3.131 \(\int \frac {(1+\log (x))^5}{x} \, dx\)

Optimal. Leaf size=10 \[ \frac {1}{6} (\log (x)+1)^6 \]

[Out]

1/6*(1+ln(x))^6

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Rubi [A]  time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2302, 30} \[ \frac {1}{6} (\log (x)+1)^6 \]

Antiderivative was successfully verified.

[In]

Int[(1 + Log[x])^5/x,x]

[Out]

(1 + Log[x])^6/6

Rule 30

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

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps

\begin {align*} \int \frac {(1+\log (x))^5}{x} \, dx &=\operatorname {Subst}\left (\int x^5 \, dx,x,1+\log (x)\right )\\ &=\frac {1}{6} (1+\log (x))^6\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \[ \frac {1}{6} (\log (x)+1)^6 \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Log[x])^5/x,x]

[Out]

(1 + Log[x])^6/6

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fricas [B]  time = 0.54, size = 31, normalized size = 3.10 \[ \frac {1}{6} \, \log \relax (x)^{6} + \log \relax (x)^{5} + \frac {5}{2} \, \log \relax (x)^{4} + \frac {10}{3} \, \log \relax (x)^{3} + \frac {5}{2} \, \log \relax (x)^{2} + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^5/x,x, algorithm="fricas")

[Out]

1/6*log(x)^6 + log(x)^5 + 5/2*log(x)^4 + 10/3*log(x)^3 + 5/2*log(x)^2 + log(x)

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giac [B]  time = 0.17, size = 31, normalized size = 3.10 \[ \frac {1}{6} \, \log \relax (x)^{6} + \log \relax (x)^{5} + \frac {5}{2} \, \log \relax (x)^{4} + \frac {10}{3} \, \log \relax (x)^{3} + \frac {5}{2} \, \log \relax (x)^{2} + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^5/x,x, algorithm="giac")

[Out]

1/6*log(x)^6 + log(x)^5 + 5/2*log(x)^4 + 10/3*log(x)^3 + 5/2*log(x)^2 + log(x)

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maple [B]  time = 0.06, size = 33, normalized size = 3.30 \[ \frac {\ln \relax (x )^{6}}{6}+\ln \relax (x )^{5}+\frac {5 \ln \relax (x )^{4}}{2}+\frac {10 \ln \relax (x )^{3}}{3}+\frac {5 \ln \relax (x )^{2}}{2}+\ln \relax (x )+\frac {1}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)+1)^5/x,x)

[Out]

1/6*ln(x)^6+ln(x)^5+5/2*ln(x)^4+10/3*ln(x)^3+5/2*ln(x)^2+ln(x)+1/6

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maxima [A]  time = 0.65, size = 8, normalized size = 0.80 \[ \frac {1}{6} \, {\left (\log \relax (x) + 1\right )}^{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+log(x))^5/x,x, algorithm="maxima")

[Out]

1/6*(log(x) + 1)^6

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mupad [B]  time = 0.39, size = 26, normalized size = 2.60 \[ \frac {\ln \relax (x)\,\left (\ln \relax (x)+2\right )\,\left ({\ln \relax (x)}^2+\ln \relax (x)+1\right )\,\left ({\ln \relax (x)}^2+3\,\ln \relax (x)+3\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) + 1)^5/x,x)

[Out]

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

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sympy [B]  time = 0.14, size = 39, normalized size = 3.90 \[ \frac {\log {\relax (x )}^{6}}{6} + \log {\relax (x )}^{5} + \frac {5 \log {\relax (x )}^{4}}{2} + \frac {10 \log {\relax (x )}^{3}}{3} + \frac {5 \log {\relax (x )}^{2}}{2} + \log {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+ln(x))**5/x,x)

[Out]

log(x)**6/6 + log(x)**5 + 5*log(x)**4/2 + 10*log(x)**3/3 + 5*log(x)**2/2 + log(x)

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