3.18.41 \(\int \frac {-x^2+\log (x) (8+x^2-4 \log (\frac {x^2}{\log ^2(5)}))-4 \log (\frac {x^2}{\log ^2(5)})}{e^5 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=26 \[ \frac {x^2+4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)} \]

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Rubi [A]  time = 1.03, antiderivative size = 31, normalized size of antiderivative = 1.19, number of steps used = 25, number of rules used = 11, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.234, Rules used = {12, 6742, 6688, 2297, 2298, 2353, 2309, 2178, 2306, 2366, 6482} \begin {gather*} \frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}+\frac {x}{e^5 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(E^5*Log[x]) + (4*Log[x^2/Log[5]^2])/(E^5*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 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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 2366

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 6482

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 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} &=\frac {\int \frac {-x^2+\log (x) \left (8+x^2-4 \log \left (\frac {x^2}{\log ^2(5)}\right )\right )-4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)} \, dx}{e^5}\\ &=\frac {\int \left (\frac {-x^2+8 \log (x)+x^2 \log (x)}{x^2 \log ^2(x)}-\frac {4 (1+\log (x)) \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)}\right ) \, dx}{e^5}\\ &=\frac {\int \frac {-x^2+8 \log (x)+x^2 \log (x)}{x^2 \log ^2(x)} \, dx}{e^5}-\frac {4 \int \frac {(1+\log (x)) \log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)} \, dx}{e^5}\\ &=\frac {\int \frac {-1+\log (x)+\frac {8 \log (x)}{x^2}}{\log ^2(x)} \, dx}{e^5}-\frac {4 \int \left (\frac {\log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)}+\frac {\log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log (x)}\right ) \, dx}{e^5}\\ &=\frac {\int \left (-\frac {1}{\log ^2(x)}+\frac {8+x^2}{x^2 \log (x)}\right ) \, dx}{e^5}-\frac {4 \int \frac {\log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log ^2(x)} \, dx}{e^5}-\frac {4 \int \frac {\log \left (\frac {x^2}{\log ^2(5)}\right )}{x^2 \log (x)} \, dx}{e^5}\\ &=\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}-\frac {\int \frac {1}{\log ^2(x)} \, dx}{e^5}+\frac {\int \frac {8+x^2}{x^2 \log (x)} \, dx}{e^5}+\frac {8 \int \frac {\text {Ei}(-\log (x))}{x} \, dx}{e^5}+\frac {8 \int \frac {-1-x \text {Ei}(-\log (x)) \log (x)}{x^2 \log (x)} \, dx}{e^5}\\ &=\frac {x}{e^5 \log (x)}+\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}+\frac {\int \left (\frac {1}{\log (x)}+\frac {8}{x^2 \log (x)}\right ) \, dx}{e^5}-\frac {\int \frac {1}{\log (x)} \, dx}{e^5}+\frac {8 \int \left (-\frac {\text {Ei}(-\log (x))}{x}-\frac {1}{x^2 \log (x)}\right ) \, dx}{e^5}+\frac {8 \operatorname {Subst}(\int \text {Ei}(-x) \, dx,x,\log (x))}{e^5}\\ &=\frac {8}{e^5 x}+\frac {x}{e^5 \log (x)}+\frac {8 \text {Ei}(-\log (x)) \log (x)}{e^5}+\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}-\frac {\text {li}(x)}{e^5}+\frac {\int \frac {1}{\log (x)} \, dx}{e^5}-\frac {8 \int \frac {\text {Ei}(-\log (x))}{x} \, dx}{e^5}\\ &=\frac {8}{e^5 x}+\frac {x}{e^5 \log (x)}+\frac {8 \text {Ei}(-\log (x)) \log (x)}{e^5}+\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}-\frac {8 \operatorname {Subst}(\int \text {Ei}(-x) \, dx,x,\log (x))}{e^5}\\ &=\frac {x}{e^5 \log (x)}+\frac {4 \log \left (\frac {x^2}{\log ^2(5)}\right )}{e^5 x \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 26, normalized size = 1.00 \begin {gather*} \frac {x^2+4 \log \left (x^2\right )-8 \log (\log (5))}{e^5 x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2 + 4*Log[x^2] - 8*Log[Log[5]])/(E^5*x*Log[x])

