3.91.97 \(\int \frac {e^{-x} (2+(-1+e^x) \log (x)-x \log ^2(x)-x \log (x) \log (\frac {100 \log ^2(x)}{x^2}))}{x \log (x)} \, dx\)

Optimal. Leaf size=22 \[ \log (x)+e^{-x} \left (\log (x)+\log \left (\frac {100 \log ^2(x)}{x^2}\right )\right ) \]

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Rubi [A]  time = 0.57, antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 3, number of rules used = 2, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {6742, 2288} \begin {gather*} \frac {e^{-x} \left (x \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \log (x)+x \log ^2(x)\right )}{x \log (x)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + (-1 + E^x)*Log[x] - x*Log[x]^2 - x*Log[x]*Log[(100*Log[x]^2)/x^2])/(E^x*x*Log[x]),x]

[Out]

Log[x] + (x*Log[x]^2 + x*Log[x]*Log[(100*Log[x]^2)/x^2])/(E^x*x*Log[x])

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{x}+\frac {e^{-x} \left (2-\log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)}\right ) \, dx\\ &=\log (x)+\int \frac {e^{-x} \left (2-\log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx\\ &=\log (x)+\frac {e^{-x} \left (x \log ^2(x)+x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 27, normalized size = 1.23 \begin {gather*} \log (x)+e^{-x} \log (x)+e^{-x} \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + (-1 + E^x)*Log[x] - x*Log[x]^2 - x*Log[x]*Log[(100*Log[x]^2)/x^2])/(E^x*x*Log[x]),x]

[Out]

Log[x] + Log[x]/E^x + Log[(100*Log[x]^2)/x^2]/E^x

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fricas [A]  time = 0.49, size = 23, normalized size = 1.05 \begin {gather*} {\left ({\left (e^{x} + 1\right )} \log \relax (x) + \log \left (\frac {100 \, \log \relax (x)^{2}}{x^{2}}\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(exp(x)-1)*log(x)+2)/x/exp(x)/log(x),x, algorithm="frica
s")

[Out]

((e^x + 1)*log(x) + log(100*log(x)^2/x^2))*e^(-x)

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giac [A]  time = 0.18, size = 23, normalized size = 1.05 \begin {gather*} e^{\left (-x\right )} \log \left (100 \, \log \relax (x)^{2}\right ) - e^{\left (-x\right )} \log \relax (x) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(exp(x)-1)*log(x)+2)/x/exp(x)/log(x),x, algorithm="giac"
)

[Out]

e^(-x)*log(100*log(x)^2) - e^(-x)*log(x) + log(x)

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maple [C]  time = 0.53, size = 235, normalized size = 10.68

method	result	size
risch	\(2 \,{\mathrm e}^{-x} \ln \left (\ln \relax (x )\right )+\frac {\left (-i \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (\frac {i}{x^{2}}\right )-i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}+2 i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-i \pi \,\mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+2 \,{\mathrm e}^{x} \ln \relax (x )+4 \ln \relax (2)+4 \ln \relax (5)-2 \ln \relax (x )\right ) {\mathrm e}^{-x}}{2}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(x)*ln(100*ln(x)^2/x^2)-x*ln(x)^2+(exp(x)-1)*ln(x)+2)/x/exp(x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

2*exp(-x)*ln(ln(x))+1/2*(-I*Pi*csgn(I/x^2*ln(x)^2)^3+I*Pi*csgn(I/x^2*ln(x)^2)^2*csgn(I/x^2)-I*Pi*csgn(I*ln(x)^
2)^3+2*I*Pi*csgn(I*ln(x))*csgn(I*ln(x)^2)^2+I*Pi*csgn(I*x^2)*csgn(I*x)^2+I*Pi*csgn(I*x^2)^3+I*Pi*csgn(I/x^2*ln
(x)^2)^2*csgn(I*ln(x)^2)-I*Pi*csgn(I/x^2*ln(x)^2)*csgn(I/x^2)*csgn(I*ln(x)^2)-I*Pi*csgn(I*ln(x))^2*csgn(I*ln(x
)^2)-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+2*exp(x)*ln(x)+4*ln(2)+4*ln(5)-2*ln(x))*exp(-x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(exp(x)-1)*log(x)+2)/x/exp(x)/log(x),x, algorithm="maxim
a")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^{-x}\,\left (x\,{\ln \relax (x)}^2-\ln \relax (x)\,\left ({\mathrm {e}}^x-1\right )+x\,\ln \relax (x)\,\ln \left (\frac {100\,{\ln \relax (x)}^2}{x^2}\right )-2\right )}{x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(x*log(x)^2 - log(x)*(exp(x) - 1) + x*log(x)*log((100*log(x)^2)/x^2) - 2))/(x*log(x)),x)

[Out]

int(-(exp(-x)*(x*log(x)^2 - log(x)*(exp(x) - 1) + x*log(x)*log((100*log(x)^2)/x^2) - 2))/(x*log(x)), x)

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sympy [A]  time = 13.09, size = 20, normalized size = 0.91 \begin {gather*} \left (\log {\relax (x )} + \log {\left (\frac {100 \log {\relax (x )}^{2}}{x^{2}} \right )}\right ) e^{- x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)*ln(100*ln(x)**2/x**2)-x*ln(x)**2+(exp(x)-1)*ln(x)+2)/x/exp(x)/ln(x),x)

[Out]

(log(x) + log(100*log(x)**2/x**2))*exp(-x) + log(x)

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