\(\int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx\) [6119]

Optimal result

Integrand size = 23, antiderivative size = 20 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=\log \left (\frac {\left (x+\frac {4 (5+x)}{x^2}\right )^2}{5 x}\right ) \]

[Out]

ln(1/5*(4*(5+x)/x^2+x)^2/x)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1608, 6874, 1601} \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=2 \log \left (x^3+4 x+20\right )-5 \log (x) \]

[In]

Int[(-100 - 12*x + x^3)/(20*x + 4*x^2 + x^4),x]

[Out]

-5*Log[x] + 2*Log[20 + 4*x + x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-100-12 x+x^3}{x \left (20+4 x+x^3\right )} \, dx \\ & = \int \left (-\frac {5}{x}+\frac {2 \left (4+3 x^2\right )}{20+4 x+x^3}\right ) \, dx \\ & = -5 \log (x)+2 \int \frac {4+3 x^2}{20+4 x+x^3} \, dx \\ & = -5 \log (x)+2 \log \left (20+4 x+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=-5 \log (x)+2 \log \left (20+4 x+x^3\right ) \]

[In]

Integrate[(-100 - 12*x + x^3)/(20*x + 4*x^2 + x^4),x]

[Out]

-5*Log[x] + 2*Log[20 + 4*x + x^3]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
default	\(2 \ln \left (x^{3}+4 x +20\right )-5 \ln \left (x \right )\)	\(17\)
norman	\(2 \ln \left (x^{3}+4 x +20\right )-5 \ln \left (x \right )\)	\(17\)
risch	\(2 \ln \left (x^{3}+4 x +20\right )-5 \ln \left (x \right )\)	\(17\)
parallelrisch	\(2 \ln \left (x^{3}+4 x +20\right )-5 \ln \left (x \right )\)	\(17\)

[In]

int((x^3-12*x-100)/(x^4+4*x^2+20*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x^3+4*x+20)-5*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=2 \, \log \left (x^{3} + 4 \, x + 20\right ) - 5 \, \log \left (x\right ) \]

[In]

integrate((x^3-12*x-100)/(x^4+4*x^2+20*x),x, algorithm="fricas")

[Out]

2*log(x^3 + 4*x + 20) - 5*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=- 5 \log {\left (x \right )} + 2 \log {\left (x^{3} + 4 x + 20 \right )} \]

[In]

integrate((x**3-12*x-100)/(x**4+4*x**2+20*x),x)

[Out]

-5*log(x) + 2*log(x**3 + 4*x + 20)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=2 \, \log \left (x^{3} + 4 \, x + 20\right ) - 5 \, \log \left (x\right ) \]

[In]

integrate((x^3-12*x-100)/(x^4+4*x^2+20*x),x, algorithm="maxima")

[Out]

2*log(x^3 + 4*x + 20) - 5*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=2 \, \log \left ({\left | x^{3} + 4 \, x + 20 \right |}\right ) - 5 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((x^3-12*x-100)/(x^4+4*x^2+20*x),x, algorithm="giac")

[Out]

2*log(abs(x^3 + 4*x + 20)) - 5*log(abs(x))

Mupad [B] (verification not implemented)

Time = 11.42 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-100-12 x+x^3}{20 x+4 x^2+x^4} \, dx=2\,\ln \left (x^3+4\,x+20\right )-5\,\ln \left (x\right ) \]

[In]

int(-(12*x - x^3 + 100)/(20*x + 4*x^2 + x^4),x)

[Out]

2*log(4*x + x^3 + 20) - 5*log(x)