3.83.28 $\int \frac{27-45x+156x^2-28x^3+e^{5-x}(-27x^2+17x^3-4x^4)+(30x^2-5x^3+e^{5-x}(-6x^2+4x^3-x^4))\log (x)}{27x-18x^2+3x^3}dx$

Optimal. Leaf size=31 $$-x+\log (x)+\frac{(-5+e^{5-x})x^2(4+\log (x))}{3(-3+x)}$$

________________________________________________________________________________________

Rubi [B]  time = 1.93, antiderivative size = 115, normalized size of antiderivative = 3.71, number of steps used = 39, number of rules used = 18, integrand size = 94, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.192, Rules used = {1594, 27, 12, 6742, 44, 43, 2351, 2314, 31, 2316, 2315, 2295, 2199, 2194, 2177, 2178, 2176, 2554} $$\begin{array}{c}\frac{4}{3}{e}^{5-x}x-\frac{23x}{3}+4e^{5-x}-\frac{12e^{5-x}}{3-x}+\frac{60}{3-x}+\frac{1}{3}{e}^{5-x}x\log (x)+\frac{5x\log (x)}{3-x}-\frac{5}{3}x\log (x)+e^{5-x}\log (x)-\frac{3e^{5-x}\log (x)}{3-x}+\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(27 - 45*x + 156*x^2 - 28*x^3 + E^(5 - x)*(-27*x^2 + 17*x^3 - 4*x^4) + (30*x^2 - 5*x^3 + E^(5 - x)*(-6*x^2
+ 4*x^3 - x^4))*Log[x])/(27*x - 18*x^2 + 3*x^3),x]

[Out]

4*E^(5 - x) + 60/(3 - x) - (12*E^(5 - x))/(3 - x) - (23*x)/3 + (4*E^(5 - x)*x)/3 + Log[x] + E^(5 - x)*Log[x] -
(3*E^(5 - x)*Log[x])/(3 - x) - (5*x*Log[x])/3 + (E^(5 - x)*x*Log[x])/3 + (5*x*Log[x])/(3 - x)

