\(\int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx\) [5882]

Optimal result

Integrand size = 25, antiderivative size = 17 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\log \left (x \left (5+x^2+x^3\right ) (x+x \log (4))\right ) \]

[Out]

ln((x^3+x^2+5)*x*(x+2*x*ln(2)))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1608, 6874, 1601} \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\log \left (x^3+x^2+5\right )+2 \log (x) \]

[In]

Int[(10 + 4*x^2 + 5*x^3)/(5*x + x^3 + x^4),x]

[Out]

2*Log[x] + Log[5 + 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 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 {10+4 x^2+5 x^3}{x \left (5+x^2+x^3\right )} \, dx \\ & = \int \left (\frac {2}{x}+\frac {x (2+3 x)}{5+x^2+x^3}\right ) \, dx \\ & = 2 \log (x)+\int \frac {x (2+3 x)}{5+x^2+x^3} \, dx \\ & = 2 \log (x)+\log \left (5+x^2+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=2 \log (x)+\log \left (5+x^2+x^3\right ) \]

[In]

Integrate[(10 + 4*x^2 + 5*x^3)/(5*x + x^3 + x^4),x]

[Out]

2*Log[x] + Log[5 + x^2 + x^3]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
default	\(2 \ln \left (x \right )+\ln \left (x^{3}+x^{2}+5\right )\)	\(15\)
norman	\(2 \ln \left (x \right )+\ln \left (x^{3}+x^{2}+5\right )\)	\(15\)
risch	\(2 \ln \left (x \right )+\ln \left (x^{3}+x^{2}+5\right )\)	\(15\)
parallelrisch	\(2 \ln \left (x \right )+\ln \left (x^{3}+x^{2}+5\right )\)	\(15\)

[In]

int((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x,method=_RETURNVERBOSE)

[Out]

2*ln(x)+ln(x^3+x^2+5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\log \left (x^{3} + x^{2} + 5\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="fricas")

[Out]

log(x^3 + x^2 + 5) + 2*log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=2 \log {\left (x \right )} + \log {\left (x^{3} + x^{2} + 5 \right )} \]

[In]

integrate((5*x**3+4*x**2+10)/(x**4+x**3+5*x),x)

[Out]

2*log(x) + log(x**3 + x**2 + 5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\log \left (x^{3} + x^{2} + 5\right ) + 2 \, \log \left (x\right ) \]

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="maxima")

[Out]

log(x^3 + x^2 + 5) + 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\log \left ({\left | x^{3} + x^{2} + 5 \right |}\right ) + 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((5*x^3+4*x^2+10)/(x^4+x^3+5*x),x, algorithm="giac")

[Out]

log(abs(x^3 + x^2 + 5)) + 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 12.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {10+4 x^2+5 x^3}{5 x+x^3+x^4} \, dx=\ln \left (x^3+x^2+5\right )+2\,\ln \left (x\right ) \]

[In]

int((4*x^2 + 5*x^3 + 10)/(5*x + x^3 + x^4),x)

[Out]

log(x^2 + x^3 + 5) + 2*log(x)