3.12 \(\int \frac{d+e x+f x^2+g x^3}{4-5 x^2+x^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac{1}{6} (d+4 f) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f) \tanh ^{-1}(x)-\frac{1}{6} (e+g) \log \left (1-x^2\right )+\frac{1}{6} (e+4 g) \log \left (4-x^2\right ) \]

[Out]

-((d + 4*f)*ArcTanh[x/2])/6 + ((d + f)*ArcTanh[x])/3 - ((e + g)*Log[1 - x^2])/6 + ((e + 4*g)*Log[4 - x^2])/6

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Rubi [A]  time = 0.0722953, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1673, 1166, 207, 1247, 632, 31} \[ -\frac{1}{6} (d+4 f) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f) \tanh ^{-1}(x)-\frac{1}{6} (e+g) \log \left (1-x^2\right )+\frac{1}{6} (e+4 g) \log \left (4-x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3)/(4 - 5*x^2 + x^4),x]

[Out]

-((d + 4*f)*ArcTanh[x/2])/6 + ((d + f)*ArcTanh[x])/3 - ((e + g)*Log[1 - x^2])/6 + ((e + 4*g)*Log[4 - x^2])/6

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(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]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{d+e x+f x^2+g x^3}{4-5 x^2+x^4} \, dx &=\int \frac{d+f x^2}{4-5 x^2+x^4} \, dx+\int \frac{x \left (e+g x^2\right )}{4-5 x^2+x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{e+g x}{4-5 x+x^2} \, dx,x,x^2\right )-\frac{1}{3} (d+f) \int \frac{1}{-1+x^2} \, dx+\frac{1}{3} (d+4 f) \int \frac{1}{-4+x^2} \, dx\\ &=-\frac{1}{6} (d+4 f) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f) \tanh ^{-1}(x)+\frac{1}{6} (-e-g) \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,x^2\right )+\frac{1}{6} (e+4 g) \operatorname{Subst}\left (\int \frac{1}{-4+x} \, dx,x,x^2\right )\\ &=-\frac{1}{6} (d+4 f) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f) \tanh ^{-1}(x)-\frac{1}{6} (e+g) \log \left (1-x^2\right )+\frac{1}{6} (e+4 g) \log \left (4-x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.032265, size = 68, normalized size = 1.19 \[ \frac{1}{12} (-2 \log (1-x) (d+e+f+g)+\log (2-x) (d+2 e+4 f+8 g)+2 \log (x+1) (d-e+f-g)-\log (x+2) (d-2 e+4 f-8 g)) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3)/(4 - 5*x^2 + x^4),x]

[Out]

(-2*(d + e + f + g)*Log[1 - x] + (d + 2*e + 4*f + 8*g)*Log[2 - x] + 2*(d - e + f - g)*Log[1 + x] - (d - 2*e +
4*f - 8*g)*Log[2 + x])/12

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Maple [B]  time = 0.01, size = 114, normalized size = 2. \begin{align*} -{\frac{\ln \left ( 2+x \right ) d}{12}}+{\frac{\ln \left ( 2+x \right ) e}{6}}-{\frac{\ln \left ( 2+x \right ) f}{3}}+{\frac{2\,\ln \left ( 2+x \right ) g}{3}}+{\frac{\ln \left ( 1+x \right ) d}{6}}-{\frac{\ln \left ( 1+x \right ) e}{6}}+{\frac{\ln \left ( 1+x \right ) f}{6}}-{\frac{\ln \left ( 1+x \right ) g}{6}}+{\frac{\ln \left ( x-2 \right ) d}{12}}+{\frac{\ln \left ( x-2 \right ) e}{6}}+{\frac{\ln \left ( x-2 \right ) f}{3}}+{\frac{2\,\ln \left ( x-2 \right ) g}{3}}-{\frac{\ln \left ( x-1 \right ) d}{6}}-{\frac{\ln \left ( x-1 \right ) e}{6}}-{\frac{\ln \left ( x-1 \right ) f}{6}}-{\frac{\ln \left ( x-1 \right ) g}{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x)

[Out]

-1/12*ln(2+x)*d+1/6*ln(2+x)*e-1/3*ln(2+x)*f+2/3*ln(2+x)*g+1/6*ln(1+x)*d-1/6*ln(1+x)*e+1/6*ln(1+x)*f-1/6*ln(1+x
)*g+1/12*ln(x-2)*d+1/6*ln(x-2)*e+1/3*ln(x-2)*f+2/3*ln(x-2)*g-1/6*ln(x-1)*d-1/6*ln(x-1)*e-1/6*ln(x-1)*f-1/6*ln(
x-1)*g

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Maxima [A]  time = 0.966077, size = 82, normalized size = 1.44 \begin{align*} -\frac{1}{12} \,{\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) + \frac{1}{6} \,{\left (d - e + f - g\right )} \log \left (x + 1\right ) - \frac{1}{6} \,{\left (d + e + f + g\right )} \log \left (x - 1\right ) + \frac{1}{12} \,{\left (d + 2 \, e + 4 \, f + 8 \, g\right )} \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="maxima")

[Out]

-1/12*(d - 2*e + 4*f - 8*g)*log(x + 2) + 1/6*(d - e + f - g)*log(x + 1) - 1/6*(d + e + f + g)*log(x - 1) + 1/1
2*(d + 2*e + 4*f + 8*g)*log(x - 2)

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Fricas [A]  time = 3.66056, size = 197, normalized size = 3.46 \begin{align*} -\frac{1}{12} \,{\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) + \frac{1}{6} \,{\left (d - e + f - g\right )} \log \left (x + 1\right ) - \frac{1}{6} \,{\left (d + e + f + g\right )} \log \left (x - 1\right ) + \frac{1}{12} \,{\left (d + 2 \, e + 4 \, f + 8 \, g\right )} \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="fricas")

[Out]

-1/12*(d - 2*e + 4*f - 8*g)*log(x + 2) + 1/6*(d - e + f - g)*log(x + 1) - 1/6*(d + e + f + g)*log(x - 1) + 1/1
2*(d + 2*e + 4*f + 8*g)*log(x - 2)

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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((g*x**3+f*x**2+e*x+d)/(x**4-5*x**2+4),x)

[Out]

Timed out

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Giac [A]  time = 1.14393, size = 93, normalized size = 1.63 \begin{align*} -\frac{1}{12} \,{\left (d + 4 \, f - 8 \, g - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac{1}{6} \,{\left (d + f - g - e\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{6} \,{\left (d + f + g + e\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac{1}{12} \,{\left (d + 4 \, f + 8 \, g + 2 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4-5*x^2+4),x, algorithm="giac")

[Out]

-1/12*(d + 4*f - 8*g - 2*e)*log(abs(x + 2)) + 1/6*(d + f - g - e)*log(abs(x + 1)) - 1/6*(d + f + g + e)*log(ab
s(x - 1)) + 1/12*(d + 4*f + 8*g + 2*e)*log(abs(x - 2))