3.48.27 \(\int \frac {3 x+2 x^2+(4 e^{2 e}+e^e (9 x+6 x^2)) \log (x)+e^{2 e} (-2+9 x+6 x^2) \log ^2(x)+e^{3 e} (-2+3 x+2 x^2) \log ^3(x)}{2 x^2+6 e^e x^2 \log (x)+6 e^{2 e} x^2 \log ^2(x)+2 e^{3 e} x^2 \log ^3(x)} \, dx\)

Optimal. Leaf size=26 \[ 2+x+\frac {x}{\left (x+\frac {e^{-e} x}{\log (x)}\right )^2}+\frac {3 \log (x)}{2} \]

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Rubi [C]  time = 1.38, antiderivative size = 162, normalized size of antiderivative = 6.23, number of steps used = 15, number of rules used = 7, integrand size = 120, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6688, 12, 6742, 14, 2306, 2309, 2178} \begin {gather*} e^{e^{-e}-2 e} \left (1-2 e^e\right ) \text {Ei}\left (-e^{-e} \left (e^e \log (x)+1\right )\right )+2 e^{e^{-e}-e} \text {Ei}\left (-e^{-e} \left (e^e \log (x)+1\right )\right )-e^{e^{-e}-2 e} \text {Ei}\left (-e^{-e} \left (e^e \log (x)+1\right )\right )+x+\frac {1}{x}+\frac {3 \log (x)}{2}-\frac {2-e^{-e}}{x \left (e^e \log (x)+1\right )}-\frac {e^{-e}}{x \left (e^e \log (x)+1\right )}+\frac {1}{x \left (e^e \log (x)+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x + 2*x^2 + (4*E^(2*E) + E^E*(9*x + 6*x^2))*Log[x] + E^(2*E)*(-2 + 9*x + 6*x^2)*Log[x]^2 + E^(3*E)*(-2
+ 3*x + 2*x^2)*Log[x]^3)/(2*x^2 + 6*E^E*x^2*Log[x] + 6*E^(2*E)*x^2*Log[x]^2 + 2*E^(3*E)*x^2*Log[x]^3),x]

[Out]

x^(-1) + x - E^(-2*E + E^(-E))*ExpIntegralEi[-((1 + E^E*Log[x])/E^E)] + 2*E^(-E + E^(-E))*ExpIntegralEi[-((1 +
E^E*Log[x])/E^E)] + E^(-2*E + E^(-E))*(1 - 2*E^E)*ExpIntegralEi[-((1 + E^E*Log[x])/E^E)] + (3*Log[x])/2 + 1/(
x*(1 + E^E*Log[x])^2) - 1/(E^E*x*(1 + E^E*Log[x])) - (2 - E^(-E))/(x*(1 + E^E*Log[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 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 2306

