\(\int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx\) [6192]

Optimal result

Integrand size = 64, antiderivative size = 24 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=1+\frac {x \left (-1+x+e^{-x} \log ^2(\log (3))\right )}{-2+\log (x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 21, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6873, 6874, 2357, 2367, 2336, 2209, 2346, 2326} \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {(1-x) x}{2-\log (x)}-\frac {e^{-x} \log ^2(\log (3)) (2 x-x \log (x))}{(2-\log (x))^2} \]

[In]

Int[(E^x*(3 - 5*x) + E^x*(-1 + 2*x)*Log[x] + (-3 + 2*x + (1 - x)*Log[x])*Log[Log[3]]^2)/(4*E^x - 4*E^x*Log[x]
+ E^x*Log[x]^2),x]

[Out]

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

Rule 2209

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

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2336

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

Rule 2346

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2357

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

Rule 2367

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

Rule 6873

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

Rule 6874

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

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

Mathematica [A] (verified)

Time = 3.93 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x \left (-1+x+e^{-x} \log ^2(\log (3))\right )}{-2+\log (x)} \]

[In]

Integrate[(E^x*(3 - 5*x) + E^x*(-1 + 2*x)*Log[x] + (-3 + 2*x + (1 - x)*Log[x])*Log[Log[3]]^2)/(4*E^x - 4*E^x*L
og[x] + E^x*Log[x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12

method	result	size
risch	\(\frac {x \left (\ln \left (\ln \left (3\right )\right )^{2}+{\mathrm e}^{x} x -{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{\ln \left (x \right )-2}\)	\(27\)
norman	\(\frac {\left (x \ln \left (\ln \left (3\right )\right )^{2}+{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{\ln \left (x \right )-2}\)	\(31\)
parallelrisch	\(\frac {\left (x \ln \left (\ln \left (3\right )\right )^{2}+{\mathrm e}^{x} x^{2}-{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{\ln \left (x \right )-2}\)	\(31\)
default	\(-\frac {x}{\ln \left (x \right )-2}+\frac {x^{2}}{\ln \left (x \right )-2}+\frac {\ln \left (\ln \left (3\right )\right )^{2} x \,{\mathrm e}^{-x}}{\ln \left (x \right )-2}\)	\(38\)
parts	\(-\frac {x}{\ln \left (x \right )-2}+\frac {x^{2}}{\ln \left (x \right )-2}+\frac {\ln \left (\ln \left (3\right )\right )^{2} x \,{\mathrm e}^{-x}}{\ln \left (x \right )-2}\)	\(38\)

[In]

int((((1-x)*ln(x)+2*x-3)*ln(ln(3))^2+(-1+2*x)*exp(x)*ln(x)+(3-5*x)*exp(x))/(exp(x)*ln(x)^2-4*exp(x)*ln(x)+4*ex
p(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x \log \left (\log \left (3\right )\right )^{2} + {\left (x^{2} - x\right )} e^{x}}{e^{x} \log \left (x\right ) - 2 \, e^{x}} \]

[In]

integrate((((1-x)*log(x)+2*x-3)*log(log(3))^2+(-1+2*x)*exp(x)*log(x)+(3-5*x)*exp(x))/(exp(x)*log(x)^2-4*exp(x)
*log(x)+4*exp(x)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x e^{- x} \log {\left (\log {\left (3 \right )} \right )}^{2}}{\log {\left (x \right )} - 2} + \frac {x^{2} - x}{\log {\left (x \right )} - 2} \]

[In]

integrate((((1-x)*ln(x)+2*x-3)*ln(ln(3))**2+(-1+2*x)*exp(x)*ln(x)+(3-5*x)*exp(x))/(exp(x)*ln(x)**2-4*exp(x)*ln
(x)+4*exp(x)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x e^{\left (-x\right )} \log \left (\log \left (3\right )\right )^{2} + x^{2} - x}{\log \left (x\right ) - 2} \]

[In]

integrate((((1-x)*log(x)+2*x-3)*log(log(3))^2+(-1+2*x)*exp(x)*log(x)+(3-5*x)*exp(x))/(exp(x)*log(x)^2-4*exp(x)
*log(x)+4*exp(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x e^{\left (-x\right )} \log \left (\log \left (3\right )\right )^{2} + x^{2} - x}{\log \left (x\right ) - 2} \]

[In]

integrate((((1-x)*log(x)+2*x-3)*log(log(3))^2+(-1+2*x)*exp(x)*log(x)+(3-5*x)*exp(x))/(exp(x)*log(x)^2-4*exp(x)
*log(x)+4*exp(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^x (3-5 x)+e^x (-1+2 x) \log (x)+(-3+2 x+(1-x) \log (x)) \log ^2(\log (3))}{4 e^x-4 e^x \log (x)+e^x \log ^2(x)} \, dx=\frac {x\,{\mathrm {e}}^{-x}\,\left ({\ln \left (\ln \left (3\right )\right )}^2-{\mathrm {e}}^x+x\,{\mathrm {e}}^x\right )}{\ln \left (x\right )-2} \]

[In]

int(-(exp(x)*(5*x - 3) + log(log(3))^2*(log(x)*(x - 1) - 2*x + 3) - exp(x)*log(x)*(2*x - 1))/(4*exp(x) - 4*exp
(x)*log(x) + exp(x)*log(x)^2),x)

[Out]

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