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

Optimal. Leaf size=76 \[ -\frac{1}{6} \tanh ^{-1}\left (\frac{x}{2}\right ) (d+4 f+16 h)+\frac{1}{3} \tanh ^{-1}(x) (d+f+h)-\frac{1}{6} \log \left (1-x^2\right ) (e+g+i)+\frac{1}{6} \log \left (4-x^2\right ) (e+4 g+16 i)+h x+\frac{i x^2}{2} \]

[Out]

h*x + (i*x^2)/2 - ((d + 4*f + 16*h)*ArcTanh[x/2])/6 + ((d + f + h)*ArcTanh[x])/3 - ((e + g + i)*Log[1 - x^2])/
6 + ((e + 4*g + 16*i)*Log[4 - x^2])/6

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Rubi [A]  time = 0.191523, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {1673, 1676, 1166, 207, 1663, 1657, 632, 31} \[ -\frac{1}{6} \tanh ^{-1}\left (\frac{x}{2}\right ) (d+4 f+16 h)+\frac{1}{3} \tanh ^{-1}(x) (d+f+h)-\frac{1}{6} \log \left (1-x^2\right ) (e+g+i)+\frac{1}{6} \log \left (4-x^2\right ) (e+4 g+16 i)+h x+\frac{i x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

h*x + (i*x^2)/2 - ((d + 4*f + 16*h)*ArcTanh[x/2])/6 + ((d + f + h)*ArcTanh[x])/3 - ((e + g + i)*Log[1 - x^2])/
6 + ((e + 4*g + 16*i)*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 1676

Int[(Pq_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^2 + c*x^4), x], x
] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x^2] && Expon[Pq, x^2] > 1

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 1663

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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+h x^4+14 x^5}{4-5 x^2+x^4} \, dx &=\int \frac{x \left (e+g x^2+14 x^4\right )}{4-5 x^2+x^4} \, dx+\int \frac{d+f x^2+h x^4}{4-5 x^2+x^4} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{e+g x+14 x^2}{4-5 x+x^2} \, dx,x,x^2\right )+\int \left (h+\frac{d-4 h+(f+5 h) x^2}{4-5 x^2+x^4}\right ) \, dx\\ &=h x+\frac{1}{2} \operatorname{Subst}\left (\int \left (14-\frac{56-e-(70+g) x}{4-5 x+x^2}\right ) \, dx,x,x^2\right )+\int \frac{d-4 h+(f+5 h) x^2}{4-5 x^2+x^4} \, dx\\ &=h x+7 x^2-\frac{1}{2} \operatorname{Subst}\left (\int \frac{56-e-(70+g) x}{4-5 x+x^2} \, dx,x,x^2\right )-\frac{1}{3} (d+f+h) \int \frac{1}{-1+x^2} \, dx+\frac{1}{3} (d+4 f+16 h) \int \frac{1}{-4+x^2} \, dx\\ &=h x+7 x^2-\frac{1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f+h) \tanh ^{-1}(x)-\frac{1}{6} (-224-e-4 g) \operatorname{Subst}\left (\int \frac{1}{-4+x} \, dx,x,x^2\right )-\frac{1}{6} (14+e+g) \operatorname{Subst}\left (\int \frac{1}{-1+x} \, dx,x,x^2\right )\\ &=h x+7 x^2-\frac{1}{6} (d+4 f+16 h) \tanh ^{-1}\left (\frac{x}{2}\right )+\frac{1}{3} (d+f+h) \tanh ^{-1}(x)-\frac{1}{6} (14+e+g) \log \left (1-x^2\right )+\frac{1}{6} (224+e+4 g) \log \left (4-x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0651369, size = 98, normalized size = 1.29 \[ \frac{1}{12} \left (-2 \log (1-x) (d+e+f+g+h+i)+\log (2-x) (d+2 e+4 (f+2 g+4 h+8 i))+2 \log (x+1) (d-e+f-g+h-i)-\log (x+2) (d-2 (e-2 f+4 g-8 h+16 i))+12 h x+6 i x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*h*x + 6*i*x^2 - 2*(d + e + f + g + h + i)*Log[1 - x] + (d + 2*e + 4*(f + 2*g + 4*h + 8*i))*Log[2 - x] + 2*
(d - e + f - g + h - i)*Log[1 + x] - (d - 2*(e - 2*f + 4*g - 8*h + 16*i))*Log[2 + x])/12

