3.18.82 \(\int \frac {2 e^3 x-x \log (x)+(-1+2 x-x^2+e^3 (-2+2 x)) \log (-\frac {\log (3)}{1+2 e^3-x})}{x+2 e^3 x-x^2} \, dx\)

Optimal. Leaf size=27 \[ x+\log (2)-(1-x+\log (x)) \log \left (\frac {\log (3)}{-1-2 e^3+x}\right ) \]

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Rubi [B]  time = 0.40, antiderivative size = 91, normalized size of antiderivative = 3.37, number of steps used = 12, number of rules used = 10, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6, 1593, 6742, 2316, 2315, 43, 2416, 2389, 2295, 2394} \begin {gather*} x-\left (2 e^3-\log \left (1+2 e^3\right )\right ) \log \left (-x+2 e^3+1\right )-\left (-x+2 e^3+1\right ) \log \left (-\frac {\log (3)}{-x+2 e^3+1}\right )-\log \left (\frac {x}{1+2 e^3}\right ) \log \left (-\frac {\log (3)}{-x+2 e^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^3*x - x*Log[x] + (-1 + 2*x - x^2 + E^3*(-2 + 2*x))*Log[-(Log[3]/(1 + 2*E^3 - x))])/(x + 2*E^3*x - x^2
),x]

[Out]

x - (2*E^3 - Log[1 + 2*E^3])*Log[1 + 2*E^3 - x] - (1 + 2*E^3 - x)*Log[-(Log[3]/(1 + 2*E^3 - x))] - Log[x/(1 +
2*E^3)]*Log[-(Log[3]/(1 + 2*E^3 - x))]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 2295

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

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 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[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 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 {2 e^3 x-x \log (x)+\left (-1+2 x-x^2+e^3 (-2+2 x)\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{\left (1+2 e^3\right ) x-x^2} \, dx\\ &=\int \frac {2 e^3 x-x \log (x)+\left (-1+2 x-x^2+e^3 (-2+2 x)\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{\left (1+2 e^3-x\right ) x} \, dx\\ &=\int \left (\frac {2 e^3-\log (x)}{1+2 e^3-x}+\frac {(-1+x) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{x}\right ) \, dx\\ &=\int \frac {2 e^3-\log (x)}{1+2 e^3-x} \, dx+\int \frac {(-1+x) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{x} \, dx\\ &=-\left (\left (2 e^3-\log \left (1+2 e^3\right )\right ) \log \left (1+2 e^3-x\right )\right )-\int \frac {\log \left (\frac {x}{1+2 e^3}\right )}{1+2 e^3-x} \, dx+\int \left (\log \left (-\frac {\log (3)}{1+2 e^3-x}\right )-\frac {\log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{x}\right ) \, dx\\ &=-\left (\left (2 e^3-\log \left (1+2 e^3\right )\right ) \log \left (1+2 e^3-x\right )\right )-\text {Li}_2\left (1-\frac {x}{1+2 e^3}\right )+\int \log \left (-\frac {\log (3)}{1+2 e^3-x}\right ) \, dx-\int \frac {\log \left (-\frac {\log (3)}{1+2 e^3-x}\right )}{x} \, dx\\ &=-\left (\left (2 e^3-\log \left (1+2 e^3\right )\right ) \log \left (1+2 e^3-x\right )\right )-\log \left (\frac {x}{1+2 e^3}\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )-\text {Li}_2\left (1-\frac {x}{1+2 e^3}\right )+\int \frac {\log \left (-\frac {x}{-1-2 e^3}\right )}{1+2 e^3-x} \, dx-\operatorname {Subst}\left (\int \log \left (-\frac {\log (3)}{x}\right ) \, dx,x,1+2 e^3-x\right )\\ &=x-\left (2 e^3-\log \left (1+2 e^3\right )\right ) \log \left (1+2 e^3-x\right )-\left (1+2 e^3-x\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )-\log \left (\frac {x}{1+2 e^3}\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.08, size = 69, normalized size = 2.56 \begin {gather*} x+\left (-2 e^3+\log \left (1+2 e^3\right )\right ) \log \left (1+2 e^3-x\right )-\left (1+2 e^3-x+\log \left (\frac {x}{1+2 e^3}\right )\right ) \log \left (-\frac {\log (3)}{1+2 e^3-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^3*x - x*Log[x] + (-1 + 2*x - x^2 + E^3*(-2 + 2*x))*Log[-(Log[3]/(1 + 2*E^3 - x))])/(x + 2*E^3*x
- x^2),x]

