\(\int \frac {1+x^2}{3 x+x^3} \, dx\) [434]

Optimal result

Integrand size = 15, antiderivative size = 12 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {1}{3} \log \left (3 x+x^3\right ) \]

[Out]

1/3*ln(x^3+3*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1601} \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {1}{3} \log \left (x^3+3 x\right ) \]

[In]

Int[(1 + x^2)/(3*x + x^3),x]

[Out]

Log[3*x + x^3]/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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \log \left (3 x+x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.42 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {\log (x)}{3}+\frac {1}{3} \log \left (3+x^2\right ) \]

[In]

Integrate[(1 + x^2)/(3*x + x^3),x]

[Out]

Log[x]/3 + Log[3 + x^2]/3

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {\ln \left (x \left (x^{2}+3\right )\right )}{3}\)	\(11\)
risch	\(\frac {\ln \left (x^{3}+3 x \right )}{3}\)	\(11\)
norman	\(\frac {\ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+3\right )}{3}\)	\(14\)
parallelrisch	\(\frac {\ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+3\right )}{3}\)	\(14\)
meijerg	\(\frac {\ln \left (1+\frac {x^{2}}{3}\right )}{3}+\frac {\ln \left (x \right )}{3}-\frac {\ln \left (3\right )}{6}\)	\(20\)

[In]

int((x^2+1)/(x^3+3*x),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(x*(x^2+3))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {1}{3} \, \log \left (x^{3} + 3 \, x\right ) \]

[In]

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

[Out]

1/3*log(x^3 + 3*x)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {\log {\left (x^{3} + 3 x \right )}}{3} \]

[In]

integrate((x**2+1)/(x**3+3*x),x)

[Out]

log(x**3 + 3*x)/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {1}{3} \, \log \left (x^{3} + 3 \, x\right ) \]

[In]

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

[Out]

1/3*log(x^3 + 3*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {1}{3} \, \log \left (3 \, {\left | \frac {1}{3} \, x^{3} + x \right |}\right ) \]

[In]

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

[Out]

1/3*log(3*abs(1/3*x^3 + x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^2}{3 x+x^3} \, dx=\frac {\ln \left (x^3+3\,x\right )}{3} \]

[In]

int((x^2 + 1)/(3*x + x^3),x)

[Out]

log(3*x + x^3)/3