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Maple [B]  time = 0.01, size = 179, normalized size = 2.4 \begin{align*} -{\frac{\ln \left ( 2+x \right ) d}{12}}+{\frac{\ln \left ( 2+x \right ) e}{6}}+{\frac{\ln \left ( 1+x \right ) d}{6}}-{\frac{\ln \left ( 1+x \right ) e}{6}}+{\frac{\ln \left ( x-2 \right ) d}{12}}+{\frac{\ln \left ( x-2 \right ) e}{6}}-{\frac{\ln \left ( x-1 \right ) d}{6}}-{\frac{\ln \left ( x-1 \right ) e}{6}}+{\frac{8\,\ln \left ( x-2 \right ) i}{3}}-{\frac{\ln \left ( x-1 \right ) i}{6}}+{\frac{8\,\ln \left ( 2+x \right ) i}{3}}-{\frac{\ln \left ( 1+x \right ) i}{6}}+{\frac{2\,\ln \left ( 2+x \right ) g}{3}}-{\frac{\ln \left ( 1+x \right ) g}{6}}+{\frac{2\,\ln \left ( x-2 \right ) g}{3}}-{\frac{\ln \left ( x-1 \right ) g}{6}}-{\frac{4\,\ln \left ( 2+x \right ) h}{3}}+{\frac{\ln \left ( 1+x \right ) h}{6}}+{\frac{4\,\ln \left ( x-2 \right ) h}{3}}-{\frac{\ln \left ( x-1 \right ) h}{6}}+{\frac{\ln \left ( x-2 \right ) f}{3}}-{\frac{\ln \left ( x-1 \right ) f}{6}}-{\frac{\ln \left ( 2+x \right ) f}{3}}+{\frac{\ln \left ( 1+x \right ) f}{6}}+{\frac{i{x}^{2}}{2}}+hx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((i*x^5+h*x^4+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+1/6*ln(1+x)*d-1/6*ln(1+x)*e+1/12*ln(x-2)*d+1/6*ln(x-2)*e-1/6*ln(x-1)*d-1/6*ln(x-
1)*e+8/3*ln(x-2)*i-1/6*ln(x-1)*i+8/3*ln(2+x)*i-1/6*ln(1+x)*i+2/3*ln(2+x)*g-1/6*ln(1+x)*g+2/3*ln(x-2)*g-1/6*ln(
x-1)*g-4/3*ln(2+x)*h+1/6*ln(1+x)*h+4/3*ln(x-2)*h-1/6*ln(x-1)*h+1/3*ln(x-2)*f-1/6*ln(x-1)*f-1/3*ln(2+x)*f+1/6*l
n(1+x)*f+1/2*i*x^2+h*x

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Maxima [A]  time = 0.953738, size = 119, normalized size = 1.57 \begin{align*} \frac{1}{2} \, i x^{2} + h x - \frac{1}{12} \,{\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac{1}{6} \,{\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac{1}{6} \,{\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac{1}{12} \,{\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*i*x^2 + h*x - 1/12*(d - 2*e + 4*f - 8*g + 16*h - 32*i)*log(x + 2) + 1/6*(d - e + f - g + h - i)*log(x + 1)
- 1/6*(d + e + f + g + h + i)*log(x - 1) + 1/12*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*log(x - 2)

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Fricas [A]  time = 55.1404, size = 279, normalized size = 3.67 \begin{align*} \frac{1}{2} \, i x^{2} + h x - \frac{1}{12} \,{\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h - 32 \, i\right )} \log \left (x + 2\right ) + \frac{1}{6} \,{\left (d - e + f - g + h - i\right )} \log \left (x + 1\right ) - \frac{1}{6} \,{\left (d + e + f + g + h + i\right )} \log \left (x - 1\right ) + \frac{1}{12} \,{\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h + 32 \, i\right )} \log \left (x - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*i*x^2 + h*x - 1/12*(d - 2*e + 4*f - 8*g + 16*h - 32*i)*log(x + 2) + 1/6*(d - e + f - g + h - i)*log(x + 1)
- 1/6*(d + e + f + g + h + i)*log(x - 1) + 1/12*(d + 2*e + 4*f + 8*g + 16*h + 32*i)*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((i*x**5+h*x**4+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 = 1.10385, size = 130, normalized size = 1.71 \begin{align*} \frac{1}{2} \, i x^{2} + h x - \frac{1}{12} \,{\left (d + 4 \, f - 8 \, g + 16 \, h - 32 \, i - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac{1}{6} \,{\left (d + f - g + h - i - e\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{6} \,{\left (d + f + g + h + i + e\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac{1}{12} \,{\left (d + 4 \, f + 8 \, g + 16 \, h + 32 \, i + 2 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*i*x^2 + h*x - 1/12*(d + 4*f - 8*g + 16*h - 32*i - 2*e)*log(abs(x + 2)) + 1/6*(d + f - g + h - i - e)*log(a
bs(x + 1)) - 1/6*(d + f + g + h + i + e)*log(abs(x - 1)) + 1/12*(d + 4*f + 8*g + 16*h + 32*i + 2*e)*log(abs(x
- 2))