[Out]

x + (-2*E^3 + Log[1 + 2*E^3])*Log[1 + 2*E^3 - x] - (1 + 2*E^3 - x + Log[x/(1 + 2*E^3)])*Log[-(Log[3]/(1 + 2*E^
3 - x))]

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fricas [A]  time = 0.84, size = 36, normalized size = 1.33 \begin {gather*} {\left (x - 1\right )} \log \left (\frac {\log \relax (3)}{x - 2 \, e^{3} - 1}\right ) - \log \relax (x) \log \left (\frac {\log \relax (3)}{x - 2 \, e^{3} - 1}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+((2*x-2)*exp(3)-x^2+2*x-1)*log(-log(3)/(2*exp(3)-x+1))+2*x*exp(3))/(2*x*exp(3)-x^2+x),x,
algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 279, normalized size = 10.33 \begin {gather*} -\frac {2 \, \pi ^{2} e^{3} \mathrm {sgn}\left (x - 2 \, e^{3} - 1\right ) \mathrm {sgn}\relax (x) - 8 \, \pi ^{2} e^{6} \mathrm {sgn}\left (x - 2 \, e^{3} - 1\right ) - 2 \, \pi ^{2} e^{3} \mathrm {sgn}\left (x - 2 \, e^{3} - 1\right ) + 6 \, \pi ^{2} e^{3} \mathrm {sgn}\relax (x) + \pi ^{2} \mathrm {sgn}\left (x - 2 \, e^{3} - 1\right ) \mathrm {sgn}\relax (x) + 8 \, \pi ^{2} e^{6} - 6 \, \pi ^{2} e^{3} + 8 \, x e^{3} \log \left ({\left | x - 2 \, e^{3} - 1 \right |}\right ) - 16 \, e^{6} \log \left ({\left | x - 2 \, e^{3} - 1 \right |}\right )^{2} - 8 \, e^{3} \log \left ({\left | x - 2 \, e^{3} - 1 \right |}\right ) \log \left ({\left | x \right |}\right ) - 8 \, x e^{3} \log \left (\log \relax (3)\right ) + 8 \, e^{3} \log \left ({\left | x \right |}\right ) \log \left (\log \relax (3)\right ) - \pi ^{2} \mathrm {sgn}\left (x - 2 \, e^{3} - 1\right ) + 3 \, \pi ^{2} \mathrm {sgn}\relax (x) - 3 \, \pi ^{2} - 8 \, x e^{3} + 4 \, x \log \left ({\left | x - 2 \, e^{3} - 1 \right |}\right ) - 4 \, \log \left ({\left | x - 2 \, e^{3} - 1 \right |}\right ) \log \left ({\left | x \right |}\right ) - 4 \, x \log \left (\log \relax (3)\right ) + 4 \, \log \left ({\left | x \right |}\right ) \log \left (\log \relax (3)\right ) - 4 \, x}{4 \, {\left (2 \, e^{3} + 1\right )}} - \frac {4 \, e^{6} \log \left (x - 2 \, e^{3} - 1\right )^{2} - 2 \, e^{3} \log \left (x - 2 \, e^{3} - 1\right ) - \log \left (x - 2 \, e^{3} - 1\right )}{2 \, e^{3} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+((2*x-2)*exp(3)-x^2+2*x-1)*log(-log(3)/(2*exp(3)-x+1))+2*x*exp(3))/(2*x*exp(3)-x^2+x),x,
algorithm="giac")

