3.82.46 \(\int \frac {-3 x-\log (4)+3 x \log (3 x \log (3))+(2 x^2-3 x^3) \log ^2(3 x \log (3))}{x \log ^2(3 x \log (3))} \, dx\)

Optimal. Leaf size=25 \[ -4+x^2-x^3+\frac {3 x+\log (4)}{\log (3 x \log (3))} \]

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Rubi [A]  time = 0.19, antiderivative size = 29, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 7, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6688, 43, 2330, 2297, 2298, 2302, 30} \begin {gather*} -x^3+x^2+\frac {3 x}{\log (x \log (27))}+\frac {\log (4)}{\log (x \log (27))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x - Log[4] + 3*x*Log[3*x*Log[3]] + (2*x^2 - 3*x^3)*Log[3*x*Log[3]]^2)/(x*Log[3*x*Log[3]]^2),x]

[Out]

x^2 - x^3 + (3*x)/Log[x*Log[27]] + Log[4]/Log[x*Log[27]]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

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]

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left ((2-3 x) x-\frac {3+\frac {\log (4)}{x}}{\log ^2(x \log (27))}+\frac {3}{\log (x \log (27))}\right ) \, dx\\ &=3 \int \frac {1}{\log (x \log (27))} \, dx+\int (2-3 x) x \, dx-\int \frac {3+\frac {\log (4)}{x}}{\log ^2(x \log (27))} \, dx\\ &=\frac {3 \text {li}(x \log (27))}{\log (27)}+\int \left (2 x-3 x^2\right ) \, dx-\int \left (\frac {3}{\log ^2(x \log (27))}+\frac {\log (4)}{x \log ^2(x \log (27))}\right ) \, dx\\ &=x^2-x^3+\frac {3 \text {li}(x \log (27))}{\log (27)}-3 \int \frac {1}{\log ^2(x \log (27))} \, dx-\log (4) \int \frac {1}{x \log ^2(x \log (27))} \, dx\\ &=x^2-x^3+\frac {3 x}{\log (x \log (27))}+\frac {3 \text {li}(x \log (27))}{\log (27)}-3 \int \frac {1}{\log (x \log (27))} \, dx-\log (4) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x \log (27))\right )\\ &=x^2-x^3+\frac {3 x}{\log (x \log (27))}+\frac {\log (4)}{\log (x \log (27))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.16 \begin {gather*} x^2-x^3+\frac {3 x}{\log (x \log (27))}+\frac {\log (4)}{\log (x \log (27))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x - Log[4] + 3*x*Log[3*x*Log[3]] + (2*x^2 - 3*x^3)*Log[3*x*Log[3]]^2)/(x*Log[3*x*Log[3]]^2),x]

[Out]

x^2 - x^3 + (3*x)/Log[x*Log[27]] + Log[4]/Log[x*Log[27]]

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fricas [A]  time = 0.70, size = 34, normalized size = 1.36 \begin {gather*} -\frac {{\left (x^{3} - x^{2}\right )} \log \left (3 \, x \log \relax (3)\right ) - 3 \, x - 2 \, \log \relax (2)}{\log \left (3 \, x \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x^3 - x^2)*log(3*x*log(3)) - 3*x - 2*log(2))/log(3*x*log(3))

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giac [A]  time = 0.19, size = 28, normalized size = 1.12 \begin {gather*} -x^{3} + x^{2} + \frac {3 \, x + 2 \, \log \relax (2)}{\log \relax (3) + \log \left (x \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x^2 + (3*x + 2*log(2))/(log(3) + log(x*log(3)))

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maple [A]  time = 0.07, size = 27, normalized size = 1.08

method	result	size
risch	\(-x^{3}+x^{2}+\frac {3 x +2 \ln \relax (2)}{\ln \left (3 x \ln \relax (3)\right )}\)	\(27\)
norman	\(\frac {x^{2} \ln \left (3 x \ln \relax (3)\right )+3 x -x^{3} \ln \left (3 x \ln \relax (3)\right )+2 \ln \relax (2)}{\ln \left (3 x \ln \relax (3)\right )}\)	\(39\)
derivativedivides	\(x^{2}-x^{3}+\frac {2 \ln \relax (2)}{\ln \left (3 x \ln \relax (3)\right )}-\frac {\expIntegralEi \left (1, -\ln \left (3 x \ln \relax (3)\right )\right )}{\ln \relax (3)}-\frac {-\frac {3 x \ln \relax (3)}{\ln \left (3 x \ln \relax (3)\right )}-\expIntegralEi \left (1, -\ln \left (3 x \ln \relax (3)\right )\right )}{\ln \relax (3)}\)	\(70\)
default	\(x^{2}-x^{3}+\frac {2 \ln \relax (2)}{\ln \left (3 x \ln \relax (3)\right )}-\frac {\expIntegralEi \left (1, -\ln \left (3 x \ln \relax (3)\right )\right )}{\ln \relax (3)}-\frac {-\frac {3 x \ln \relax (3)}{\ln \left (3 x \ln \relax (3)\right )}-\expIntegralEi \left (1, -\ln \left (3 x \ln \relax (3)\right )\right )}{\ln \relax (3)}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.39, size = 49, normalized size = 1.96 \begin {gather*} -x^{3} + x^{2} + \frac {{\rm Ei}\left (\log \left (3 \, x \log \relax (3)\right )\right )}{\log \relax (3)} - \frac {\Gamma \left (-1, -\log \left (3 \, x \log \relax (3)\right )\right )}{\log \relax (3)} + \frac {2 \, \log \relax (2)}{\log \left (3 \, x \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^3 + x^2 + Ei(log(3*x*log(3)))/log(3) - gamma(-1, -log(3*x*log(3)))/log(3) + 2*log(2)/log(3*x*log(3))

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mupad [B]  time = 5.93, size = 24, normalized size = 0.96 \begin {gather*} \frac {3\,x+\ln \relax (4)}{\ln \left (3\,x\,\ln \relax (3)\right )}+x^2-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 2*log(2) - log(3*x*log(3))^2*(2*x^2 - 3*x^3) - 3*x*log(3*x*log(3)))/(x*log(3*x*log(3))^2),x)

[Out]

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

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sympy [A]  time = 0.12, size = 22, normalized size = 0.88 \begin {gather*} - x^{3} + x^{2} + \frac {3 x + 2 \log {\relax (2 )}}{\log {\left (3 x \log {\relax (3 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**3+2*x**2)*ln(3*x*ln(3))**2+3*x*ln(3*x*ln(3))-2*ln(2)-3*x)/x/ln(3*x*ln(3))**2,x)

[Out]

-x**3 + x**2 + (3*x + 2*log(2))/log(3*x*log(3))

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