### 3.29 $$\int \frac{x^4 (A+B x+C x^2)}{(a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=412 $\frac{\left (-\frac{A c \left (4 a c+b^2\right )+b C \left (b^2-8 a c\right )}{\sqrt{b^2-4 a c}}+C \left (b^2-6 a c\right )+A b c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (\frac{A c \left (4 a c+b^2\right )+b C \left (b^2-8 a c\right )}{\sqrt{b^2-4 a c}}+C \left (b^2-6 a c\right )+A b c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} c^{3/2} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{x^3 \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{x (2 A c-b C)}{2 c \left (b^2-4 a c\right )}+\frac{B x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{2 a B \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}$

[Out]

((2*A*c - b*C)*x)/(2*c*(b^2 - 4*a*c)) + (B*x^2*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (x^3*(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)) + ((A*b*c + (b^2 - 6*a*c)*C - (A*c*(b^2
+ 4*a*c) + b*(b^2 - 8*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*S
qrt[2]*c^(3/2)*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((A*b*c + (b^2 - 6*a*c)*C + (A*c*(b^2 + 4*a*c) + b
*(b^2 - 8*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]*c^(3/
2)*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*a*B*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)
^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 1.33379, antiderivative size = 412, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.357, Rules used = {1662, 1275, 1279, 1166, 205, 12, 1114, 722, 618, 206} $\frac{\left (-\frac{A c \left (4 a c+b^2\right )+b C \left (b^2-8 a c\right )}{\sqrt{b^2-4 a c}}+C \left (b^2-6 a c\right )+A b c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (\frac{A c \left (4 a c+b^2\right )+b C \left (b^2-8 a c\right )}{\sqrt{b^2-4 a c}}+C \left (b^2-6 a c\right )+A b c\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} c^{3/2} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{x^3 \left (-2 a C+x^2 (2 A c-b C)+A b\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{x (2 A c-b C)}{2 c \left (b^2-4 a c\right )}+\frac{B x^2 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{2 a B \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*A*c - b*C)*x)/(2*c*(b^2 - 4*a*c)) + (B*x^2*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - (x^3*(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)) + ((A*b*c + (b^2 - 6*a*c)*C - (A*c*(b^2
+ 4*a*c) + b*(b^2 - 8*a*c)*C)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(2*S
qrt[2]*c^(3/2)*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((A*b*c + (b^2 - 6*a*c)*C + (A*c*(b^2 + 4*a*c) + b
*(b^2 - 8*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]*c^(3/
2)*(b^2 - 4*a*c)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) + (2*a*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 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[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 722

