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

Optimal. Leaf size=22 \[ 4+\frac {5}{2} \left (-9+\frac {e+x}{256 x \log (x)}\right ) \]

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Rubi [A]  time = 0.28, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {12, 6741, 6742, 2353, 2306, 2309, 2178, 2302, 30} \begin {gather*} \frac {5 e}{512 x \log (x)}+\frac {5}{512 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[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 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 2302

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 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 2353

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 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} &=\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) \operatorname {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} \operatorname {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) \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 14, normalized size = 0.64 \begin {gather*} \frac {5 (e+x)}{512 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.55, size = 13, normalized size = 0.59 \begin {gather*} \frac {5 \, {\left (x + e\right )}}{512 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.30, size = 13, normalized size = 0.59 \begin {gather*} \frac {5 \, {\left (x + e\right )}}{512 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.02, size = 14, normalized size = 0.64

method	result	size
risch	\(\frac {\frac {5 x}{512}+\frac {5 \,{\mathrm e}}{512}}{x \ln \relax (x )}\)	\(14\)
norman	\(\frac {\frac {5 x}{512}+\frac {5 \,{\mathrm e}}{512}}{x \ln \relax (x )}\)	\(17\)
default	\(\frac {5 \,{\mathrm e} \expIntegralEi \left (1, \ln \relax (x )\right )}{512}-\frac {5 \,{\mathrm e} \left (-\frac {1}{x \ln \relax (x )}+\expIntegralEi \left (1, \ln \relax (x )\right )\right )}{512}+\frac {5}{512 \ln \relax (x )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [C]  time = 0.36, size = 24, normalized size = 1.09 \begin {gather*} -\frac {5}{512} \, {\rm Ei}\left (-\log \relax (x)\right ) e + \frac {5}{512} \, e \Gamma \left (-1, \log \relax (x)\right ) + \frac {5}{512 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 2.95, size = 20, normalized size = 0.91 \begin {gather*} \frac {5\,x^2+5\,\mathrm {e}\,x}{512\,x^2\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.09, size = 14, normalized size = 0.64 \begin {gather*} \frac {5 x + 5 e}{512 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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