3.63.7 \(\int \frac {-x+(-3-5 x+2 x^2+e^x (-3 x+x^2)) \log (-3+x)+(-3+x) \log (-3+x) \log (\frac {4 e^{-e^x} \log (-3+x)}{x})}{(-6 x^2+2 x^3) \log (-3+x)} \, dx\)

Optimal. Leaf size=27 \[ \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x} \]

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Rubi [A]  time = 1.47, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 30, number of rules used = 13, integrand size = 75, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.173, Rules used = {1593, 6688, 12, 14, 2178, 6742, 43, 2416, 2390, 2302, 29, 30, 2555} \begin {gather*} \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (x-3)}{x}\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x + (-3 - 5*x + 2*x^2 + E^x*(-3*x + x^2))*Log[-3 + x] + (-3 + x)*Log[-3 + x]*Log[(4*Log[-3 + x])/(E^E^x*
x)])/((-6*x^2 + 2*x^3)*Log[-3 + x]),x]

[Out]

Log[x] - Log[(4*Log[-3 + x])/(E^E^x*x)]/(2*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 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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 1593

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

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 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 2390

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

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2555

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

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x+\left (-3-5 x+2 x^2+e^x \left (-3 x+x^2\right )\right ) \log (-3+x)+(-3+x) \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2 (-6+2 x) \log (-3+x)} \, dx\\ &=\int \frac {1+\left (2+e^x\right ) x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x^2} \, dx\\ &=\frac {1}{2} \int \frac {1+\left (2+e^x\right ) x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {e^x}{x}+\frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)-3 \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )+x \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{(-3+x) x^2 \log (-3+x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^x}{x} \, dx+\frac {1}{2} \int \frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)-3 \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )+x \log (-3+x) \log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{(-3+x) x^2 \log (-3+x)} \, dx\\ &=\frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \frac {1+2 x-\frac {x}{(-3+x) \log (-3+x)}+\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx\\ &=\frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \left (\frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)}{(-3+x) x^2 \log (-3+x)}+\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2}\right ) \, dx\\ &=\frac {\text {Ei}(x)}{2}+\frac {1}{2} \int \frac {-x-3 \log (-3+x)-5 x \log (-3+x)+2 x^2 \log (-3+x)}{(-3+x) x^2 \log (-3+x)} \, dx+\frac {1}{2} \int \frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x^2} \, dx\\ &=\frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \left (\frac {1+2 x}{x^2}-\frac {1}{(-3+x) x \log (-3+x)}\right ) \, dx-\frac {1}{2} \int \frac {1+e^x x-\frac {x}{(-3+x) \log (-3+x)}}{x^2} \, dx\\ &=\frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \frac {1+2 x}{x^2} \, dx-\frac {1}{2} \int \frac {1}{(-3+x) x \log (-3+x)} \, dx-\frac {1}{2} \int \left (\frac {e^x}{x}+\frac {-x-3 \log (-3+x)+x \log (-3+x)}{(-3+x) x^2 \log (-3+x)}\right ) \, dx\\ &=\frac {\text {Ei}(x)}{2}-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{2} \int \left (\frac {1}{x^2}+\frac {2}{x}\right ) \, dx-\frac {1}{2} \int \frac {e^x}{x} \, dx-\frac {1}{2} \int \left (\frac {1}{3 (-3+x) \log (-3+x)}-\frac {1}{3 x \log (-3+x)}\right ) \, dx-\frac {1}{2} \int \frac {-x-3 \log (-3+x)+x \log (-3+x)}{(-3+x) x^2 \log (-3+x)} \, dx\\ &=-\frac {1}{2 x}+\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}-\frac {1}{6} \int \frac {1}{(-3+x) \log (-3+x)} \, dx+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{2} \int \frac {1-\frac {x}{(-3+x) \log (-3+x)}}{x^2} \, dx\\ &=-\frac {1}{2 x}+\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3+x\right )-\frac {1}{2} \int \left (\frac {1}{x^2}-\frac {1}{(-3+x) x \log (-3+x)}\right ) \, dx\\ &=\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3+x)\right )+\frac {1}{2} \int \frac {1}{(-3+x) x \log (-3+x)} \, dx\\ &=\log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{x \log (-3+x)} \, dx+\frac {1}{2} \int \left (\frac {1}{3 (-3+x) \log (-3+x)}-\frac {1}{3 x \log (-3+x)}\right ) \, dx\\ &=\log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \int \frac {1}{(-3+x) \log (-3+x)} \, dx\\ &=\log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-3+x\right )\\ &=\log (x)-\frac {1}{6} \log (\log (-3+x))-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-3+x)\right )\\ &=\log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 31, normalized size = 1.15 \begin {gather*} \frac {1}{2} \left (2 \log (x)-\frac {\log \left (\frac {4 e^{-e^x} \log (-3+x)}{x}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x + (-3 - 5*x + 2*x^2 + E^x*(-3*x + x^2))*Log[-3 + x] + (-3 + x)*Log[-3 + x]*Log[(4*Log[-3 + x])/(
E^E^x*x)])/((-6*x^2 + 2*x^3)*Log[-3 + x]),x]

[Out]

