3.104.33 \(\int (1+\log (\frac {x+e x (i \pi +\log (5-\log (5)))}{e})) \, dx\)

Optimal. Leaf size=23 \[ 4+x \log \left (x \left (\frac {1}{e}+i \pi +\log (5-\log (5))\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2444, 2295} \begin {gather*} x \log \left (\frac {x (1+e (\log (5-\log (5))+i \pi ))}{e}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1 + Log[(x + E*x*(I*Pi + Log[5 - Log[5]]))/E],x]

[Out]

x*Log[(x*(1 + E*(I*Pi + Log[5 - Log[5]])))/E]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x+\int \log \left (\frac {x+e x (i \pi +\log (5-\log (5)))}{e}\right ) \, dx\\ &=x+\int \log \left (\frac {x (1+e (i \pi +\log (5-\log (5))))}{e}\right ) \, dx\\ &=x \log \left (\frac {x (1+e (i \pi +\log (5-\log (5))))}{e}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.09 \begin {gather*} -x+x \log (x+e x (i \pi +\log (5-\log (5)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1 + Log[(x + E*x*(I*Pi + Log[5 - Log[5]]))/E],x]

[Out]

-x + x*Log[x + E*x*(I*Pi + Log[5 - Log[5]])]

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fricas [A]  time = 0.57, size = 17, normalized size = 0.74 \begin {gather*} x \log \left ({\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log((x*e*log(log(5) - 5) + x)*e^(-1))

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giac [B]  time = 0.17, size = 62, normalized size = 2.70 \begin {gather*} x + \frac {{\left ({\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )} \log \left ({\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )}\right ) - {\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )}\right )} e}{e \log \left (\log \relax (5) - 5\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 18, normalized size = 0.78

method	result	size
risch	\(x \ln \left (x \left ({\mathrm e} \ln \left (\ln \relax (5)-5\right )+1\right ) {\mathrm e}^{-1}\right )\)	\(18\)
derivativedivides	\(x \ln \left (x \left ({\mathrm e} \ln \left (\ln \relax (5)-5\right )+1\right ) {\mathrm e}^{-1}\right )\)	\(20\)
norman	\(x \ln \left (\left (x \,{\mathrm e} \ln \left (\ln \relax (5)-5\right )+x \right ) {\mathrm e}^{-1}\right )\)	\(20\)
default	\(x +\frac {{\mathrm e} x \ln \left (x \left ({\mathrm e} \ln \left (\ln \relax (5)-5\right )+1\right ) {\mathrm e}^{-1}\right ) \ln \left (\ln \relax (5)-5\right )}{{\mathrm e} \ln \left (\ln \relax (5)-5\right )+1}+\frac {x \ln \left (x \left ({\mathrm e} \ln \left (\ln \relax (5)-5\right )+1\right ) {\mathrm e}^{-1}\right )}{{\mathrm e} \ln \left (\ln \relax (5)-5\right )+1}-\frac {{\mathrm e} x \ln \left (\ln \relax (5)-5\right )}{{\mathrm e} \ln \left (\ln \relax (5)-5\right )+1}-\frac {x}{{\mathrm e} \ln \left (\ln \relax (5)-5\right )+1}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*ln(x*(exp(1)*ln(ln(5)-5)+1)*exp(-1))

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maxima [B]  time = 0.35, size = 62, normalized size = 2.70 \begin {gather*} x + \frac {{\left ({\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )} \log \left ({\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )}\right ) - {\left (x e \log \left (\log \relax (5) - 5\right ) + x\right )} e^{\left (-1\right )}\right )} e}{e \log \left (\log \relax (5) - 5\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.06, size = 17, normalized size = 0.74 \begin {gather*} x\,\left (\ln \left (x\,\left (\ln \left (\ln \relax (5)-5\right )\,\mathrm {e}+1\right )\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(log(x*(log(log(5) - 5)*exp(1) + 1)) - 1)

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sympy [A]  time = 0.11, size = 26, normalized size = 1.13 \begin {gather*} x \log {\left (x + e x \log {\left (5 - \log {\relax (5 )} \right )} + e i \pi x \right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(x + E*x*log(5 - log(5)) + E*I*pi*x) - x

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