(4*a*c-b^2)/(1
6*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*(-4*a*c+b^2)^(1/2)*b^4+32/a*c^2/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(
2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*b-16/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*b^3+
12/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*(-4*a*c+b^2)^(1/2)*b^2-6/a*c/(4*a*c-b^2
)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*C*(-4*a*c+b^2)^(1/2)*b-12/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^2)*l
n(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*A*(-4*a*c+b^2)^(1/2)*b^2+6/a*c/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c
+b^2)^(1/2)-b)*C*(-4*a*c+b^2)^(1/2)*b+8/a*c/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(-2*c*x^2+(-4*a*c+b^2)^(1/2)-b)*C*b^2
-16/a^2*c/(4*a*c-b^2)/(16*a*c-4*b^2)*ln(2*c*x^2+(-4*a*c+b^2)^(1/2)+b)*A*b^3+40*c^3/(4*a*c-b^2)/(16*a*c-4*b^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))*B

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Maxima [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)^2,x, algorithm="maxima")

[Out]

Timed out

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

[Out]

Exception raised: NotImplementedError