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fricas [A]  time = 0.89, size = 42, normalized size = 1.62 \begin {gather*} \frac {2 \, {\left (x^{2} + 4 \, \log \left (\frac {x^{2}}{\log \relax (5)^{2}}\right )\right )}}{x e^{5} \log \left (\log \relax (5)^{2}\right ) + x e^{5} \log \left (\frac {x^{2}}{\log \relax (5)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^2/exp(5)/log(x)^2,x, algorithm="fric
as")

[Out]

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

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giac [A]  time = 0.25, size = 26, normalized size = 1.00 \begin {gather*} {\left (\frac {8}{x} + \frac {x^{2} - 8 \, \log \left (\log \relax (5)\right )}{x \log \relax (x)}\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8/x + (x^2 - 8*log(log(5)))/(x*log(x)))*e^(-5)

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maple [C]  time = 0.07, size = 80, normalized size = 3.08

method	result	size
risch	\(\frac {8 \,{\mathrm e}^{-5}}{x}-\frac {{\mathrm e}^{-5} \left (2 i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}-4 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-x^{2}+8 \ln \left (\ln \relax (5)\right )\right )}{x \ln \relax (x )}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/x*exp(-5)-exp(-5)/x*(2*I*Pi*csgn(I*x)^2*csgn(I*x^2)-4*I*Pi*csgn(I*x)*csgn(I*x^2)^2+2*I*Pi*csgn(I*x^2)^3-x^2+
8*ln(ln(5)))/ln(x)

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maxima [C]  time = 0.49, size = 68, normalized size = 2.62 \begin {gather*} {\left (8 \, {\rm Ei}\left (-\log \relax (x)\right ) \log \relax (x) - 8 \, \Gamma \left (-1, \log \relax (x)\right ) \log \relax (x) - 4 \, {\rm Ei}\left (-\log \relax (x)\right ) \log \left (\frac {x^{2}}{\log \relax (5)^{2}}\right ) + 4 \, \Gamma \left (-1, \log \relax (x)\right ) \log \left (\frac {x^{2}}{\log \relax (5)^{2}}\right ) + \frac {8}{x} + {\rm Ei}\left (\log \relax (x)\right ) - \Gamma \left (-1, -\log \relax (x)\right )\right )} e^{\left (-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(x^2/log(5)^2)+x^2+8)*log(x)-4*log(x^2/log(5)^2)-x^2)/x^2/exp(5)/log(x)^2,x, algorithm="maxi
ma")

[Out]

(8*Ei(-log(x))*log(x) - 8*gamma(-1, log(x))*log(x) - 4*Ei(-log(x))*log(x^2/log(5)^2) + 4*gamma(-1, log(x))*log
(x^2/log(5)^2) + 8/x + Ei(log(x)) - gamma(-1, -log(x)))*e^(-5)

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mupad [B]  time = 1.15, size = 25, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {e}}^{-5}\,\left (4\,\ln \left (x^2\right )-8\,\ln \left (\ln \relax (5)\right )+x^2\right )}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(-5)*(4*log(x^2) - 8*log(log(5)) + x^2))/(x*log(x))

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sympy [A]  time = 0.25, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^{2} - 8 \log {\left (\log {\relax (5 )} \right )}}{x e^{5} \log {\relax (x )}} + \frac {8}{x e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*ln(x**2/ln(5)**2)+x**2+8)*ln(x)-4*ln(x**2/ln(5)**2)-x**2)/x**2/exp(5)/ln(x)**2,x)

[Out]

(x**2 - 8*log(log(5)))*exp(-5)/(x*log(x)) + 8*exp(-5)/x

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