### 3.22 $$\int \frac{x^3 (A+B x+C x^2)}{a+b x^2+c x^4} \, dx$$

Optimal. Leaf size=278 $\frac{\left (2 a c C+A b c+b^2 (-C)\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}+\frac{(A c-b C) \log \left (a+b x^2+c x^4\right )}{4 c^2}-\frac{B \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} c^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{B x}{c}+\frac{C x^2}{2 c}$

[Out]

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

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Rubi [A]  time = 0.465966, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.393, Rules used = {1662, 1251, 773, 634, 618, 206, 628, 12, 1122, 1166, 205} $\frac{\left (2 a c C+A b c+b^2 (-C)\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}+\frac{(A c-b C) \log \left (a+b x^2+c x^4\right )}{4 c^2}-\frac{B \left (b-\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \left (\frac{b^2-2 a c}{\sqrt{b^2-4 a c}}+b\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} c^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}+\frac{B x}{c}+\frac{C x^2}{2 c}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(B*x)/c + (C*x^2)/(2*c) - (B*(b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^
2 - 4*a*c]]])/(Sqrt[2]*c^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (B*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*ArcTan[
(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*c^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + ((A*b*c - b^
2*C + 2*a*c*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c^2*Sqrt[b^2 - 4*a*c]) + ((A*c - b*C)*Log[a + b*x^
2 + c*x^4])/(4*c^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 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*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]

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 1122

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d^3*(d*x)^(m - 3)*(a + b*
x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 1)), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && 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{x^3 \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx &=\int \frac{B x^4}{a+b x^2+c x^4} \, dx+\int \frac{x^3 \left (A+C x^2\right )}{a+b x^2+c x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (A+C x)}{a+b x+c x^2} \, dx,x,x^2\right )+B \int \frac{x^4}{a+b x^2+c x^4} \, dx\\ &=\frac{B x}{c}+\frac{C x^2}{2 c}+\frac{\operatorname{Subst}\left (\int \frac{-a C+(A c-b C) x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 c}-\frac{B \int \frac{a+b x^2}{a+b x^2+c x^4} \, dx}{c}\\ &=\frac{B x}{c}+\frac{C x^2}{2 c}-\frac{\left (B \left (b-\frac{b^2-2 a 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}{2 c}-\frac{\left (B \left (b+\frac{b^2-2 a 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}{2 c}+\frac{(A c-b C) \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}-\frac{\left (A b c-b^2 C+2 a c C\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{B x}{c}+\frac{C x^2}{2 c}-\frac{B \left (b-\frac{b^2-2 a 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 )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \left (b+\frac{b^2-2 a 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 )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{(A c-b C) \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac{\left (A b c-b^2 C+2 a c C\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{2 c^2}\\ &=\frac{B x}{c}+\frac{C x^2}{2 c}-\frac{B \left (b-\frac{b^2-2 a 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 )}{\sqrt{2} c^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{B \left (b+\frac{b^2-2 a 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 )}{\sqrt{2} c^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{\left (A b c-b^2 C+2 a c C\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 c^2 \sqrt{b^2-4 a c}}+\frac{(A c-b C) \log \left (a+b x^2+c x^4\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.45116, size = 377, normalized size = 1.36 $\frac{\frac{\left (A c \left (\sqrt{b^2-4 a c}-b\right )+C \left (-b \sqrt{b^2-4 a c}-2 a c+b^2\right )\right ) \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\sqrt{b^2-4 a c}}-\frac{\left (C \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right )-A c \left (\sqrt{b^2-4 a c}+b\right )\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\sqrt{b^2-4 a c}}-\frac{2 \sqrt{2} B \sqrt{c} \left (b \sqrt{b^2-4 a c}+2 a c-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{b^2-4 a c} \sqrt{b-\sqrt{b^2-4 a c}}}-\frac{2 \sqrt{2} B \sqrt{c} \left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}+4 B c x+2 c C x^2}{4 c^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 1171, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError