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

Optimal. Leaf size=64 \[ -\frac {1}{6} \tanh ^{-1}\left (\frac {x}{2}\right ) (d+4 f+16 h)+\frac {1}{3} \tanh ^{-1}(x) (d+f+h)-\frac {1}{6} (e+g) \log \left (1-x^2\right )+\frac {1}{6} (e+4 g) \log \left (4-x^2\right )+h x \]

[Out]

h*x-1/6*(d+4*f+16*h)*arctanh(1/2*x)+1/3*(d+f+h)*arctanh(x)-1/6*(e+g)*ln(-x^2+1)+1/6*(e+4*g)*ln(-x^2+4)

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Rubi [A]  time = 0.15, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1673, 1676, 1166, 207, 1247, 632, 31} \[ -\frac {1}{6} \tanh ^{-1}\left (\frac {x}{2}\right ) (d+4 f+16 h)+\frac {1}{3} \tanh ^{-1}(x) (d+f+h)-\frac {1}{6} (e+g) \log \left (1-x^2\right )+\frac {1}{6} (e+4 g) \log \left (4-x^2\right )+h x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 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 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 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 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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 81, normalized size = 1.27 \[ \frac {1}{12} (-2 \log (1-x) (d+e+f+g+h)+\log (2-x) (d+2 (e+2 f+4 g+8 h))+2 \log (x+1) (d-e+f-g+h)-\log (x+2) (d-2 e+4 f-8 g+16 h)+12 h x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 4.82, size = 72, normalized size = 1.12 \[ h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f + g + h\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h\right )} \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 80, normalized size = 1.25 \[ h x - \frac {1}{12} \, {\left (d + 4 \, f - 8 \, g + 16 \, h - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) + \frac {1}{6} \, {\left (d + f - g + h - e\right )} \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{6} \, {\left (d + f + g + h + e\right )} \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{12} \, {\left (d + 4 \, f + 8 \, g + 16 \, h + 2 \, e\right )} \log \left ({\left | x - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 145, normalized size = 2.27 \[ -\frac {d \ln \left (x +2\right )}{12}+\frac {d \ln \left (x -2\right )}{12}-\frac {d \ln \left (x -1\right )}{6}+\frac {d \ln \left (x +1\right )}{6}+\frac {e \ln \left (x +2\right )}{6}+\frac {e \ln \left (x -2\right )}{6}-\frac {e \ln \left (x -1\right )}{6}-\frac {e \ln \left (x +1\right )}{6}-\frac {f \ln \left (x +2\right )}{3}+\frac {f \ln \left (x -2\right )}{3}-\frac {f \ln \left (x -1\right )}{6}+\frac {f \ln \left (x +1\right )}{6}+\frac {2 g \ln \left (x +2\right )}{3}+\frac {2 g \ln \left (x -2\right )}{3}-\frac {g \ln \left (x -1\right )}{6}-\frac {g \ln \left (x +1\right )}{6}+h x -\frac {4 h \ln \left (x +2\right )}{3}+\frac {4 h \ln \left (x -2\right )}{3}-\frac {h \ln \left (x -1\right )}{6}+\frac {h \ln \left (x +1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.24, size = 72, normalized size = 1.12 \[ h x - \frac {1}{12} \, {\left (d - 2 \, e + 4 \, f - 8 \, g + 16 \, h\right )} \log \left (x + 2\right ) + \frac {1}{6} \, {\left (d - e + f - g + h\right )} \log \left (x + 1\right ) - \frac {1}{6} \, {\left (d + e + f + g + h\right )} \log \left (x - 1\right ) + \frac {1}{12} \, {\left (d + 2 \, e + 4 \, f + 8 \, g + 16 \, h\right )} \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.81, size = 90, normalized size = 1.41 \[ h\,x-\ln \left (x-1\right )\,\left (\frac {d}{6}+\frac {e}{6}+\frac {f}{6}+\frac {g}{6}+\frac {h}{6}\right )+\ln \left (x+1\right )\,\left (\frac {d}{6}-\frac {e}{6}+\frac {f}{6}-\frac {g}{6}+\frac {h}{6}\right )+\ln \left (x-2\right )\,\left (\frac {d}{12}+\frac {e}{6}+\frac {f}{3}+\frac {2\,g}{3}+\frac {4\,h}{3}\right )-\ln \left (x+2\right )\,\left (\frac {d}{12}-\frac {e}{6}+\frac {f}{3}-\frac {2\,g}{3}+\frac {4\,h}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

h*x - log(x - 1)*(d/6 + e/6 + f/6 + g/6 + h/6) + log(x + 1)*(d/6 - e/6 + f/6 - g/6 + h/6) + log(x - 2)*(d/12 +
e/6 + f/3 + (2*g)/3 + (4*h)/3) - log(x + 2)*(d/12 - e/6 + f/3 - (2*g)/3 + (4*h)/3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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