[Out]

-1/4*(2*pi^2*e^3*sgn(x - 2*e^3 - 1)*sgn(x) - 8*pi^2*e^6*sgn(x - 2*e^3 - 1) - 2*pi^2*e^3*sgn(x - 2*e^3 - 1) + 6
*pi^2*e^3*sgn(x) + pi^2*sgn(x - 2*e^3 - 1)*sgn(x) + 8*pi^2*e^6 - 6*pi^2*e^3 + 8*x*e^3*log(abs(x - 2*e^3 - 1))
- 16*e^6*log(abs(x - 2*e^3 - 1))^2 - 8*e^3*log(abs(x - 2*e^3 - 1))*log(abs(x)) - 8*x*e^3*log(log(3)) + 8*e^3*l
og(abs(x))*log(log(3)) - pi^2*sgn(x - 2*e^3 - 1) + 3*pi^2*sgn(x) - 3*pi^2 - 8*x*e^3 + 4*x*log(abs(x - 2*e^3 -
1)) - 4*log(abs(x - 2*e^3 - 1))*log(abs(x)) - 4*x*log(log(3)) + 4*log(abs(x))*log(log(3)) - 4*x)/(2*e^3 + 1) -
(4*e^6*log(x - 2*e^3 - 1)^2 - 2*e^3*log(x - 2*e^3 - 1) - log(x - 2*e^3 - 1))/(2*e^3 + 1)

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maple [B]  time = 0.55, size = 59, normalized size = 2.19

method	result	size
norman	\(x -\ln \left (-\frac {\ln \relax (3)}{2 \,{\mathrm e}^{3}-x +1}\right )+\ln \left (-\frac {\ln \relax (3)}{2 \,{\mathrm e}^{3}-x +1}\right ) x -\ln \relax (x ) \ln \left (-\frac {\ln \relax (3)}{2 \,{\mathrm e}^{3}-x +1}\right )\)	\(59\)
default	\(\left (x -1-2 \,{\mathrm e}^{3}\right ) \ln \left (\frac {1}{x -1-2 \,{\mathrm e}^{3}}\right )+x -1-2 \,{\mathrm e}^{3}-\frac {2 \,{\mathrm e}^{3} \dilog \left (1+\frac {2 \,{\mathrm e}^{3}+1}{x -1-2 \,{\mathrm e}^{3}}\right )}{2 \,{\mathrm e}^{3}+1}-\frac {2 \,{\mathrm e}^{3} \ln \left (\frac {1}{x -1-2 \,{\mathrm e}^{3}}\right ) \ln \left (1+\frac {2 \,{\mathrm e}^{3}+1}{x -1-2 \,{\mathrm e}^{3}}\right )}{2 \,{\mathrm e}^{3}+1}-\frac {\dilog \left (1+\frac {2 \,{\mathrm e}^{3}+1}{x -1-2 \,{\mathrm e}^{3}}\right )}{2 \,{\mathrm e}^{3}+1}-\frac {\ln \left (\frac {1}{x -1-2 \,{\mathrm e}^{3}}\right ) \ln \left (1+\frac {2 \,{\mathrm e}^{3}+1}{x -1-2 \,{\mathrm e}^{3}}\right )}{2 \,{\mathrm e}^{3}+1}+\frac {\ln \left (\frac {1}{x -1-2 \,{\mathrm e}^{3}}\right )^{2}}{2}-\ln \left (\ln \relax (3)\right ) \ln \relax (x )+\ln \left (\ln \relax (3)\right ) x -2 \,{\mathrm e}^{3} \ln \left (2 \,{\mathrm e}^{3}-x +1\right )+\left (\ln \relax (x )-\ln \left (\frac {x}{2 \,{\mathrm e}^{3}+1}\right )\right ) \ln \left (\frac {2 \,{\mathrm e}^{3}-x +1}{2 \,{\mathrm e}^{3}+1}\right )-\dilog \left (\frac {x}{2 \,{\mathrm e}^{3}+1}\right )\)	\(255\)
risch	\(\left (\ln \relax (x )-x \right ) \ln \left ({\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}\right )-i \pi x \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+2 \pi -2 i \ln \left (\ln \relax (3)\right )+2 i \ln \relax (2)-2 i\right ) x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+i \pi x -i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+2 \pi -2 i \ln \left (\ln \relax (3)\right )+2 i \ln \relax (2)-2 i\right ) x \right )-i \pi \ln \left (\left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+2 \pi -2 i \ln \left (\ln \relax (3)\right )+2 i \ln \relax (2)-2 i\right ) x \right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+\ln \left (\ln \relax (3)\right ) x -x \ln \relax (2)+x +i \pi x \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}-\ln \left (\ln \relax (3)\right ) \ln \left (\left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+2 \pi -2 i \ln \left (\ln \relax (3)\right )+2 i \ln \relax (2)-2 i\right ) x \right )+\ln \relax (2) \ln \left (\left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-\frac {x}{2}+\frac {1}{2}}\right )^{3}+2 \pi -2 i \ln \left (\ln \relax (3)\right )+2 i \ln \relax (2)-2 i\right ) x \right )+\ln \left (2 \,{\mathrm e}^{3}-x +1\right )\)	\(413\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)+((2*x-2)*exp(3)-x^2+2*x-1)*ln(-ln(3)/(2*exp(3)-x+1))+2*x*exp(3))/(2*x*exp(3)-x^2+x),x,method=_RE
TURNVERBOSE)

