### 3.30 $$\int \frac{x^3 (A+B x+C x^2)}{(a+b x^2+c x^4)^2} \, dx$$

Optimal. Leaf size=347 $\frac{x^2 \left (2 a c C+A b c+b^2 (-C)\right )+a (2 A c-b C)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a C) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{B x \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{B \left (b-\frac{4 a c+b^2}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{B \left (b \sqrt{b^2-4 a c}+4 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}$

[Out]

(B*x*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (a*(2*A*c - b*C) + (A*b*c - b^2*C + 2*a*c*C)*x^2)/
(2*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (B*(b - (b^2 + 4*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]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (B*(b^2 + 4*a*c
+ b*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*
c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((A*b - 2*a*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 = 0.617503, antiderivative size = 347, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.321, Rules used = {1662, 1251, 777, 618, 206, 12, 1120, 1166, 205} $\frac{x^2 \left (2 a c C+A b c+b^2 (-C)\right )+a (2 A c-b C)}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{(A b-2 a C) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{B x \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac{B \left (b-\frac{4 a c+b^2}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{B \left (b \sqrt{b^2-4 a c}+4 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(B*x*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (a*(2*A*c - b*C) + (A*b*c - b^2*C + 2*a*c*C)*x^2)/
(2*c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (B*(b - (b^2 + 4*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]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (B*(b^2 + 4*a*c
+ b*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(2*Sqrt[2]*Sqrt[c]*(b^2 - 4*a*
c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) - ((A*b - 2*a*C)*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 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 777

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

Rule 12

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

Rule 1120

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(d^3*(d*x)^(m - 3)*(2*a +
b*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*(p + 1)*(b^2 - 4*a*c)), x] + Dist[d^4/(2*(p + 1)*(b^2 - 4*a*c)), Int[(
d*x)^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 3] && 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 )}{\left (a+b x^2+c x^4\right )^2} \, dx &=\int \frac{B x^4}{\left (a+b x^2+c x^4\right )^2} \, dx+\int \frac{x^3 \left (A+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (A+C x)}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )+B \int \frac{x^4}{\left (a+b x^2+c x^4\right )^2} \, dx\\ &=\frac{B x \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{a (2 A c-b C)+\left (A b c-b^2 C+2 a c C\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{B \int \frac{2 a-b x^2}{a+b x^2+c x^4} \, dx}{2 \left (b^2-4 a c\right )}+\frac{(A b-2 a C) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )}\\ &=\frac{B x \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{a (2 A c-b C)+\left (A b c-b^2 C+2 a c C\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (B \left (b^2+4 a c-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}{4 \left (b^2-4 a c\right )^{3/2}}+\frac{\left (B \left (b^2+4 a c+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}{4 \left (b^2-4 a c\right )^{3/2}}-\frac{(A b-2 a 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 x \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{a (2 A c-b C)+\left (A b c-b^2 C+2 a c C\right ) x^2}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{B \left (b^2+4 a c-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 )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{B \left (b^2+4 a c+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 )}{2 \sqrt{2} \sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{b+\sqrt{b^2-4 a c}}}-\frac{(A b-2 a 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 = 0.995371, size = 358, normalized size = 1.03 $\frac{1}{4} \left (-\frac{2 \left (a (2 A c-b C+2 c x (B+C x))+b x^2 (A c-b C+B c x)\right )}{c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )}+\frac{2 (A b-2 a C) \log \left (\sqrt{b^2-4 a c}-b-2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{2 (A b-2 a C) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\sqrt{2} B \left (b \sqrt{b^2-4 a c}-4 a c-b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\sqrt{2} B \left (b \sqrt{b^2-4 a c}+4 a c+b^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{c} \left (b^2-4 a c\right )^{3/2} \sqrt{\sqrt{b^2-4 a c}+b}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-2*(b*x^2*(A*c - b*C + B*c*x) + a*(2*A*c - b*C + 2*c*x*(B + C*x))))/(c*(-b^2 + 4*a*c)*(a + b*x^2 + c*x^4)) +
(Sqrt[2]*B*(-b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqr
t[c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (Sqrt[2]*B*(b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*ArcTan[
(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[c]*(b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]) +
(2*(A*b - 2*a*C)*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2])/(b^2 - 4*a*c)^(3/2) - (2*(A*b - 2*a*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 [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \text{hanged} \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)^2,x)

[Out]

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

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

[Out]

Exception raised: NotImplementedError