3.243 \(\int x \log (\frac {1+x}{x^2}) \, dx\)

Optimal. Leaf size=36 \[ \frac {x^2}{4}+\frac {1}{2} x^2 \log \left (\frac {x+1}{x^2}\right )+\frac {x}{2}-\frac {1}{2} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2495, 30, 43} \[ \frac {x^2}{4}+\frac {1}{2} x^2 \log \left (\frac {x+1}{x^2}\right )+\frac {x}{2}-\frac {1}{2} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[x*Log[(1 + x)/x^2],x]

[Out]

x/2 + x^2/4 - Log[1 + x]/2 + (x^2*Log[(1 + x)/x^2])/2

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \log \left (\frac {1+x}{x^2}\right ) \, dx &=\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right )-\frac {1}{2} \int \frac {x^2}{1+x} \, dx+\int x \, dx\\ &=\frac {x^2}{2}+\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right )-\frac {1}{2} \int \left (-1+x+\frac {1}{1+x}\right ) \, dx\\ &=\frac {x}{2}+\frac {x^2}{4}-\frac {1}{2} \log (1+x)+\frac {1}{2} x^2 \log \left (\frac {1+x}{x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.75 \[ \frac {1}{4} \left (x \left (2 x \log \left (\frac {x+1}{x^2}\right )+x+2\right )-2 \log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[(1 + x)/x^2],x]

[Out]

(-2*Log[1 + x] + x*(2 + x + 2*x*Log[(1 + x)/x^2]))/4

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fricas [A]  time = 0.43, size = 28, normalized size = 0.78 \[ \frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log((x + 1)/x^2) + 1/4*x^2 + 1/2*x - 1/2*log(x + 1)

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giac [A]  time = 0.17, size = 29, normalized size = 0.81 \[ \frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log((x + 1)/x^2) + 1/4*x^2 + 1/2*x - 1/2*log(abs(x + 1))

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maple [A]  time = 0.08, size = 39, normalized size = 1.08 \[ \frac {x^{2} \ln \left (\frac {\frac {1}{x}+1}{x}\right )}{2}+\frac {x^{2}}{4}+\frac {x}{2}+\frac {\ln \left (\frac {1}{x}\right )}{2}-\frac {\ln \left (\frac {1}{x}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln((x+1)/x^2),x)

[Out]

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

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maxima [A]  time = 0.77, size = 28, normalized size = 0.78 \[ \frac {1}{2} \, x^{2} \log \left (\frac {x + 1}{x^{2}}\right ) + \frac {1}{4} \, x^{2} + \frac {1}{2} \, x - \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log((x + 1)/x^2) + 1/4*x^2 + 1/2*x - 1/2*log(x + 1)

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mupad [B]  time = 0.49, size = 40, normalized size = 1.11 \[ \frac {x}{2}-\frac {\ln \left (x\,\left (x+1\right )\right )}{3}-\frac {\ln \left (\frac {x+1}{x^2}\right )}{6}+\frac {x^2\,\ln \left (\frac {x+1}{x^2}\right )}{2}+\frac {x^2}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log((x + 1)/x^2),x)

[Out]

x/2 - log(x*(x + 1))/3 - log((x + 1)/x^2)/6 + (x^2*log((x + 1)/x^2))/2 + x^2/4

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sympy [A]  time = 0.15, size = 27, normalized size = 0.75 \[ \frac {x^{2} \log {\left (\frac {x + 1}{x^{2}} \right )}}{2} + \frac {x^{2}}{4} + \frac {x}{2} - \frac {\log {\left (x + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*log((x + 1)/x**2)/2 + x**2/4 + x/2 - log(x + 1)/2

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