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

Optimal. Leaf size=20 \[ \frac {\left (e^x+\frac {2+\log (\log (2))}{x}\right ) \log (\log (x))}{x^2} \]

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Rubi [A]  time = 0.81, antiderivative size = 23, normalized size of antiderivative = 1.15, number of steps used = 11, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6741, 6742, 2309, 2178, 2522, 2288} \begin {gather*} \frac {(2+\log (\log (2))) \log (\log (x))}{x^3}+\frac {e^x \log (\log (x))}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 2522

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

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

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Mathematica [A]  time = 0.17, size = 17, normalized size = 0.85 \begin {gather*} \frac {\left (2+e^x x+\log (\log (2))\right ) \log (\log (x))}{x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.40, size = 16, normalized size = 0.80 \begin {gather*} \frac {{\left (x e^{x} + \log \left (\log \relax (2)\right ) + 2\right )} \log \left (\log \relax (x)\right )}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 24, normalized size = 1.20 \begin {gather*} \frac {x e^{x} \log \left (\log \relax (x)\right ) + \log \left (\log \relax (2)\right ) \log \left (\log \relax (x)\right ) + 2 \, \log \left (\log \relax (x)\right )}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 17, normalized size = 0.85

method	result	size
risch	\(\frac {\left ({\mathrm e}^{x} x +\ln \left (\ln \relax (2)\right )+2\right ) \ln \left (\ln \relax (x )\right )}{x^{3}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*ln(x)*ln(ln(2))+((x^2-2*x)*exp(x)-6)*ln(x))*ln(ln(x))+ln(ln(2))+exp(x)*x+2)/x^4/ln(x),x,method=_RETUR
NVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -{\left (\log \left (\log \relax (2)\right ) + 2\right )} \int \frac {1}{x^{4} \log \relax (x)}\,{d x} + {\rm Ei}\left (-3 \, \log \relax (x)\right ) \log \left (\log \relax (2)\right ) + \frac {{\left (x e^{x} + \log \left (\log \relax (2)\right ) + 2\right )} \log \left (\log \relax (x)\right )}{x^{3}} + 2 \, {\rm Ei}\left (-3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(log(2)) + 2)*integrate(1/(x^4*log(x)), x) + Ei(-3*log(x))*log(log(2)) + (x*e^x + log(log(2)) + 2)*log(lo
g(x))/x^3 + 2*Ei(-3*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {\ln \left (\ln \relax (2)\right )-\ln \left (\ln \relax (x)\right )\,\left (3\,\ln \left (\ln \relax (2)\right )\,\ln \relax (x)+\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (2\,x-x^2\right )+6\right )\right )+x\,{\mathrm {e}}^x+2}{x^4\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.46, size = 26, normalized size = 1.30 \begin {gather*} \frac {e^{x} \log {\left (\log {\relax (x )} \right )}}{x^{2}} + \frac {\left (\log {\left (\log {\relax (2 )} \right )} + 2\right ) \log {\left (\log {\relax (x )} \right )}}{x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*log(log(x))/x**2 + (log(log(2)) + 2)*log(log(x))/x**3

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