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

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

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

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 2314

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{27-45x+156x^2-28x^3+e^{5-x}(-27x^2+17x^3-4x^4)+(30x^2-5x^3+e^{5-x}(-6x^2+4x^3-x^4))\log (x)}{x(27-18x+3x^2)}dx\\ & =\int \frac{27-45x+156x^2-28x^3+e^{5-x}(-27x^2+17x^3-4x^4)+(30x^2-5x^3+e^{5-x}(-6x^2+4x^3-x^4))\log (x)}{3(-3+x{)}^{2}x}dx\\ & =\frac{1}{3}\int \frac{27-45x+156x^2-28x^3+e^{5-x}(-27x^2+17x^3-4x^4)+(30x^2-5x^3+e^{5-x}(-6x^2+4x^3-x^4))\log (x)}{(-3+x{)}^{2}x}dx\\ & =\frac{1}{3}\int (-\frac{45}{(-3+x{)}^2}+\frac{27}{(-3+x{)}^{2}x}+\frac{156x}{(-3+x{)}^2}-\frac{28x^2}{(-3+x{)}^2}+\frac{30x\log (x)}{(-3+x{)}^2}-\frac{5x^2\log (x)}{(-3+x{)}^2}-\frac{e^{5-x}x(27-17x+4x^2+6\log (x)-4x\log (x)+x^2\log (x))}{(-3+x{)}^2})dx\\ & =-\frac{15}{3-x}-\frac{1}{3}\int \frac{e^{5-x}x(27-17x+4x^2+6\log (x)-4x\log (x)+x^2\log (x))}{(-3+x{)}^2}dx-\frac{5}{3}\int \frac{x^2\log (x)}{(-3+x{)}^2}dx+9\int \frac{1}{(-3+x{)}^{2}x}dx-\frac{28}{3}\int \frac{x^2}{(-3+x{)}^2}dx+10\int \frac{x\log (x)}{(-3+x{)}^2}dx+52\int \frac{x}{(-3+x{)}^2}dx\\ & =-\frac{15}{3-x}-\frac{1}{3}\int (\frac{e^{5-x}x(27-17x+4x^2)}{(-3+x{)}^2}+\frac{e^{5-x}x(6-4x+x^2)\log (x)}{(-3+x{)}^2})dx-\frac{5}{3}\int (\log (x)+\frac{9\log (x)}{(-3+x{)}^2}+\frac{6\log (x)}{-3+x})dx+9\int (\frac{1}{3(-3+x{)}^2}-\frac{1}{9(-3+x)}+\frac{1}{9x})dx-\frac{28}{3}\int (1+\frac{9}{(-3+x{)}^2}+\frac{6}{-3+x})dx+10\int (\frac{3\log (x)}{(-3+x{)}^2}+\frac{\log (x)}{-3+x})dx+52\int (\frac{3}{(-3+x{)}^2}+\frac{1}{-3+x})dx\\ & =\frac{60}{3-x}-\frac{28x}{3}-5\log (3-x)+\log (x)-\frac{1}{3}\int \frac{e^{5-x}x(27-17x+4x^2)}{(-3+x{)}^2}dx-\frac{1}{3}\int \frac{e^{5-x}x(6-4x+x^2)\log (x)}{(-3+x{)}^2}dx-\frac{5}{3}\int \log (x)dx-15\int \frac{\log (x)}{(-3+x{)}^2}dx+30\int \frac{\log (x)}{(-3+x{)}^2}dx\\ & =\frac{60}{3-x}-\frac{23x}{3}-5\log (3-x)+\log (x)+e^{5-x}\log (x)-\frac{3e^{5-x}\log (x)}{3-x}-\frac{5}{3}x\log (x)+\frac{1}{3}{e}^{5-x}x\log (x)+\frac{5x\log (x)}{3-x}+\frac{1}{3}\int \frac{e^{5-x}x}{3-x}dx-\frac{1}{3}\int (7e^{5-x}+\frac{36e^{5-x}}{(-3+x{)}^2}+\frac{33e^{5-x}}{-3+x}+4e^{5-x}x)dx-5\int \frac{1}{-3+x}dx+10\int \frac{1}{-3+x}dx\\ & =\frac{60}{3-x}-\frac{23x}{3}+\log (x)+e^{5-x}\log (x)-\frac{3e^{5-x}\log (x)}{3-x}-\frac{5}{3}x\log (x)+\frac{1}{3}{e}^{5-x}x\log (x)+\frac{5x\log (x)}{3-x}+\frac{1}{3}\int (-e^{5-x}-\frac{3e^{5-x}}{-3+x})dx-\frac{4}{3}\int e^{5-x}xdx-\frac{7}{3}\int e^{5-x}dx-11\int \frac{e^{5-x}}{-3+x}dx-12\int \frac{e^{5-x}}{(-3+x{)}^2}dx\\ & =\frac{7e^{5-x}}{3}+\frac{60}{3-x}-\frac{12e^{5-x}}{3-x}-\frac{23x}{3}+\frac{4}{3}{e}^{5-x}x-11e^2\text{Ei}(3-x)+\log (x)+e^{5-x}\log (x)-\frac{3e^{5-x}\log (x)}{3-x}-\frac{5}{3}x\log (x)+\frac{1}{3}{e}^{5-x}x\log (x)+\frac{5x\log (x)}{3-x}-\frac{1}{3}\int e^{5-x}dx-\frac{4}{3}\int e^{5-x}dx+12\int \frac{e^{5-x}}{-3+x}dx-\int \frac{e^{5-x}}{-3+x}dx\\ & =4e^{5-x}+\frac{60}{3-x}-\frac{12e^{5-x}}{3-x}-\frac{23x}{3}+\frac{4}{3}{e}^{5-x}x+\log (x)+e^{5-x}\log (x)-\frac{3e^{5-x}\log (x)}{3-x}-\frac{5}{3}x\log (x)+\frac{1}{3}{e}^{5-x}x\log (x)+\frac{5x\log (x)}{3-x}\end{array}\end{array}$$

________________________________________________________________________________________

Mathematica [A]  time = 0.66, size = 62, normalized size = 2.00 $$\begin{array}{c}\frac{e^{-x}(4e^5{x}^2+e^x(-180+69x-23x^2)+(e^5{x}^2+e^x(-9+3x-5x^2))\log (x))}{3(-3+x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(27 - 45*x + 156*x^2 - 28*x^3 + E^(5 - x)*(-27*x^2 + 17*x^3 - 4*x^4) + (30*x^2 - 5*x^3 + E^(5 - x)*(
-6*x^2 + 4*x^3 - x^4))*Log[x])/(27*x - 18*x^2 + 3*x^3),x]

[Out]

(4*E^5*x^2 + E^x*(-180 + 69*x - 23*x^2) + (E^5*x^2 + E^x*(-9 + 3*x - 5*x^2))*Log[x])/(3*E^x*(-3 + x))

________________________________________________________________________________________

fricas [A]  time = 0.97, size = 51, normalized size = 1.65 $$\begin{array}{c}\frac{4x^2{e}^{(-x+5)}-23x^2+(x^2{e}^{(-x+5)}-5x^2+3x-9)\log (x)+69x-180}{3(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-6*x^2)*exp(5-x)-5*x^3+30*x^2)*log(x)+(-4*x^4+17*x^3-27*x^2)*exp(5-x)-28*x^3+156*x^2-45
*x+27)/(3*x^3-18*x^2+27*x),x, algorithm="fricas")

[Out]

1/3*(4*x^2*e^(-x + 5) - 23*x^2 + (x^2*e^(-x + 5) - 5*x^2 + 3*x - 9)*log(x) + 69*x - 180)/(x - 3)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 56, normalized size = 1.81 $$\begin{array}{c}\frac{x^2{e}^{(-x+5)}\log (x)+4x^2{e}^{(-x+5)}-5x^2\log (x)-23x^2+3x\log (x)+69x-9\log (x)-180}{3(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-6*x^2)*exp(5-x)-5*x^3+30*x^2)*log(x)+(-4*x^4+17*x^3-27*x^2)*exp(5-x)-28*x^3+156*x^2-45
*x+27)/(3*x^3-18*x^2+27*x),x, algorithm="giac")

