\(\int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx\) [4221]

Optimal result

Integrand size = 23, antiderivative size = 22 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=4+\frac {5}{2} \left (-9+\frac {e+x}{256 x \log (x)}\right ) \]

[Out]

-37/2+5/512*(x+exp(1))/x/ln(x)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {12, 6873, 6874, 2395, 2343, 2346, 2209, 2339, 30} \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5 e}{512 x \log (x)}+\frac {5}{512 \log (x)} \]

[In]

Int[(-5*E - 5*x - 5*E*Log[x])/(512*x^2*Log[x]^2),x]

[Out]

5/(512*Log[x]) + (5*E)/(512*x*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && 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}& = \frac {1}{512} \int \frac {-5 e-5 x-5 e \log (x)}{x^2 \log ^2(x)} \, dx \\ & = \frac {1}{512} \int \frac {5 (-e-x-e \log (x))}{x^2 \log ^2(x)} \, dx \\ & = \frac {5}{512} \int \frac {-e-x-e \log (x)}{x^2 \log ^2(x)} \, dx \\ & = \frac {5}{512} \int \left (\frac {-e-x}{x^2 \log ^2(x)}-\frac {e}{x^2 \log (x)}\right ) \, dx \\ & = \frac {5}{512} \int \frac {-e-x}{x^2 \log ^2(x)} \, dx-\frac {1}{512} (5 e) \int \frac {1}{x^2 \log (x)} \, dx \\ & = \frac {5}{512} \int \left (-\frac {e}{x^2 \log ^2(x)}-\frac {1}{x \log ^2(x)}\right ) \, dx-\frac {1}{512} (5 e) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right ) \\ & = -\frac {5}{512} e \text {Ei}(-\log (x))-\frac {5}{512} \int \frac {1}{x \log ^2(x)} \, dx-\frac {1}{512} (5 e) \int \frac {1}{x^2 \log ^2(x)} \, dx \\ & = -\frac {5}{512} e \text {Ei}(-\log (x))+\frac {5 e}{512 x \log (x)}-\frac {5}{512} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )+\frac {1}{512} (5 e) \int \frac {1}{x^2 \log (x)} \, dx \\ & = -\frac {5}{512} e \text {Ei}(-\log (x))+\frac {5}{512 \log (x)}+\frac {5 e}{512 x \log (x)}+\frac {1}{512} (5 e) \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log (x)\right ) \\ & = \frac {5}{512 \log (x)}+\frac {5 e}{512 x \log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5 (e+x)}{512 x \log (x)} \]

[In]

Integrate[(-5*E - 5*x - 5*E*Log[x])/(512*x^2*Log[x]^2),x]

[Out]

(5*(E + x))/(512*x*Log[x])

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {\frac {5 x}{512}+\frac {5 \,{\mathrm e}}{512}}{x \ln \left (x \right )}\)	\(14\)
norman	\(\frac {\frac {5 x}{512}+\frac {5 \,{\mathrm e}}{512}}{x \ln \left (x \right )}\)	\(17\)
parallelrisch	\(\frac {5 \,{\mathrm e}+5 x}{512 x \ln \left (x \right )}\)	\(18\)
default	\(\frac {5 \,{\mathrm e} \,\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )}{512}-\frac {5 \,{\mathrm e} \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )\right )}{512}+\frac {5}{512 \ln \left (x \right )}\)	\(34\)
parts	\(\frac {5 \,{\mathrm e} \,\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )}{512}-\frac {5 \,{\mathrm e} \left (-\frac {1}{x \ln \left (x \right )}+\operatorname {Ei}_{1}\left (\ln \left (x \right )\right )\right )}{512}+\frac {5}{512 \ln \left (x \right )}\)	\(34\)

[In]

int(1/512*(-5*exp(1)*ln(x)-5*exp(1)-5*x)/x^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

5/512*(x+exp(1))/x/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5 \, {\left (x + e\right )}}{512 \, x \log \left (x\right )} \]

[In]

integrate(1/512*(-5*exp(1)*log(x)-5*exp(1)-5*x)/x^2/log(x)^2,x, algorithm="fricas")

[Out]

5/512*(x + e)/(x*log(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5 x + 5 e}{512 x \log {\left (x \right )}} \]

[In]

integrate(1/512*(-5*exp(1)*ln(x)-5*exp(1)-5*x)/x**2/ln(x)**2,x)

[Out]

(5*x + 5*E)/(512*x*log(x))

Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=-\frac {5}{512} \, {\rm Ei}\left (-\log \left (x\right )\right ) e + \frac {5}{512} \, e \Gamma \left (-1, \log \left (x\right )\right ) + \frac {5}{512 \, \log \left (x\right )} \]

[In]

integrate(1/512*(-5*exp(1)*log(x)-5*exp(1)-5*x)/x^2/log(x)^2,x, algorithm="maxima")

[Out]

-5/512*Ei(-log(x))*e + 5/512*e*gamma(-1, log(x)) + 5/512/log(x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.59 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5 \, {\left (x + e\right )}}{512 \, x \log \left (x\right )} \]

[In]

integrate(1/512*(-5*exp(1)*log(x)-5*exp(1)-5*x)/x^2/log(x)^2,x, algorithm="giac")

[Out]

5/512*(x + e)/(x*log(x))

Mupad [B] (verification not implemented)

Time = 10.59 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-5 e-5 x-5 e \log (x)}{512 x^2 \log ^2(x)} \, dx=\frac {5\,x^2+5\,\mathrm {e}\,x}{512\,x^2\,\ln \left (x\right )} \]

[In]

int(-((5*x)/512 + (5*exp(1))/512 + (5*exp(1)*log(x))/512)/(x^2*log(x)^2),x)

[Out]

(5*x*exp(1) + 5*x^2)/(512*x^2*log(x))