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

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^4 \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx &=\int \frac{B x^5}{\left (a+b x^2+c x^4\right )^2} \, dx+\int \frac{x^4 \left (A+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=-\frac{x^3 \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^5}{\left (a+b x^2+c x^4\right )^2} \, dx+\frac{\int \frac{x^2 \left (3 (A b-2 a C)+(2 A c-b C) x^2\right )}{a+b x^2+c x^4} \, dx}{2 \left (b^2-4 a c\right )}\\ &=\frac{(2 A c-b C) x}{2 c \left (b^2-4 a c\right )}-\frac{x^3 \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^2}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )-\frac{\int \frac{a (2 A c-b C)+\left (-A b c-\left (b^2-6 a c\right ) C\right ) x^2}{a+b x^2+c x^4} \, dx}{2 c \left (b^2-4 a c\right )}\\ &=\frac{(2 A c-b C) x}{2 c \left (b^2-4 a c\right )}+\frac{B x^2 \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^3 \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{(a B) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{b^2-4 a c}+\frac{\left (A b c+\left (b^2-6 a c\right ) C-\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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 c \left (b^2-4 a c\right )}+\frac{\left (A b c+\left (b^2-6 a c\right ) C+\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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 c \left (b^2-4 a c\right )}\\ &=\frac{(2 A c-b C) x}{2 c \left (b^2-4 a c\right )}+\frac{B x^2 \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^3 \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 (A b c+\left (b^2-6 a c\right ) C-\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (A b c+\left (b^2-6 a c\right ) C+\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{(2 a 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{(2 A c-b C) x}{2 c \left (b^2-4 a c\right )}+\frac{B x^2 \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^3 \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 (A b c+\left (b^2-6 a c\right ) C-\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (A b c+\left (b^2-6 a c\right ) C+\frac{A c \left (b^2+4 a c\right )+b \left (b^2-8 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} c^{3/2} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}+\frac{2 a 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 = 1.56148, size = 444, normalized size = 1.08 $\frac{1}{4} \left (\frac{2 \left (a (b (B+C x)-2 c x (A+x (B+C x)))+b x^2 (b (B+C x)-A c x)\right )}{c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )}+\frac{\sqrt{2} \left (C \left (b^2 \sqrt{b^2-4 a c}-6 a c \sqrt{b^2-4 a c}+8 a b c-b^3\right )-A c \left (-b \sqrt{b^2-4 a c}+4 a c+b^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} \left (A c \left (b \sqrt{b^2-4 a c}+4 a c+b^2\right )+C \left (b^2 \sqrt{b^2-4 a c}-6 a c \sqrt{b^2-4 a c}-8 a b c+b^3\right )\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{4 a B \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{4 a B \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((2*(b*x^2*(-(A*c*x) + b*(B + C*x)) + a*(b*(B + C*x) - 2*c*x*(A + x*(B + C*x)))))/(c*(-b^2 + 4*a*c)*(a + b*x^2
+ c*x^4)) + (Sqrt[2]*(-(A*c*(b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])) + (-b^3 + 8*a*b*c + b^2*Sqrt[b^2 - 4*a*c] -
6*a*c*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(c^(3/2)*(b^2 - 4*a*c)^(3
/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*(A*c*(b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c]) + (b^3 - 8*a*b*c + b^2*Sq
rt[b^2 - 4*a*c] - 6*a*c*Sqrt[b^2 - 4*a*c])*C)*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(c^(3/2
)*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - (4*a*B*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/(b^2 - 4*a*
c)^(3/2) + (4*a*B*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/4

________________________________________________________________________________________

Maple [B]  time = 0.042, size = 1429, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/2*(A*b*c+2*C*a*c-C*b^2)/(4*a*c-b^2)/c*x^3-1/2*B*(2*a*c-b^2)/(4*a*c-b^2)/c*x^2-1/2*a*(2*A*c-C*b)/(4*a*c-b^2
)/c*x+1/2*B*a*b/c/(4*a*c-b^2))/(c*x^4+b*x^2+a)-1/(4*a*c-b^2)^2*B*(-4*a*c+b^2)^(1/2)*a*ln(-2*c*x^2+(-4*a*c+b^2)
^(1/2)-b)+1/(4*a*c-b^2)^2*c*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)*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))*A*(-4*a*c+b^2)^(1/2)*b^2+1/(4*a*c-b^2)^2*c*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*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))*A*b^3-2/(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))*C*(-4*a*c+b^2)^(1/2)*a*b+1/
4/(4*a*c-b^2)^2/c*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^3-6/(4*a*c-b^2)^2*c*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^2+5/2/(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))*C*a*b^2-1/4/(4*a*c-b^2)^2/c*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2)*a
rctanh(c*x*2^(1/2)/(((-4*a*c+b^2)^(1/2)-b)*c)^(1/2))*C*b^4+1/(4*a*c-b^2)^2*B*(-4*a*c+b^2)^(1/2)*a*ln(2*c*x^2+(
-4*a*c+b^2)^(1/2)+b)+1/(4*a*c-b^2)^2*c*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)*a+1/4/(4*a*c-b^2)^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))*A*(-4*a*c+b^2)^(1/2)*b^2-1/(4*a*c-b^2)^2*c*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*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))*A*b^3-2/(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))*C*(-4*a*c+b^2)^(1/2)*
a*b+1/4/(4*a*c-b^2)^2/c*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^3+6/(4*a*c-b^2)^2*c*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^2-5/2/(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))*C*a*b^2+1/4/(4*a*c-b^2)^2/c*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^4

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Exception raised: NotImplementedError