[Out]

1/3*(x^2*e^(-x + 5)*log(x) + 4*x^2*e^(-x + 5) - 5*x^2*log(x) - 23*x^2 + 3*x*log(x) + 69*x - 9*log(x) - 180)/(x
- 3)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 48, normalized size = 1.55

method	result	size
norman	$\frac{-\frac{23x^2}{3}+\frac{4x^2{\mathrm{e}}^{5-x}}{3}-\frac{5x^2\ln (x)}{3}+9+\frac{\ln (x){\mathrm{e}}^{5-x}{x}^2}{3}}{x-3}+\ln (x)$	$48$
default	$\frac{\ln (x){\mathrm{e}}^{5-x}{x}^2+4x^2{\mathrm{e}}^{5-x}}{3x-9}-\frac{23x}{3}+\ln (x)-\frac{60}{x-3}-\frac{5x\ln (x)}{3}-\frac{5\ln (x)x}{x-3}$	$60$
risch	$\frac{(x^2{\mathrm{e}}^{5-x}-5x^2+15x-45)\ln (x)}{3x-9}-\frac{-4x^2{\mathrm{e}}^{5-x}+12x\ln (x)+23x^2-36\ln (x)-69x+180}{3(x-3)}$	$68$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^4+4*x^3-6*x^2)*exp(5-x)-5*x^3+30*x^2)*ln(x)+(-4*x^4+17*x^3-27*x^2)*exp(5-x)-28*x^3+156*x^2-45*x+27)/
(3*x^3-18*x^2+27*x),x,method=_RETURNVERBOSE)

[Out]

(-23/3*x^2+4/3*x^2*exp(5-x)-5/3*x^2*ln(x)+9+1/3*ln(x)*exp(5-x)*x^2)/(x-3)+ln(x)

________________________________________________________________________________________

maxima [B]  time = 0.42, size = 64, normalized size = 2.06 $$\begin{array}{c}-\frac{28}{3}x+\frac{5x^2+(x^2{e}^5\log (x)+4x^2{e}^5)e^{(-x)}-5(x^2-3x+9)\log (x)-15x}{3(x-3)}-\frac{60}{x-3}-4\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^4+4*x^3-6*x^2)*exp(5-x)-5*x^3+30*x^2)*log(x)+(-4*x^4+17*x^3-27*x^2)*exp(5-x)-28*x^3+156*x^2-45
*x+27)/(3*x^3-18*x^2+27*x),x, algorithm="maxima")

[Out]

-28/3*x + 1/3*(5*x^2 + (x^2*e^5*log(x) + 4*x^2*e^5)*e^(-x) - 5*(x^2 - 3*x + 9)*log(x) - 15*x)/(x - 3) - 60/(x
- 3) - 4*log(x)

________________________________________________________________________________________

mupad [B]  time = 5.10, size = 64, normalized size = 2.06 $$\begin{array}{c}\ln (x)-\frac{23x}{3}-\frac{60}{x-3}+\frac{4x^2{\mathrm{e}}^{5-x}}{3(x-3)}-\frac{5x^2\ln (x)}{3(x-3)}+\frac{x^2{\mathrm{e}}^{5-x}\ln (x)}{3(x-3)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(45*x + log(x)*(exp(5 - x)*(6*x^2 - 4*x^3 + x^4) - 30*x^2 + 5*x^3) - 156*x^2 + 28*x^3 + exp(5 - x)*(27*x^
2 - 17*x^3 + 4*x^4) - 27)/(27*x - 18*x^2 + 3*x^3),x)

[Out]

log(x) - (23*x)/3 - 60/(x - 3) + (4*x^2*exp(5 - x))/(3*(x - 3)) - (5*x^2*log(x))/(3*(x - 3)) + (x^2*exp(5 - x)
*log(x))/(3*(x - 3))

________________________________________________________________________________________

sympy [B]  time = 0.45, size = 54, normalized size = 1.74 $$\begin{array}{c}-\frac{23x}{3}-4\log (x)+\frac{(x^2\log (x)+4x^2)e^{5-x}}{3x-9}+\frac{(-5x^2+15x-45)\log (x)}{3x-9}-\frac{60}{x-3}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**4+4*x**3-6*x**2)*exp(5-x)-5*x**3+30*x**2)*ln(x)+(-4*x**4+17*x**3-27*x**2)*exp(5-x)-28*x**3+15
6*x**2-45*x+27)/(3*x**3-18*x**2+27*x),x)

[Out]

-23*x/3 - 4*log(x) + (x**2*log(x) + 4*x**2)*exp(5 - x)/(3*x - 9) + (-5*x**2 + 15*x - 45)*log(x)/(3*x - 9) - 60
/(x - 3)

________________________________________________________________________________________