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

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 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 (3+2 x)+e^e \left (4 e^e+3 x (3+2 x)\right ) \log (x)+e^{2 e} \left (-2+9 x+6 x^2\right ) \log ^2(x)+e^{3 e} \left (-2+3 x+2 x^2\right ) \log ^3(x)}{2 x^2 \left (1+e^e \log (x)\right )^3} \, dx\\ &=\frac {1}{2} \int \frac {x (3+2 x)+e^e \left (4 e^e+3 x (3+2 x)\right ) \log (x)+e^{2 e} \left (-2+9 x+6 x^2\right ) \log ^2(x)+e^{3 e} \left (-2+3 x+2 x^2\right ) \log ^3(x)}{x^2 \left (1+e^e \log (x)\right )^3} \, dx\\ &=\frac {1}{2} \int \left (\frac {-2+3 x+2 x^2}{x^2}-\frac {4 e^e}{x^2 \left (1+e^e \log (x)\right )^3}+\frac {2 \left (-1+2 e^e\right )}{x^2 \left (1+e^e \log (x)\right )^2}+\frac {4}{x^2 \left (1+e^e \log (x)\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {-2+3 x+2 x^2}{x^2} \, dx+2 \int \frac {1}{x^2 \left (1+e^e \log (x)\right )} \, dx-\left (2 e^e\right ) \int \frac {1}{x^2 \left (1+e^e \log (x)\right )^3} \, dx+\left (-1+2 e^e\right ) \int \frac {1}{x^2 \left (1+e^e \log (x)\right )^2} \, dx\\ &=\frac {1}{x \left (1+e^e \log (x)\right )^2}+\frac {e^{-e} \left (1-2 e^e\right )}{x \left (1+e^e \log (x)\right )}+\frac {1}{2} \int \left (2-\frac {2}{x^2}+\frac {3}{x}\right ) \, dx+2 \operatorname {Subst}\left (\int \frac {e^{-x}}{1+e^e x} \, dx,x,\log (x)\right )+\left (-2+e^{-e}\right ) \int \frac {1}{x^2 \left (1+e^e \log (x)\right )} \, dx+\int \frac {1}{x^2 \left (1+e^e \log (x)\right )^2} \, dx\\ &=\frac {1}{x}+x+2 e^{-e+e^{-e}} \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+\frac {3 \log (x)}{2}+\frac {1}{x \left (1+e^e \log (x)\right )^2}-\frac {e^{-e}}{x \left (1+e^e \log (x)\right )}+\frac {e^{-e} \left (1-2 e^e\right )}{x \left (1+e^e \log (x)\right )}-e^{-e} \int \frac {1}{x^2 \left (1+e^e \log (x)\right )} \, dx+\left (-2+e^{-e}\right ) \operatorname {Subst}\left (\int \frac {e^{-x}}{1+e^e x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}+x+2 e^{-e+e^{-e}} \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+e^{-2 e+e^{-e}} \left (1-2 e^e\right ) \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+\frac {3 \log (x)}{2}+\frac {1}{x \left (1+e^e \log (x)\right )^2}-\frac {e^{-e}}{x \left (1+e^e \log (x)\right )}+\frac {e^{-e} \left (1-2 e^e\right )}{x \left (1+e^e \log (x)\right )}-e^{-e} \operatorname {Subst}\left (\int \frac {e^{-x}}{1+e^e x} \, dx,x,\log (x)\right )\\ &=\frac {1}{x}+x-e^{-2 e+e^{-e}} \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+2 e^{-e+e^{-e}} \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+e^{-2 e+e^{-e}} \left (1-2 e^e\right ) \text {Ei}\left (-e^{-e} \left (1+e^e \log (x)\right )\right )+\frac {3 \log (x)}{2}+\frac {1}{x \left (1+e^e \log (x)\right )^2}-\frac {e^{-e}}{x \left (1+e^e \log (x)\right )}+\frac {e^{-e} \left (1-2 e^e\right )}{x \left (1+e^e \log (x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 42, normalized size = 1.62 \begin {gather*} \frac {1}{2} \left (3 \log (x)+2 \left (x-\frac {2}{x+e^e x \log (x)}+\frac {1+\frac {1}{\left (1+e^e \log (x)\right )^2}}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x + 2*x^2 + (4*E^(2*E) + E^E*(9*x + 6*x^2))*Log[x] + E^(2*E)*(-2 + 9*x + 6*x^2)*Log[x]^2 + E^(3*E
)*(-2 + 3*x + 2*x^2)*Log[x]^3)/(2*x^2 + 6*E^E*x^2*Log[x] + 6*E^(2*E)*x^2*Log[x]^2 + 2*E^(3*E)*x^2*Log[x]^3),x]

[Out]

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

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fricas [B]  time = 0.77, size = 82, normalized size = 3.15 \begin {gather*} \frac {3 \, x e^{\left (2 \, e\right )} \log \relax (x)^{3} + 2 \, {\left ({\left (x^{2} + 1\right )} e^{\left (2 \, e\right )} + 3 \, x e^{e}\right )} \log \relax (x)^{2} + 2 \, x^{2} + {\left (4 \, x^{2} e^{e} + 3 \, x\right )} \log \relax (x)}{2 \, {\left (x e^{\left (2 \, e\right )} \log \relax (x)^{2} + 2 \, x e^{e} \log \relax (x) + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3*x-2)*exp(exp(1))^3*log(x)^3+(6*x^2+9*x-2)*exp(exp(1))^2*log(x)^2+(4*exp(exp(1))^2+(6*x^2+9
*x)*exp(exp(1)))*log(x)+2*x^2+3*x)/(2*x^2*exp(exp(1))^3*log(x)^3+6*x^2*exp(exp(1))^2*log(x)^2+6*x^2*exp(exp(1)
)*log(x)+2*x^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.16, size = 93, normalized size = 3.58 \begin {gather*} \frac {2 \, x^{2} e^{\left (2 \, e\right )} \log \relax (x)^{2} + 3 \, x e^{\left (2 \, e\right )} \log \relax (x)^{3} + 4 \, x^{2} e^{e} \log \relax (x) + 6 \, x e^{e} \log \relax (x)^{2} + 2 \, e^{\left (2 \, e\right )} \log \relax (x)^{2} + 2 \, x^{2} + 3 \, x \log \relax (x)}{2 \, {\left (x e^{\left (2 \, e\right )} \log \relax (x)^{2} + 2 \, x e^{e} \log \relax (x) + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3*x-2)*exp(exp(1))^3*log(x)^3+(6*x^2+9*x-2)*exp(exp(1))^2*log(x)^2+(4*exp(exp(1))^2+(6*x^2+9
*x)*exp(exp(1)))*log(x)+2*x^2+3*x)/(2*x^2*exp(exp(1))^3*log(x)^3+6*x^2*exp(exp(1))^2*log(x)^2+6*x^2*exp(exp(1)
)*log(x)+2*x^2),x, algorithm="giac")

