3.3.29 \(\int x \log (\sqrt [3]{1+3 x}) \, dx\) [229]

Optimal. Leaf size=40 \[ \frac {x}{18}-\frac {x^2}{12}+\frac {1}{2} x^2 \log \left (\sqrt [3]{1+3 x}\right )-\frac {1}{54} \log (1+3 x) \]

[Out]

1/18*x-1/12*x^2+1/6*x^2*ln(1+3*x)-1/54*ln(1+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2442, 45} \begin {gather*} -\frac {x^2}{12}+\frac {1}{2} x^2 \log \left (\sqrt [3]{3 x+1}\right )+\frac {x}{18}-\frac {1}{54} \log (3 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Log[(1 + 3*x)^(1/3)],x]

[Out]

x/18 - x^2/12 + (x^2*Log[(1 + 3*x)^(1/3)])/2 - Log[1 + 3*x]/54

Rule 45

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 2442

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

Rubi steps

\begin {align*} \int x \log \left (\sqrt [3]{1+3 x}\right ) \, dx &=\frac {1}{2} x^2 \log \left (\sqrt [3]{1+3 x}\right )-\frac {1}{2} \int \frac {x^2}{1+3 x} \, dx\\ &=\frac {1}{2} x^2 \log \left (\sqrt [3]{1+3 x}\right )-\frac {1}{2} \int \left (-\frac {1}{9}+\frac {x}{3}+\frac {1}{9 (1+3 x)}\right ) \, dx\\ &=\frac {x}{18}-\frac {x^2}{12}+\frac {1}{2} x^2 \log \left (\sqrt [3]{1+3 x}\right )-\frac {1}{54} \log (1+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 40, normalized size = 1.00 \begin {gather*} \frac {1}{3} \left (\frac {x}{6}-\frac {x^2}{4}-\frac {1}{18} \log (1+3 x)+\frac {1}{2} x^2 \log (1+3 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Log[(1 + 3*x)^(1/3)],x]

[Out]

(x/6 - x^2/4 - Log[1 + 3*x]/18 + (x^2*Log[1 + 3*x])/2)/3

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Maple [A]
time = 0.03, size = 43, normalized size = 1.08

method	result	size
meijerg	\(\frac {x \left (-9 x +6\right )}{108}-\frac {\left (-27 x^{2}+3\right ) \ln \left (1+3 x \right )}{162}\)	\(25\)
norman	\(\frac {x}{18}-\frac {x^{2}}{12}+\frac {x^{2} \ln \left (1+3 x \right )}{6}-\frac {\ln \left (1+3 x \right )}{54}\)	\(29\)
risch	\(\frac {x}{18}-\frac {x^{2}}{12}+\frac {x^{2} \ln \left (1+3 x \right )}{6}-\frac {\ln \left (1+3 x \right )}{54}\)	\(29\)
derivativedivides	\(-\frac {\ln \left (1+3 x \right ) \left (1+3 x \right )}{27}+\frac {1}{27}+\frac {x}{9}+\frac {\left (1+3 x \right )^{2} \ln \left (1+3 x \right )}{54}-\frac {\left (1+3 x \right )^{2}}{108}\)	\(43\)
default	\(-\frac {\ln \left (1+3 x \right ) \left (1+3 x \right )}{27}+\frac {1}{27}+\frac {x}{9}+\frac {\left (1+3 x \right )^{2} \ln \left (1+3 x \right )}{54}-\frac {\left (1+3 x \right )^{2}}{108}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*ln(1+3*x)*(1+3*x)+1/27+1/9*x+1/54*(1+3*x)^2*ln(1+3*x)-1/108*(1+3*x)^2

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Maxima [A]
time = 0.27, size = 28, normalized size = 0.70 \begin {gather*} \frac {1}{6} \, x^{2} \log \left (3 \, x + 1\right ) - \frac {1}{12} \, x^{2} + \frac {1}{18} \, x - \frac {1}{54} \, \log \left (3 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^2*log(3*x + 1) - 1/12*x^2 + 1/18*x - 1/54*log(3*x + 1)

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Fricas [A]
time = 0.39, size = 24, normalized size = 0.60 \begin {gather*} -\frac {1}{12} \, x^{2} + \frac {1}{54} \, {\left (9 \, x^{2} - 1\right )} \log \left (3 \, x + 1\right ) + \frac {1}{18} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*x^2 + 1/54*(9*x^2 - 1)*log(3*x + 1) + 1/18*x

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Sympy [A]
time = 0.10, size = 27, normalized size = 0.68 \begin {gather*} \frac {x^{2} \log {\left (3 x + 1 \right )}}{6} - \frac {x^{2}}{12} + \frac {x}{18} - \frac {\log {\left (3 x + 1 \right )}}{54} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*log(3*x + 1)/6 - x**2/12 + x/18 - log(3*x + 1)/54

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Giac [A]
time = 7.32, size = 42, normalized size = 1.05 \begin {gather*} \frac {1}{54} \, {\left (3 \, x + 1\right )}^{2} \log \left (3 \, x + 1\right ) - \frac {1}{108} \, {\left (3 \, x + 1\right )}^{2} - \frac {1}{27} \, {\left (3 \, x + 1\right )} \log \left (3 \, x + 1\right ) + \frac {1}{9} \, x + \frac {1}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(3*x + 1)^2*log(3*x + 1) - 1/108*(3*x + 1)^2 - 1/27*(3*x + 1)*log(3*x + 1) + 1/9*x + 1/27

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Mupad [B]
time = 0.38, size = 22, normalized size = 0.55 \begin {gather*} \frac {x}{18}+\frac {\ln \left (3\,x+1\right )\,\left (x^2-\frac {1}{9}\right )}{6}-\frac {x^2}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(3*x + 1))/3,x)

[Out]

x/18 + (log(3*x + 1)*(x^2 - 1/9))/6 - x^2/12

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