3.94.56 \(\int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx\)

Optimal. Leaf size=14 \[ \left (\frac {20}{9}+x\right ) (4-\log (x)) \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 27, normalized size of antiderivative = 1.93, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {12, 14, 43, 2346, 2301, 2295, 2296} \begin {gather*} -x \log ^2(x)-\frac {20 \log ^2(x)}{9}+4 x \log (x)+\frac {80 \log (x)}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(80 + 36*x + (-40 + 18*x)*Log[x] - 9*x*Log[x]^2)/(9*x),x]

[Out]

(80*Log[x])/9 + 4*x*Log[x] - (20*Log[x]^2)/9 - x*Log[x]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{x} \, dx\\ &=\frac {1}{9} \int \left (\frac {4 (20+9 x)}{x}+\frac {2 (-20+9 x) \log (x)}{x}-9 \log ^2(x)\right ) \, dx\\ &=\frac {2}{9} \int \frac {(-20+9 x) \log (x)}{x} \, dx+\frac {4}{9} \int \frac {20+9 x}{x} \, dx-\int \log ^2(x) \, dx\\ &=-x \log ^2(x)+\frac {4}{9} \int \left (9+\frac {20}{x}\right ) \, dx+2 (2 \int \log (x) \, dx)-\frac {40}{9} \int \frac {\log (x)}{x} \, dx\\ &=4 x+\frac {80 \log (x)}{9}-\frac {20 \log ^2(x)}{9}-x \log ^2(x)+2 (-2 x+2 x \log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.93 \begin {gather*} \frac {80 \log (x)}{9}+4 x \log (x)-\frac {20 \log ^2(x)}{9}-x \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(80 + 36*x + (-40 + 18*x)*Log[x] - 9*x*Log[x]^2)/(9*x),x]

[Out]

(80*Log[x])/9 + 4*x*Log[x] - (20*Log[x]^2)/9 - x*Log[x]^2

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fricas [A]  time = 0.84, size = 21, normalized size = 1.50 \begin {gather*} -\frac {1}{9} \, {\left (9 \, x + 20\right )} \log \relax (x)^{2} + \frac {4}{9} \, {\left (9 \, x + 20\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-9*x*log(x)^2+(18*x-40)*log(x)+36*x+80)/x,x, algorithm="fricas")

[Out]

-1/9*(9*x + 20)*log(x)^2 + 4/9*(9*x + 20)*log(x)

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giac [A]  time = 0.14, size = 21, normalized size = 1.50 \begin {gather*} -\frac {1}{9} \, {\left (9 \, x + 20\right )} \log \relax (x)^{2} + 4 \, x \log \relax (x) + \frac {80}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-9*x*log(x)^2+(18*x-40)*log(x)+36*x+80)/x,x, algorithm="giac")

[Out]

-1/9*(9*x + 20)*log(x)^2 + 4*x*log(x) + 80/9*log(x)

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maple [A]  time = 0.02, size = 22, normalized size = 1.57

method	result	size
risch	\(\frac {\left (-9 x -20\right ) \ln \relax (x )^{2}}{9}+4 x \ln \relax (x )+\frac {80 \ln \relax (x )}{9}\)	\(22\)
default	\(-x \ln \relax (x )^{2}+4 x \ln \relax (x )-\frac {20 \ln \relax (x )^{2}}{9}+\frac {80 \ln \relax (x )}{9}\)	\(24\)
norman	\(-x \ln \relax (x )^{2}+4 x \ln \relax (x )-\frac {20 \ln \relax (x )^{2}}{9}+\frac {80 \ln \relax (x )}{9}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(-9*x*ln(x)^2+(18*x-40)*ln(x)+36*x+80)/x,x,method=_RETURNVERBOSE)

[Out]

1/9*(-9*x-20)*ln(x)^2+4*x*ln(x)+80/9*ln(x)

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maxima [B]  time = 0.44, size = 32, normalized size = 2.29 \begin {gather*} -{\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + 2 \, x \log \relax (x) - \frac {20}{9} \, \log \relax (x)^{2} + 2 \, x + \frac {80}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-9*x*log(x)^2+(18*x-40)*log(x)+36*x+80)/x,x, algorithm="maxima")

[Out]

-(log(x)^2 - 2*log(x) + 2)*x + 2*x*log(x) - 20/9*log(x)^2 + 2*x + 80/9*log(x)

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mupad [B]  time = 7.30, size = 13, normalized size = 0.93 \begin {gather*} -\frac {\ln \relax (x)\,\left (9\,x+20\right )\,\left (\ln \relax (x)-4\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - x*log(x)^2 + (log(x)*(18*x - 40))/9 + 80/9)/x,x)

[Out]

-(log(x)*(9*x + 20)*(log(x) - 4))/9

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sympy [A]  time = 0.12, size = 24, normalized size = 1.71 \begin {gather*} 4 x \log {\relax (x )} + \left (- x - \frac {20}{9}\right ) \log {\relax (x )}^{2} + \frac {80 \log {\relax (x )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(-9*x*ln(x)**2+(18*x-40)*ln(x)+36*x+80)/x,x)

[Out]

4*x*log(x) + (-x - 20/9)*log(x)**2 + 80*log(x)/9

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