[Out]

1/2*(2*x^2*e^(2*e)*log(x)^2 + 3*x*e^(2*e)*log(x)^3 + 4*x^2*e^e*log(x) + 6*x*e^e*log(x)^2 + 2*e^(2*e)*log(x)^2
+ 2*x^2 + 3*x*log(x))/(x*e^(2*e)*log(x)^2 + 2*x*e^e*log(x) + x)

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maple [A]  time = 0.11, size = 43, normalized size = 1.65

method	result	size
risch	\(\frac {3 x \ln \relax (x )+2 x^{2}+2}{2 x}-\frac {2 \,{\mathrm e}^{{\mathrm e}} \ln \relax (x )+1}{x \left ({\mathrm e}^{{\mathrm e}} \ln \relax (x )+1\right )^{2}}\)	\(43\)
norman	\(\frac {x^{2}+{\mathrm e}^{2 \,{\mathrm e}} \ln \relax (x )^{2}+x^{2} {\mathrm e}^{2 \,{\mathrm e}} \ln \relax (x )^{2}+\frac {3 \,{\mathrm e}^{2 \,{\mathrm e}} x \ln \relax (x )^{3}}{2}+\frac {3 x \ln \relax (x )}{2}+3 \,{\mathrm e}^{{\mathrm e}} x \ln \relax (x )^{2}+2 x^{2} {\mathrm e}^{{\mathrm e}} \ln \relax (x )}{x \left ({\mathrm e}^{{\mathrm e}} \ln \relax (x )+1\right )^{2}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+3*x-2)*exp(exp(1))^3*ln(x)^3+(6*x^2+9*x-2)*exp(exp(1))^2*ln(x)^2+(4*exp(exp(1))^2+(6*x^2+9*x)*exp(
exp(1)))*ln(x)+2*x^2+3*x)/(2*x^2*exp(exp(1))^3*ln(x)^3+6*x^2*exp(exp(1))^2*ln(x)^2+6*x^2*exp(exp(1))*ln(x)+2*x
^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.40, size = 108, normalized size = 4.15 \begin {gather*} \frac {8 \, x^{2} e^{\left (2 \, e\right )} \log \relax (x) + 4 \, x^{2} e^{e} + 4 \, {\left (x^{2} e^{\left (3 \, e\right )} + e^{\left (3 \, e\right )}\right )} \log \relax (x)^{2} + 3 \, x}{4 \, {\left (x e^{\left (3 \, e\right )} \log \relax (x)^{2} + 2 \, x e^{\left (2 \, e\right )} \log \relax (x) + x e^{e}\right )}} - \frac {3}{4 \, {\left (e^{\left (3 \, e\right )} \log \relax (x)^{2} + 2 \, e^{\left (2 \, e\right )} \log \relax (x) + e^{e}\right )}} + \frac {3}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+3*x-2)*exp(exp(1))^3*log(x)^3+(6*x^2+9*x-2)*exp(exp(1))^2*log(x)^2+(4*exp(exp(1))^2+(6*x^2+9
*x)*exp(exp(1)))*log(x)+2*x^2+3*x)/(2*x^2*exp(exp(1))^3*log(x)^3+6*x^2*exp(exp(1))^2*log(x)^2+6*x^2*exp(exp(1)
)*log(x)+2*x^2),x, algorithm="maxima")

[Out]

