3.475 \(\int \frac {-2+x^2}{x (2+x^2)} \, dx\)

Optimal. Leaf size=11 \[ \log \left (x^2+2\right )-\log (x) \]

[Out]

-ln(x)+ln(x^2+2)

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {446, 72} \[ \log \left (x^2+2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Int[(-2 + x^2)/(x*(2 + x^2)),x]

[Out]

-Log[x] + Log[2 + x^2]

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {-2+x^2}{x \left (2+x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-2+x}{x (2+x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{x}+\frac {2}{2+x}\right ) \, dx,x,x^2\right )\\ &=-\log (x)+\log \left (2+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 1.00 \[ \log \left (x^2+2\right )-\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(-2 + x^2)/(x*(2 + x^2)),x]

[Out]

-Log[x] + Log[2 + x^2]

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fricas [A]  time = 0.45, size = 11, normalized size = 1.00 \[ \log \left (x^{2} + 2\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2)/x/(x^2+2),x, algorithm="fricas")

[Out]

log(x^2 + 2) - log(x)

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giac [A]  time = 0.29, size = 13, normalized size = 1.18 \[ \log \left (x^{2} + 2\right ) - \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2)/x/(x^2+2),x, algorithm="giac")

[Out]

log(x^2 + 2) - 1/2*log(x^2)

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maple [A]  time = 0.00, size = 12, normalized size = 1.09 \[ -\ln \relax (x )+\ln \left (x^{2}+2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-2)/x/(x^2+2),x)

[Out]

-ln(x)+ln(x^2+2)

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maxima [A]  time = 0.69, size = 13, normalized size = 1.18 \[ \log \left (x^{2} + 2\right ) - \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-2)/x/(x^2+2),x, algorithm="maxima")

[Out]

log(x^2 + 2) - 1/2*log(x^2)

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mupad [B]  time = 0.06, size = 11, normalized size = 1.00 \[ \ln \left (x^2+2\right )-\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 2)/(x*(x^2 + 2)),x)

[Out]

log(x^2 + 2) - log(x)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.73 \[ - \log {\relax (x )} + \log {\left (x^{2} + 2 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-2)/x/(x**2+2),x)

[Out]

-log(x) + log(x**2 + 2)

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