\(\int \frac {(2-3 x+x^2) (d+e x+f x^2)}{4-5 x^2+x^4} \, dx\) [75]

Optimal result

Integrand size = 31, antiderivative size = 29 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x+(d-e+f) \log (1+x)-(d-2 e+4 f) \log (2+x) \]

[Out]

f*x+(d-e+f)*ln(1+x)-(d-2*e+4*f)*ln(2+x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1600, 1671, 646, 31} \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=\log (x+1) (d-e+f)-\log (x+2) (d-2 e+4 f)+f x \]

[In]

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

[Out]

f*x + (d - e + f)*Log[1 + x] - (d - 2*e + 4*f)*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}{2+3 x+x^2} \, dx \\ & = \int \left (f+\frac {d-2 f+(e-3 f) x}{2+3 x+x^2}\right ) \, dx \\ & = f x+\int \frac {d-2 f+(e-3 f) x}{2+3 x+x^2} \, dx \\ & = f x+(d-e+f) \int \frac {1}{1+x} \, dx-(d-2 e+4 f) \int \frac {1}{2+x} \, dx \\ & = f x+(d-e+f) \log (1+x)-(d-2 e+4 f) \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x+(d-e+f) \log (1+x)+(-d+2 e-4 f) \log (2+x) \]

[In]

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

[Out]

f*x + (d - e + f)*Log[1 + x] + (-d + 2*e - 4*f)*Log[2 + x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07

method	result	size
default	\(f x +\left (-d +2 e -4 f \right ) \ln \left (x +2\right )+\left (d -e +f \right ) \ln \left (x +1\right )\)	\(31\)
norman	\(f x +\left (-d +2 e -4 f \right ) \ln \left (x +2\right )+\left (d -e +f \right ) \ln \left (x +1\right )\)	\(31\)
parallelrisch	\(\ln \left (x +1\right ) d -\ln \left (x +1\right ) e +\ln \left (x +1\right ) f -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f +f x\)	\(45\)
risch	\(f x +\ln \left (-x -1\right ) d -\ln \left (-x -1\right ) e +\ln \left (-x -1\right ) f -\ln \left (x +2\right ) d +2 \ln \left (x +2\right ) e -4 \ln \left (x +2\right ) f\)	\(51\)

[In]

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

[Out]

f*x+(-d+2*e-4*f)*ln(x+2)+(d-e+f)*ln(x+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x - {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) + {\left (d - e + f\right )} \log \left (x + 1\right ) \]

[In]

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

[Out]

f*x - (d - 2*e + 4*f)*log(x + 2) + (d - e + f)*log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x + \left (- d + 2 e - 4 f\right ) \log {\left (x + \frac {4 d - 6 e + 10 f}{2 d - 3 e + 5 f} \right )} + \left (d - e + f\right ) \log {\left (x + 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x - {\left (d - 2 \, e + 4 \, f\right )} \log \left (x + 2\right ) + {\left (d - e + f\right )} \log \left (x + 1\right ) \]

[In]

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

[Out]

f*x - (d - 2*e + 4*f)*log(x + 2) + (d - e + f)*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f x - {\left (d - 2 \, e + 4 \, f\right )} \log \left ({\left | x + 2 \right |}\right ) + {\left (d - e + f\right )} \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

f*x - (d - 2*e + 4*f)*log(abs(x + 2)) + (d - e + f)*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2-3 x+x^2\right ) \left (d+e x+f x^2\right )}{4-5 x^2+x^4} \, dx=f\,x+\ln \left (x+1\right )\,\left (d-e+f\right )-\ln \left (x+2\right )\,\left (d-2\,e+4\,f\right ) \]

[In]

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

[Out]

f*x + log(x + 1)*(d - e + f) - log(x + 2)*(d - 2*e + 4*f)