1/4*(8*x^2*e^(2*e)*log(x) + 4*x^2*e^e + 4*(x^2*e^(3*e) + e^(3*e))*log(x)^2 + 3*x)/(x*e^(3*e)*log(x)^2 + 2*x*e^
(2*e)*log(x) + x*e^e) - 3/4/(e^(3*e)*log(x)^2 + 2*e^(2*e)*log(x) + e^e) + 3/2*log(x)

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mupad [B]  time = 3.55, size = 178, normalized size = 6.85 \begin {gather*} x+\frac {3\,\ln \relax (x)}{2}-\frac {\frac {{\mathrm {e}}^{-3\,\mathrm {e}}}{2\,x}+\frac {{\mathrm {e}}^{-\mathrm {e}}\,{\ln \relax (x)}^2}{x}+\frac {{\mathrm {e}}^{-2\,\mathrm {e}}\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^{\mathrm {e}}+3\right )}{2\,x}}{{\ln \relax (x)}^2+2\,{\mathrm {e}}^{-\mathrm {e}}\,\ln \relax (x)+{\mathrm {e}}^{-2\,\mathrm {e}}}-\frac {\frac {{\mathrm {e}}^{-3\,\mathrm {e}}\,\left (2\,{\mathrm {e}}^{2\,\mathrm {e}}+3\,{\mathrm {e}}^{\mathrm {e}}-1\right )}{2\,x}-\frac {{\mathrm {e}}^{-\mathrm {e}}\,{\ln \relax (x)}^2}{x}+\frac {{\mathrm {e}}^{-2\,\mathrm {e}}\,\ln \relax (x)\,\left (2\,{\mathrm {e}}^{\mathrm {e}}-3\right )}{2\,x}}{{\mathrm {e}}^{-\mathrm {e}}+\ln \relax (x)}-\frac {{\mathrm {e}}^{-\mathrm {e}}\,\ln \relax (x)}{x}+\frac {{\mathrm {e}}^{-2\,\mathrm {e}}\,\left (2\,{\mathrm {e}}^{2\,\mathrm {e}}+4\,{\mathrm {e}}^{\mathrm {e}}-1\right )}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + log(x)*(4*exp(2*exp(1)) + exp(exp(1))*(9*x + 6*x^2)) + 2*x^2 + exp(3*exp(1))*log(x)^3*(3*x + 2*x^2
- 2) + exp(2*exp(1))*log(x)^2*(9*x + 6*x^2 - 2))/(2*x^2 + 6*x^2*exp(2*exp(1))*log(x)^2 + 2*x^2*exp(3*exp(1))*l
og(x)^3 + 6*x^2*exp(exp(1))*log(x)),x)

[Out]

x + (3*log(x))/2 - (exp(-3*exp(1))/(2*x) + (exp(-exp(1))*log(x)^2)/x + (exp(-2*exp(1))*log(x)*(2*exp(exp(1)) +
3))/(2*x))/(exp(-2*exp(1)) + log(x)^2 + 2*exp(-exp(1))*log(x)) - ((exp(-3*exp(1))*(2*exp(2*exp(1)) + 3*exp(ex
p(1)) - 1))/(2*x) - (exp(-exp(1))*log(x)^2)/x + (exp(-2*exp(1))*log(x)*(2*exp(exp(1)) - 3))/(2*x))/(exp(-exp(1
)) + log(x)) - (exp(-exp(1))*log(x))/x + (exp(-2*exp(1))*(2*exp(2*exp(1)) + 4*exp(exp(1)) - 1))/(2*x)

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sympy [B]  time = 0.19, size = 51, normalized size = 1.96 \begin {gather*} x + \frac {- 2 e^{e} \log {\relax (x )} - 1}{x e^{2 e} \log {\relax (x )}^{2} + 2 x e^{e} \log {\relax (x )} + x} + \frac {3 \log {\relax (x )}}{2} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+3*x-2)*exp(exp(1))**3*ln(x)**3+(6*x**2+9*x-2)*exp(exp(1))**2*ln(x)**2+(4*exp(exp(1))**2+(6*
x**2+9*x)*exp(exp(1)))*ln(x)+2*x**2+3*x)/(2*x**2*exp(exp(1))**3*ln(x)**3+6*x**2*exp(exp(1))**2*ln(x)**2+6*x**2
*exp(exp(1))*ln(x)+2*x**2),x)

[Out]

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

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