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

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fricas [A]  time = 1.17, size = 28, normalized size = 1.04 \begin {gather*} \frac {2 \, x \log \relax (x) - \log \left (\frac {4 \, e^{\left (-e^{x}\right )} \log \left (x - 3\right )}{x}\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*log(x-3)*log(4*log(x-3)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x^2-5*x-3)*log(x-3)-x)/(2*x^3-6*x^
2)/log(x-3),x, algorithm="fricas")

[Out]

1/2*(2*x*log(x) - log(4*e^(-e^x)*log(x - 3)/x))/x

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giac [A]  time = 0.26, size = 24, normalized size = 0.89 \begin {gather*} \frac {2 \, x \log \relax (x) + e^{x} + \log \relax (x) - \log \left (4 \, \log \left (x - 3\right )\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*log(x-3)*log(4*log(x-3)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x^2-5*x-3)*log(x-3)-x)/(2*x^3-6*x^
2)/log(x-3),x, algorithm="giac")

[Out]

1/2*(2*x*log(x) + e^x + log(x) - log(4*log(x - 3)))/x

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maple [C]  time = 0.64, size = 271, normalized size = 10.04

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )}{2 x}+\frac {i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (x -3\right )}{x}\right )^{3}+i \pi \,\mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (x -3\right )}{x}\right ) \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i \ln \left (x -3\right )\right ) \mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )-i \pi \,\mathrm {csgn}\left (i \ln \left (x -3\right )\right ) \mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}-i \pi \,\mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (x -3\right )}{x}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-{\mathrm e}^{x}} \ln \left (x -3\right )}{x}\right )^{2} \mathrm {csgn}\left (\frac {i}{x}\right )+i \pi \mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{3}-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) \mathrm {csgn}\left (i \ln \left (x -3\right ) {\mathrm e}^{-{\mathrm e}^{x}}\right )^{2}+4 x \ln \relax (x )-4 \ln \relax (2)+2 \ln \relax (x )-2 \ln \left (\ln \left (x -3\right )\right )}{4 x}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-3)*ln(x-3)*ln(4*ln(x-3)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x^2-5*x-3)*ln(x-3)-x)/(2*x^3-6*x^2)/ln(x-3)
,x,method=_RETURNVERBOSE)

[Out]

1/2/x*ln(exp(exp(x)))+1/4*(I*Pi*csgn(I/x*exp(-exp(x))*ln(x-3))^3+I*Pi*csgn(I*ln(x-3)*exp(-exp(x)))*csgn(I/x*ex
p(-exp(x))*ln(x-3))*csgn(I/x)+I*Pi*csgn(I*exp(-exp(x)))*csgn(I*ln(x-3))*csgn(I*ln(x-3)*exp(-exp(x)))-I*Pi*csgn
(I*ln(x-3))*csgn(I*ln(x-3)*exp(-exp(x)))^2-I*Pi*csgn(I*ln(x-3)*exp(-exp(x)))*csgn(I/x*exp(-exp(x))*ln(x-3))^2-
I*Pi*csgn(I/x*exp(-exp(x))*ln(x-3))^2*csgn(I/x)+I*Pi*csgn(I*ln(x-3)*exp(-exp(x)))^3-I*Pi*csgn(I*exp(-exp(x)))*
csgn(I*ln(x-3)*exp(-exp(x)))^2+4*x*ln(x)-4*ln(2)+2*ln(x)-2*ln(ln(x-3)))/x

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maxima [B]  time = 0.51, size = 52, normalized size = 1.93 \begin {gather*} \frac {x \log \left (x - 3\right ) + {\left (5 \, x + 3\right )} \log \relax (x) + 3 \, e^{x} - 6 \, \log \relax (2) - 3 \, \log \left (\log \left (x - 3\right )\right ) + 3}{6 \, x} - \frac {1}{2 \, x} - \frac {1}{6} \, \log \left (x - 3\right ) + \frac {1}{6} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*log(x-3)*log(4*log(x-3)/x/exp(exp(x)))+((x^2-3*x)*exp(x)+2*x^2-5*x-3)*log(x-3)-x)/(2*x^3-6*x^
2)/log(x-3),x, algorithm="maxima")

[Out]

1/6*(x*log(x - 3) + (5*x + 3)*log(x) + 3*e^x - 6*log(2) - 3*log(log(x - 3)) + 3)/x - 1/2/x - 1/6*log(x - 3) +
1/6*log(x)

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mupad [B]  time = 4.43, size = 31, normalized size = 1.15 \begin {gather*} \ln \relax (x)+\frac {{\mathrm {e}}^x}{2\,x}-\frac {\ln \left (\frac {\ln \left (x-3\right )}{x}\right )}{2\,x}-\frac {\ln \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(x - 3)*(5*x + exp(x)*(3*x - x^2) - 2*x^2 + 3) - log(x - 3)*log((4*log(x - 3)*exp(-exp(x)))/x)*(x
- 3))/(log(x - 3)*(6*x^2 - 2*x^3)),x)

[Out]

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

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sympy [A]  time = 4.58, size = 20, normalized size = 0.74 \begin {gather*} \log {\relax (x )} - \frac {\log {\left (\frac {4 e^{- e^{x}} \log {\left (x - 3 \right )}}{x} \right )}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*ln(x-3)*ln(4*ln(x-3)/x/exp(exp(x)))+((x**2-3*x)*exp(x)+2*x**2-5*x-3)*ln(x-3)-x)/(2*x**3-6*x**
2)/ln(x-3),x)

[Out]

log(x) - log(4*exp(-exp(x))*log(x - 3)/x)/(2*x)

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