[Out]

x-ln(-ln(3)/(2*exp(3)-x+1))+ln(-ln(3)/(2*exp(3)-x+1))*x-ln(x)*ln(-ln(3)/(2*exp(3)-x+1))

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maxima [A]  time = 0.59, size = 48, normalized size = 1.78 \begin {gather*} x {\left (\log \left (\log \relax (3)\right ) + 1\right )} - {\left (x - 2 \, e^{3} - \log \relax (x) - 1\right )} \log \left (x - 2 \, e^{3} - 1\right ) - 2 \, e^{3} \log \left (x - 2 \, e^{3} - 1\right ) - \log \relax (x) \log \left (\log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+((2*x-2)*exp(3)-x^2+2*x-1)*log(-log(3)/(2*exp(3)-x+1))+2*x*exp(3))/(2*x*exp(3)-x^2+x),x,
algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.66, size = 33, normalized size = 1.22 \begin {gather*} x+\ln \left (x-2\,{\mathrm {e}}^3-1\right )+\ln \left (-\frac {\ln \relax (3)}{2\,{\mathrm {e}}^3-x+1}\right )\,\left (x-\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*exp(3) + log(-log(3)/(2*exp(3) - x + 1))*(2*x - x^2 + exp(3)*(2*x - 2) - 1) - x*log(x))/(x + 2*x*exp(
3) - x^2),x)

[Out]

x + log(x - 2*exp(3) - 1) + log(-log(3)/(2*exp(3) - x + 1))*(x - log(x))

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sympy [A]  time = 0.39, size = 31, normalized size = 1.15 \begin {gather*} x + \left (x - \log {\relax (x )}\right ) \log {\left (- \frac {\log {\relax (3 )}}{- x + 1 + 2 e^{3}} \right )} + \log {\left (x - 2 e^{3} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)+((2*x-2)*exp(3)-x**2+2*x-1)*ln(-ln(3)/(2*exp(3)-x+1))+2*x*exp(3))/(2*x*exp(3)-x**2+x),x)

[Out]

x + (x - log(x))*log(-log(3)/(-x + 1 + 2*exp(3))) + log(x - 2*exp(3) - 1)

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