\(\int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx\) [10218]

Optimal result

Integrand size = 33, antiderivative size = 27 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=4 \left (5-e^{10-x}+\frac {x^2}{5 e^3 \log (x)}\right ) \]

[Out]

20+4/5*x^2/ln(x)/exp(3)-4*exp(10-x)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.11, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {12, 6874, 2225, 2343, 2346, 2209, 2413, 6617} \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {8 \operatorname {ExpIntegralEi}(2 \log (x))}{5 e^3}-\frac {8 (1-2 \log (x)) \operatorname {ExpIntegralEi}(2 \log (x))}{5 e^3}-\frac {16 \log (x) \operatorname {ExpIntegralEi}(2 \log (x))}{5 e^3}+\frac {8 x^2}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-4 e^{10-x} \]

[In]

Int[(-4*x + 8*x*Log[x] + 20*E^(13 - x)*Log[x]^2)/(5*E^3*Log[x]^2),x]

[Out]

-4*E^(10 - x) + (8*x^2)/(5*E^3) + (8*ExpIntegralEi[2*Log[x]])/(5*E^3) - (8*ExpIntegralEi[2*Log[x]]*(1 - 2*Log[
x]))/(5*E^3) + (4*x^2*(1 - 2*Log[x]))/(5*E^3*Log[x]) - (16*ExpIntegralEi[2*Log[x]]*Log[x])/(5*E^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2343

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 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 2413

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 6617

Int[ExpIntegralEi[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(a + b*x)*(ExpIntegralEi[a + b*x]/b), x] - Simp[E^(a
+ b*x)/b, x] /; FreeQ[{a, b}, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{\log ^2(x)} \, dx}{5 e^3} \\ & = \frac {\int \left (20 e^{13-x}+\frac {4 x (-1+2 \log (x))}{\log ^2(x)}\right ) \, dx}{5 e^3} \\ & = \frac {4 \int \frac {x (-1+2 \log (x))}{\log ^2(x)} \, dx}{5 e^3}+\frac {4 \int e^{13-x} \, dx}{e^3} \\ & = -4 e^{10-x}-\frac {8 \operatorname {ExpIntegralEi}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-\frac {8 \int \left (\frac {2 \operatorname {ExpIntegralEi}(2 \log (x))}{x}-\frac {x}{\log (x)}\right ) \, dx}{5 e^3} \\ & = -4 e^{10-x}-\frac {8 \operatorname {ExpIntegralEi}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}+\frac {8 \int \frac {x}{\log (x)} \, dx}{5 e^3}-\frac {16 \int \frac {\operatorname {ExpIntegralEi}(2 \log (x))}{x} \, dx}{5 e^3} \\ & = -4 e^{10-x}-\frac {8 \operatorname {ExpIntegralEi}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}+\frac {8 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )}{5 e^3}-\frac {16 \text {Subst}(\int \operatorname {ExpIntegralEi}(2 x) \, dx,x,\log (x))}{5 e^3} \\ & = -4 e^{10-x}+\frac {8 x^2}{5 e^3}+\frac {8 \operatorname {ExpIntegralEi}(2 \log (x))}{5 e^3}-\frac {8 \operatorname {ExpIntegralEi}(2 \log (x)) (1-2 \log (x))}{5 e^3}+\frac {4 x^2 (1-2 \log (x))}{5 e^3 \log (x)}-\frac {16 \operatorname {ExpIntegralEi}(2 \log (x)) \log (x)}{5 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {-20 e^{13-x}+\frac {4 x^2}{\log (x)}}{5 e^3} \]

[In]

Integrate[(-4*x + 8*x*Log[x] + 20*E^(13 - x)*Log[x]^2)/(5*E^3*Log[x]^2),x]

[Out]

(-20*E^(13 - x) + (4*x^2)/Log[x])/(5*E^3)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
risch	\(-4 \,{\mathrm e}^{10-x}+\frac {4 x^{2} {\mathrm e}^{-3}}{5 \ln \left (x \right )}\)	\(21\)
default	\(\frac {4 \,{\mathrm e}^{-3} \left (\frac {x^{2}}{\ln \left (x \right )}-5 \,{\mathrm e}^{3} {\mathrm e}^{10-x}\right )}{5}\)	\(26\)
norman	\(\frac {\frac {4 \,{\mathrm e}^{-3} x^{2}}{5}-4 \,{\mathrm e}^{10-x} \ln \left (x \right )}{\ln \left (x \right )}\)	\(26\)
parallelrisch	\(-\frac {{\mathrm e}^{-3} \left (20 \,{\mathrm e}^{3} \ln \left (x \right ) {\mathrm e}^{10-x}-4 x^{2}\right )}{5 \ln \left (x \right )}\)	\(29\)
parts	\(-\frac {4 \,{\mathrm e}^{-3} \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{5}-\frac {8 \,{\mathrm e}^{-3} \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )}{5}-4 \,{\mathrm e}^{10-x}\)	\(46\)

[In]

int(1/5*(20*exp(3)*exp(10-x)*ln(x)^2+8*x*ln(x)-4*x)/exp(3)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-4*exp(10-x)+4/5*x^2/ln(x)*exp(-3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {4 \, {\left (x^{2} - 5 \, e^{\left (-x + 13\right )} \log \left (x\right )\right )} e^{\left (-3\right )}}{5 \, \log \left (x\right )} \]

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="fricas")

[Out]

4/5*(x^2 - 5*e^(-x + 13)*log(x))*e^(-3)/log(x)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {4 x^{2}}{5 e^{3} \log {\left (x \right )}} - 4 e^{10 - x} \]

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*ln(x)**2+8*x*ln(x)-4*x)/exp(3)/ln(x)**2,x)

[Out]

4*x**2*exp(-3)/(5*log(x)) - 4*exp(10 - x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {4}{5} \, {\left (2 \, {\rm Ei}\left (2 \, \log \left (x\right )\right ) - 5 \, e^{\left (-x + 13\right )} - 2 \, \Gamma \left (-1, -2 \, \log \left (x\right )\right )\right )} e^{\left (-3\right )} \]

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="maxima")

[Out]

4/5*(2*Ei(2*log(x)) - 5*e^(-x + 13) - 2*gamma(-1, -2*log(x)))*e^(-3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {4 \, {\left (x^{2} - 5 \, e^{\left (-x + 13\right )} \log \left (x\right )\right )} e^{\left (-3\right )}}{5 \, \log \left (x\right )} \]

[In]

integrate(1/5*(20*exp(3)*exp(10-x)*log(x)^2+8*x*log(x)-4*x)/exp(3)/log(x)^2,x, algorithm="giac")

[Out]

4/5*(x^2 - 5*e^(-x + 13)*log(x))*e^(-3)/log(x)

Mupad [B] (verification not implemented)

Time = 15.60 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-4 x+8 x \log (x)+20 e^{13-x} \log ^2(x)}{5 e^3 \log ^2(x)} \, dx=\frac {4\,x^2\,{\mathrm {e}}^{-3}}{5\,\ln \left (x\right )}-4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{10} \]

[In]

int((exp(-3)*((8*x*log(x))/5 - (4*x)/5 + 4*exp(3)*exp(10 - x)*log(x)^2))/log(x)^2,x)

[Out]

(4*x^2*exp(-3))/(5*log(x)) - 4*exp(-x)*exp(10)