3.44.68 \(\int \frac {e^x \log (5)+e^x \log (5) \log (x)+e^x (-1+x) \log (5) \log (x) \log (x \log (x))}{x^2 \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \log (5) \left (9+\frac {e^x \log (x \log (x))}{x}\right ) \]

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Rubi [A]  time = 0.81, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 13, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {6688, 12, 6742, 2177, 2178, 2197, 2555} \begin {gather*} \frac {e^x \log (5) \log (x \log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*Log[5] + E^x*Log[5]*Log[x] + E^x*(-1 + x)*Log[5]*Log[x]*Log[x*Log[x]])/(x^2*Log[x]),x]

[Out]

(E^x*Log[5]*Log[x*Log[x]])/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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

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

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Mathematica [A]  time = 0.11, size = 14, normalized size = 0.82 \begin {gather*} \frac {e^x \log (5) \log (x \log (x))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*Log[5] + E^x*Log[5]*Log[x] + E^x*(-1 + x)*Log[5]*Log[x]*Log[x*Log[x]])/(x^2*Log[x]),x]

[Out]

(E^x*Log[5]*Log[x*Log[x]])/x

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fricas [A]  time = 0.68, size = 13, normalized size = 0.76 \begin {gather*} \frac {e^{x} \log \relax (5) \log \left (x \log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*log(5)*exp(x)*log(x)*log(x*log(x))+log(5)*exp(x)*log(x)+exp(x)*log(5))/x^2/log(x),x, algorith
m="fricas")

[Out]

e^x*log(5)*log(x*log(x))/x

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giac [A]  time = 0.20, size = 20, normalized size = 1.18 \begin {gather*} \frac {e^{x} \log \relax (5) \log \relax (x) + e^{x} \log \relax (5) \log \left (\log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*log(5)*exp(x)*log(x)*log(x*log(x))+log(5)*exp(x)*log(x)+exp(x)*log(5))/x^2/log(x),x, algorith
m="giac")

[Out]

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

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maple [C]  time = 0.08, size = 99, normalized size = 5.82

method	result	size
risch	\(\frac {\ln \relax (5) {\mathrm e}^{x} \ln \left (\ln \relax (x )\right )}{x}+\frac {{\mathrm e}^{x} \ln \relax (5) \left (-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i x \ln \relax (x )\right )^{2}-i \pi \mathrm {csgn}\left (i x \ln \relax (x )\right )^{3}+2 \ln \relax (x )\right )}{2 x}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(5)/x*exp(x)*ln(ln(x))+1/2*exp(x)*ln(5)*(-I*Pi*csgn(I*x)*csgn(I*ln(x))*csgn(I*x*ln(x))+I*Pi*csgn(I*x)*csgn(I
*x*ln(x))^2+I*Pi*csgn(I*ln(x))*csgn(I*x*ln(x))^2-I*Pi*csgn(I*x*ln(x))^3+2*ln(x))/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \Gamma \left (-1, -x\right ) \log \relax (5) - \int \frac {e^{x}}{x^{2}}\,{d x} \log \relax (5) + \frac {e^{x} \log \relax (5) \log \relax (x) + e^{x} \log \relax (5) \log \left (\log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-1)*log(5)*exp(x)*log(x)*log(x*log(x))+log(5)*exp(x)*log(x)+exp(x)*log(5))/x^2/log(x),x, algorith
m="maxima")

[Out]

gamma(-1, -x)*log(5) - integrate(e^x/x^2, x)*log(5) + (e^x*log(5)*log(x) + e^x*log(5)*log(log(x)))/x

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mupad [B]  time = 3.44, size = 13, normalized size = 0.76 \begin {gather*} \frac {\ln \left (x\,\ln \relax (x)\right )\,{\mathrm {e}}^x\,\ln \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(x*log(x))*exp(x)*log(5))/x

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sympy [A]  time = 0.34, size = 14, normalized size = 0.82 \begin {gather*} \frac {e^{x} \log {\relax (5 )} \log {\left (x \log {\relax (x )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)*log(5)*log(x*log(x))/x

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