\(\int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx\) [9166]

Optimal result

Integrand size = 34, antiderivative size = 18 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=-4+\log \left (-e^5-\frac {1}{x}+x+x^2\right ) \]

[Out]

ln(x-1/x+x^2-exp(5))-4

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2099, 1601} \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=\log \left (-x^3-x^2+e^5 x+1\right )-\log (x) \]

[In]

Int[(-1 - x^2 - 2*x^3)/(x + E^5*x^2 - x^3 - x^4),x]

[Out]

-Log[x] + Log[1 + E^5*x - x^2 - 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 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{x}+\frac {e^5-2 x-3 x^2}{1+e^5 x-x^2-x^3}\right ) \, dx \\ & = -\log (x)+\int \frac {e^5-2 x-3 x^2}{1+e^5 x-x^2-x^3} \, dx \\ & = -\log (x)+\log \left (1+e^5 x-x^2-x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=-\log (x)+\log \left (1+e^5 x-x^2-x^3\right ) \]

[In]

Integrate[(-1 - x^2 - 2*x^3)/(x + E^5*x^2 - x^3 - x^4),x]

[Out]

-Log[x] + Log[1 + E^5*x - x^2 - x^3]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
default	\(-\ln \left (x \right )+\ln \left (x^{3}-x \,{\mathrm e}^{5}+x^{2}-1\right )\)	\(20\)
risch	\(-\ln \left (x \right )+\ln \left (x^{3}-x \,{\mathrm e}^{5}+x^{2}-1\right )\)	\(20\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (x^{3}-x \,{\mathrm e}^{5}+x^{2}-1\right )\)	\(20\)
norman	\(-\ln \left (x \right )+\ln \left (-x^{3}+x \,{\mathrm e}^{5}-x^{2}+1\right )\)	\(23\)

[In]

int((-2*x^3-x^2-1)/(x^2*exp(5)-x^4-x^3+x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(x^3-x*exp(5)+x^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=\log \left (x^{3} + x^{2} - x e^{5} - 1\right ) - \log \left (x\right ) \]

[In]

integrate((-2*x^3-x^2-1)/(x^2*exp(5)-x^4-x^3+x),x, algorithm="fricas")

[Out]

log(x^3 + x^2 - x*e^5 - 1) - log(x)

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=- \log {\left (x \right )} + \log {\left (x^{3} + x^{2} - x e^{5} - 1 \right )} \]

[In]

integrate((-2*x**3-x**2-1)/(x**2*exp(5)-x**4-x**3+x),x)

[Out]

-log(x) + log(x**3 + x**2 - x*exp(5) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=\log \left (x^{3} + x^{2} - x e^{5} - 1\right ) - \log \left (x\right ) \]

[In]

integrate((-2*x^3-x^2-1)/(x^2*exp(5)-x^4-x^3+x),x, algorithm="maxima")

[Out]

log(x^3 + x^2 - x*e^5 - 1) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=\log \left ({\left | x^{3} + x^{2} - x e^{5} - 1 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-2*x^3-x^2-1)/(x^2*exp(5)-x^4-x^3+x),x, algorithm="giac")

[Out]

log(abs(x^3 + x^2 - x*e^5 - 1)) - log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {-1-x^2-2 x^3}{x+e^5 x^2-x^3-x^4} \, dx=\ln \left (x^3+x^2-{\mathrm {e}}^5\,x-1\right )-\ln \left (x\right ) \]

[In]

int(-(x^2 + 2*x^3 + 1)/(x + x^2*exp(5) - x^3 - x^4),x)

[Out]

log(x^2 - x*exp(5) + x^3 - 1) - log(x)