\(\int \frac {(d+e x) (2-x-2 x^2+x^3)}{4-5 x^2+x^4} \, dx\) [68]

Optimal result

Integrand size = 31, antiderivative size = 14 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e x+(d-2 e) \log (2+x) \]

[Out]

e*x+(d-2*e)*ln(2+x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1600, 45} \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=(d-2 e) \log (x+2)+e x \]

[In]

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

[Out]

e*x + (d - 2*e)*Log[2 + x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {d+e x}{2+x} \, dx \\ & = \int \left (e+\frac {d-2 e}{2+x}\right ) \, dx \\ & = e x+(d-2 e) \log (2+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e (2+x)+(d-2 e) \log (2+x) \]

[In]

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

[Out]

e*(2 + x) + (d - 2*e)*Log[2 + x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
default	\(e x +\left (d -2 e \right ) \ln \left (x +2\right )\)	\(15\)
norman	\(e x +\left (d -2 e \right ) \ln \left (x +2\right )\)	\(15\)
risch	\(e x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e\)	\(18\)
parallelrisch	\(e x +\ln \left (x +2\right ) d -2 \ln \left (x +2\right ) e\)	\(18\)

[In]

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

[Out]

e*x+(d-2*e)*ln(x+2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e x + {\left (d - 2 \, e\right )} \log \left (x + 2\right ) \]

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e x + \left (d - 2 e\right ) \log {\left (x + 2 \right )} \]

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e x + {\left (d - 2 \, e\right )} \log \left (x + 2\right ) \]

[In]

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

[Out]

e*x + (d - 2*e)*log(x + 2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=e x + {\left (d - 2 \, e\right )} \log \left ({\left | x + 2 \right |}\right ) \]

[In]

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

[Out]

e*x + (d - 2*e)*log(abs(x + 2))

Mupad [B] (verification not implemented)

Time = 7.86 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (2-x-2 x^2+x^3\right )}{4-5 x^2+x^4} \, dx=\ln \left (x+2\right )\,\left (d-2\,e\right )+e\,x \]

[In]

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

[Out]

log(x + 2)*(d - 2*e) + e*x