\(\int \frac {(2-3 x+x^2) (d+e x+f x^2+g x^3)}{4-5 x^2+x^4} \, dx\) [76]

Optimal result

Integrand size = 36, antiderivative size = 47 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=(f-3 g) x+\frac {g x^2}{2}+(d-e+f-g) \log (1+x)-(d-2 e+4 f-8 g) \log (2+x) \]

[Out]

(f-3*g)*x+1/2*g*x^2+(d-e+f-g)*ln(1+x)-(d-2*e+4*f-8*g)*ln(2+x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1600, 1671, 646, 31} \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\log (x+1) (d-e+f-g)-\log (x+2) (d-2 e+4 f-8 g)+x (f-3 g)+\frac {g x^2}{2} \]

[In]

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

[Out]

(f - 3*g)*x + (g*x^2)/2 + (d - e + f - g)*Log[1 + x] - (d - 2*e + 4*f - 8*g)*Log[2 + x]

Rule 31

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

Rule 646

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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1671

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x+f x^2+g x^3}{2+3 x+x^2} \, dx \\ & = \int \left (f-3 g+g x+\frac {d-2 f+6 g+(e-3 f+7 g) x}{2+3 x+x^2}\right ) \, dx \\ & = (f-3 g) x+\frac {g x^2}{2}+\int \frac {d-2 f+6 g+(e-3 f+7 g) x}{2+3 x+x^2} \, dx \\ & = (f-3 g) x+\frac {g x^2}{2}-(d-2 e+4 f-8 g) \int \frac {1}{2+x} \, dx+(d-e+f-g) \int \frac {1}{1+x} \, dx \\ & = (f-3 g) x+\frac {g x^2}{2}+(d-e+f-g) \log (1+x)-(d-2 e+4 f-8 g) \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=f x+\frac {1}{2} g (-6+x) x+(d-e+f-g) \log (1+x)-(d-2 e+4 f-8 g) \log (2+x) \]

[In]

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

[Out]

f*x + (g*(-6 + x)*x)/2 + (d - e + f - g)*Log[1 + x] - (d - 2*e + 4*f - 8*g)*Log[2 + x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {g \,x^{2}}{2}+f x -3 g x +\left (-d +2 e -4 f +8 g \right ) \ln \left (x +2\right )+\left (d -e +f -g \right ) \ln \left (x +1\right )\)	\(47\)
norman	\(\left (f -3 g \right ) x +\frac {g \,x^{2}}{2}+\left (-d +2 e -4 f +8 g \right ) \ln \left (x +2\right )+\left (d -e +f -g \right ) \ln \left (x +1\right )\)	\(47\)
parallelrisch	\(\frac {g \,x^{2}}{2}+f x -3 g x +\ln \left (x +1\right ) d -\ln \left (x +1\right ) e +\ln \left (x +1\right ) f -\ln \left (x +1\right ) g -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f +8 \ln \left (x +2\right ) g\)	\(69\)
risch	\(\frac {g \,x^{2}}{2}+f x -3 g x +\ln \left (-x -1\right ) d -\ln \left (-x -1\right ) e +\ln \left (-x -1\right ) f -\ln \left (-x -1\right ) g -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f +8 \ln \left (x +2\right ) g\)	\(77\)

[In]

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

[Out]

1/2*g*x^2+f*x-3*g*x+(-d+2*e-4*f+8*g)*ln(x+2)+(d-e+f-g)*ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, g x^{2} + {\left (f - 3 \, g\right )} x - {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) + {\left (d - e + f - g\right )} \log \left (x + 1\right ) \]

[In]

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

[Out]

1/2*g*x^2 + (f - 3*g)*x - (d - 2*e + 4*f - 8*g)*log(x + 2) + (d - e + f - g)*log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\frac {g x^{2}}{2} + x \left (f - 3 g\right ) + \left (- d + 2 e - 4 f + 8 g\right ) \log {\left (x + \frac {4 d - 6 e + 10 f - 18 g}{2 d - 3 e + 5 f - 9 g} \right )} + \left (d - e + f - g\right ) \log {\left (x + 1 \right )} \]

[In]

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

[Out]

g*x**2/2 + x*(f - 3*g) + (-d + 2*e - 4*f + 8*g)*log(x + (4*d - 6*e + 10*f - 18*g)/(2*d - 3*e + 5*f - 9*g)) + (
d - e + f - g)*log(x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, g x^{2} + {\left (f - 3 \, g\right )} x - {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left (x + 2\right ) + {\left (d - e + f - g\right )} \log \left (x + 1\right ) \]

[In]

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

[Out]

1/2*g*x^2 + (f - 3*g)*x - (d - 2*e + 4*f - 8*g)*log(x + 2) + (d - e + f - g)*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\frac {1}{2} \, g x^{2} + f x - 3 \, g x - {\left (d - 2 \, e + 4 \, f - 8 \, g\right )} \log \left ({\left | x + 2 \right |}\right ) + {\left (d - e + f - g\right )} \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

1/2*g*x^2 + f*x - 3*g*x - (d - 2*e + 4*f - 8*g)*log(abs(x + 2)) + (d - e + f - g)*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2+g x^3\right )}{4-5 x^2+x^4} \, dx=\ln \left (x+1\right )\,\left (d-e+f-g\right )+x\,\left (f-3\,g\right )+\frac {g\,x^2}{2}-\ln \left (x+2\right )\,\left (d-2\,e+4\,f-8\,g\right ) \]

[In]

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

[Out]

log(x + 1)*(d - e + f - g) + x*(f - 3*g) + (g*x^2)/2 - log(x + 2)*(d - 2*e + 4*f - 8*g)