3.4.24 \(\int \frac {e^{-x} (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} (-20 \log (2)+e^x \log (\log (x)))} (e^x+20 x \log (2) \log (x)))}{5 x \log (x)} \, dx\)

Optimal. Leaf size=19 \[ 2^{-4 e^{-x}} \sqrt [5]{\log (x)}+\log (x) \]

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Rubi [A]  time = 1.38, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {12, 6742, 2288} \begin {gather*} \log (x)+2^{-4 e^{-x}} \sqrt [5]{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5*E^x*Log[x] + E^((-20*Log[2] + E^x*Log[Log[x]])/(5*E^x))*(E^x + 20*x*Log[2]*Log[x]))/(5*E^x*x*Log[x]),x]

[Out]

Log[x]^(1/5)/2^(4/E^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 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} &=\frac {1}{5} \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{x \log (x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {5}{x}+\frac {2^{-4 e^{-x}} e^{-x} \left (e^x+20 x \log (2) \log (x)\right )}{x \log ^{\frac {4}{5}}(x)}\right ) \, dx\\ &=\log (x)+\frac {1}{5} \int \frac {2^{-4 e^{-x}} e^{-x} \left (e^x+20 x \log (2) \log (x)\right )}{x \log ^{\frac {4}{5}}(x)} \, dx\\ &=2^{-4 e^{-x}} \sqrt [5]{\log (x)}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 33, normalized size = 1.74 \begin {gather*} \frac {1}{5} \left (\frac {2^{1-4 e^{-x}} \log (32) \sqrt [5]{\log (x)}}{\log (4)}+5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^x*Log[x] + E^((-20*Log[2] + E^x*Log[Log[x]])/(5*E^x))*(E^x + 20*x*Log[2]*Log[x]))/(5*E^x*x*Log[
x]),x]

[Out]

((2^(1 - 4/E^x)*Log[32]*Log[x]^(1/5))/Log[4] + 5*Log[x])/5

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fricas [A]  time = 0.78, size = 21, normalized size = 1.11 \begin {gather*} e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (\log \relax (x)\right ) - 20 \, \log \relax (2)\right )} e^{\left (-x\right )}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20*log(2))/exp(x))+5*exp(x)*log(x))/x/e
xp(x)/log(x),x, algorithm="fricas")

[Out]

e^(1/5*(e^x*log(log(x)) - 20*log(2))*e^(-x)) + log(x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (20 \, x \log \relax (2) \log \relax (x) + e^{x}\right )} e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (\log \relax (x)\right ) - 20 \, \log \relax (2)\right )} e^{\left (-x\right )}\right )} + 5 \, e^{x} \log \relax (x)\right )} e^{\left (-x\right )}}{5 \, x \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20*log(2))/exp(x))+5*exp(x)*log(x))/x/e
xp(x)/log(x),x, algorithm="giac")

[Out]

integrate(1/5*((20*x*log(2)*log(x) + e^x)*e^(1/5*(e^x*log(log(x)) - 20*log(2))*e^(-x)) + 5*e^x*log(x))*e^(-x)/
(x*log(x)), x)

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maple [A]  time = 0.07, size = 15, normalized size = 0.79

method	result	size
risch	\(\ln \relax (x )+\ln \relax (x )^{\frac {1}{5}} \left (\frac {1}{16}\right )^{{\mathrm e}^{-x}}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((20*x*ln(2)*ln(x)+exp(x))*exp(1/5*(exp(x)*ln(ln(x))-20*ln(2))/exp(x))+5*exp(x)*ln(x))/x/exp(x)/ln(x),
x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(x)^(1/5)*(1/16)^exp(-x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{5} \, \int \frac {{\left (20 \, x \log \relax (2) \log \relax (x) + e^{x}\right )} e^{\left (-4 \, e^{\left (-x\right )} \log \relax (2) - x\right )}}{x \log \relax (x)^{\frac {4}{5}}}\,{d x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20*log(2))/exp(x))+5*exp(x)*log(x))/x/e
xp(x)/log(x),x, algorithm="maxima")

[Out]

1/5*integrate((20*x*log(2)*log(x) + e^x)*e^(-4*e^(-x)*log(2) - x)/(x*log(x)^(4/5)), x) + log(x)

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mupad [B]  time = 0.50, size = 18, normalized size = 0.95 \begin {gather*} \ln \relax (x)+\frac {{\ln \relax (x)}^{1/5}}{2^{4\,{\mathrm {e}}^{-x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x)*log(x) + (exp(-exp(-x)*(4*log(2) - (log(log(x))*exp(x))/5))*(exp(x) + 20*x*log(2)*log(x))
)/5))/(x*log(x)),x)

[Out]

log(x) + log(x)^(1/5)/2^(4*exp(-x))

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sympy [A]  time = 0.65, size = 22, normalized size = 1.16 \begin {gather*} e^{\left (\frac {e^{x} \log {\left (\log {\relax (x )} \right )}}{5} - 4 \log {\relax (2 )}\right ) e^{- x}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((20*x*ln(2)*ln(x)+exp(x))*exp(1/5*(exp(x)*ln(ln(x))-20*ln(2))/exp(x))+5*exp(x)*ln(x))/x/exp(x)/
ln(x),x)

[Out]

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

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