\(\int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx\) [9356]

Optimal result

Integrand size = 27, antiderivative size = 14 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=\left (\frac {20}{9}+x\right ) (4-\log (x)) \log (x) \]

[Out]

ln(x)*(-ln(x)+4)*(20/9+x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {12, 14, 45, 2388, 2338, 2332, 2333} \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=-x \log ^2(x)-\frac {20 \log ^2(x)}{9}+4 x \log (x)+\frac {80 \log (x)}{9} \]

[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 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 2332

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

Rule 2333

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 2338

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 2388

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=\frac {80 \log (x)}{9}+4 x \log (x)-\frac {20 \log ^2(x)}{9}-x \log ^2(x) \]

[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

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=-\frac {1}{9} \, {\left (9 \, x + 20\right )} \log \left (x\right )^{2} + \frac {4}{9} \, {\left (9 \, x + 20\right )} \log \left (x\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.71 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=4 x \log {\left (x \right )} + \left (- x - \frac {20}{9}\right ) \log {\left (x \right )}^{2} + \frac {80 \log {\left (x \right )}}{9} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (13) = 26\).

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 2.29 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=-{\left (\log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 2\right )} x + 2 \, x \log \left (x\right ) - \frac {20}{9} \, \log \left (x\right )^{2} + 2 \, x + \frac {80}{9} \, \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.50 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=-\frac {1}{9} \, {\left (9 \, x + 20\right )} \log \left (x\right )^{2} + 4 \, x \log \left (x\right ) + \frac {80}{9} \, \log \left (x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 12.37 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {80+36 x+(-40+18 x) \log (x)-9 x \log ^2(x)}{9 x} \, dx=-\frac {\ln \left (x\right )\,\left (9\,x+20\right )\,\left (\ln \left (x\right )-4\right )}{9} \]

[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