3.44.68 \(\int \frac {e^x \log (5)+e^x \log (5) \log (x)+e^x (-1+x) \log (5) \log (x) \log (x \log (x))}{x^2 \log (x)} \, dx\) [4368]

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

[Out]

(9+exp(x)/x*ln(x*ln(x)))*ln(5)

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Rubi [A]
time = 0.87, 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 = {6820, 12, 6874, 2208, 2209, 2228, 2635} \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 2208

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] &&  !TrueQ[$UseGamm
a]

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 2228

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 2635

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

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 = 0.04, 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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.31, size = 99, normalized size = 5.82

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

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*ln(x))*csgn(I*x)*csgn(I*x*ln(x))+I*Pi*csgn(I*ln(x))*cs
gn(I*x*ln(x))^2+I*Pi*csgn(I*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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)*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, algorit
hm="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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Fricas [A]
time = 0.45, size = 13, normalized size = 0.76 \begin {gather*} \frac {e^{x} \log \left (5\right ) \log \left (x \log \left (x\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)*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, algorit
hm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)*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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Giac [A]
time = 0.43, size = 20, normalized size = 1.18 \begin {gather*} \frac {e^{x} \log \left (5\right ) \log \left (x\right ) + e^{x} \log \left (5\right ) \log \left (\log \left (x\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-1+x)*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, algorit
hm="giac")

[Out]

(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 \left (x\right )\right )\,{\mathrm {e}}^x\,\ln \left (5\right )}{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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