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

Optimal. Leaf size=24 \[ 1+\frac {x \left (-1+x+e^{-x} \log ^2(\log (3))\right )}{-2+\log (x)} \]

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Rubi [A]  time = 0.91, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 21, number of rules used = 8, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6741, 6742, 2320, 2330, 2299, 2178, 2309, 2288} \begin {gather*} \frac {(1-x) x}{2-\log (x)}-\frac {e^{-x} \log ^2(\log (3)) (2 x-x \log (x))}{(2-\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[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 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 2288

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 2299

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 2309

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 2320

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 2330

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 6741

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

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 {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-\operatorname {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 \operatorname {Subst}\left (\int \frac {e^{2 x}}{-2+x} \, dx,x,\log (x)\right )-\operatorname {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 \operatorname {Subst}\left (\int \frac {e^x}{-2+x} \, dx,x,\log (x)\right )-2 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 22, normalized size = 0.92 \begin {gather*} \frac {x \left (-1+x+e^{-x} \log ^2(\log (3))\right )}{-2+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.92, size = 31, normalized size = 1.29 \begin {gather*} \frac {x \log \left (\log \relax (3)\right )^{2} + {\left (x^{2} - x\right )} e^{x}}{e^{x} \log \relax (x) - 2 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 0.16, size = 25, normalized size = 1.04 \begin {gather*} \frac {x e^{\left (-x\right )} \log \left (\log \relax (3)\right )^{2} + x^{2} - x}{\log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.14, size = 27, normalized size = 1.12

method	result	size
risch	\(\frac {x \left (\ln \left (\ln \relax (3)\right )^{2}+{\mathrm e}^{x} x -{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{\ln \relax (x )-2}\)	\(27\)
norman	\(\frac {\left ({\mathrm e}^{x} x^{2}+\ln \left (\ln \relax (3)\right )^{2} x -{\mathrm e}^{x} x \right ) {\mathrm e}^{-x}}{\ln \relax (x )-2}\)	\(31\)
default	\(\frac {x^{2}-x}{\ln \relax (x )-2}+\frac {\ln \left (\ln \relax (3)\right )^{2} x \,{\mathrm e}^{-x}}{\ln \relax (x )-2}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 25, normalized size = 1.04 \begin {gather*} \frac {x e^{\left (-x\right )} \log \left (\log \relax (3)\right )^{2} + x^{2} - x}{\log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 4.38, size = 26, normalized size = 1.08 \begin {gather*} \frac {x\,{\mathrm {e}}^{-x}\,\left ({\ln \left (\ln \relax (3)\right )}^2-{\mathrm {e}}^x+x\,{\mathrm {e}}^x\right )}{\ln \relax (x)-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.31, size = 26, normalized size = 1.08 \begin {gather*} \frac {x e^{- x} \log {\left (\log {\relax (3 )} \right )}^{2}}{\log {\relax (x )} - 2} + \frac {x^{2} - x}{\log {\relax (x )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x+1)*ln(x)+2*x-3)*ln(ln(3))**2+(2*x-1)*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)

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