3.56.69 \(\int \frac {e^{-1-e^5-e^8+2 x} (1-x)-x+x^2+(x-2 e^{-1-e^5-e^8+2 x} x) \log (\frac {3 e^x}{x})}{x} \, dx\)

Optimal. Leaf size=31 \[ \left (-e^{-1-e^5-e^8+2 x}+x\right ) \log \left (\frac {3 e^x}{x}\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 40, normalized size of antiderivative = 1.29, number of steps used = 5, number of rules used = 3, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {14, 2548, 2288} \begin {gather*} x \log \left (\frac {3 e^x}{x}\right )-e^{2 x-e^8-e^5-1} \log \left (\frac {3 e^x}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-1 - E^5 - E^8 + 2*x)*(1 - x) - x + x^2 + (x - 2*E^(-1 - E^5 - E^8 + 2*x)*x)*Log[(3*E^x)/x])/x,x]

[Out]

-(E^(-1 - E^5 - E^8 + 2*x)*Log[(3*E^x)/x]) + x*Log[(3*E^x)/x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+x+\log \left (\frac {3 e^x}{x}\right )-\frac {e^{-1-e^5-e^8+2 x} \left (-1+x+2 x \log \left (\frac {3 e^x}{x}\right )\right )}{x}\right ) \, dx\\ &=-x+\frac {x^2}{2}+\int \log \left (\frac {3 e^x}{x}\right ) \, dx-\int \frac {e^{-1-e^5-e^8+2 x} \left (-1+x+2 x \log \left (\frac {3 e^x}{x}\right )\right )}{x} \, dx\\ &=-x+\frac {x^2}{2}-e^{-1-e^5-e^8+2 x} \log \left (\frac {3 e^x}{x}\right )+x \log \left (\frac {3 e^x}{x}\right )-\int (-1+x) \, dx\\ &=-e^{-1-e^5-e^8+2 x} \log \left (\frac {3 e^x}{x}\right )+x \log \left (\frac {3 e^x}{x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 31, normalized size = 1.00 \begin {gather*} \left (-e^{-1-e^5-e^8+2 x}+x\right ) \log \left (\frac {3 e^x}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-1 - E^5 - E^8 + 2*x)*(1 - x) - x + x^2 + (x - 2*E^(-1 - E^5 - E^8 + 2*x)*x)*Log[(3*E^x)/x])/x,x
]

[Out]

(-E^(-1 - E^5 - E^8 + 2*x) + x)*Log[(3*E^x)/x]

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fricas [B]  time = 0.63, size = 69, normalized size = 2.23 \begin {gather*} -{\left (x e^{\left (e^{8} + e^{5} + 1\right )} \log \relax (x) + {\left (x^{3} + x^{2} \log \relax (3) - x^{2} \log \relax (x)\right )} e^{\left (2 \, x - 2 \, \log \relax (x)\right )} - {\left (x^{2} + x \log \relax (3)\right )} e^{\left (e^{8} + e^{5} + 1\right )}\right )} e^{\left (-e^{8} - e^{5} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(-exp(8)-exp(5)+2*x-1)+x)*log(3*exp(x-log(x)))+(-x+1)*exp(-exp(8)-exp(5)+2*x-1)+x^2-x)/x,x
, algorithm="fricas")

[Out]

-(x*e^(e^8 + e^5 + 1)*log(x) + (x^3 + x^2*log(3) - x^2*log(x))*e^(2*x - 2*log(x)) - (x^2 + x*log(3))*e^(e^8 +
e^5 + 1))*e^(-e^8 - e^5 - 1)

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giac [B]  time = 0.66, size = 65, normalized size = 2.10 \begin {gather*} x^{2} - x e^{\left (2 \, x - e^{8} - e^{5} - 1\right )} + x \log \relax (3) - e^{\left (2 \, x - e^{8} - e^{5} - 1\right )} \log \relax (3) - x \log \relax (x) + e^{\left (2 \, x - e^{8} - e^{5} - 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(-exp(8)-exp(5)+2*x-1)+x)*log(3*exp(x-log(x)))+(-x+1)*exp(-exp(8)-exp(5)+2*x-1)+x^2-x)/x,x
, algorithm="giac")

[Out]

x^2 - x*e^(2*x - e^8 - e^5 - 1) + x*log(3) - e^(2*x - e^8 - e^5 - 1)*log(3) - x*log(x) + e^(2*x - e^8 - e^5 -
1)*log(x)

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maple [B]  time = 0.24, size = 76, normalized size = 2.45

method	result	size
risch	\(\ln \left (\frac {{\mathrm e}^{x}}{x}\right ) \left ({\mathrm e}^{{\mathrm e}^{8}+{\mathrm e}^{5}+1}-\frac {{\mathrm e}^{2 x}}{x}\right ) x \,{\mathrm e}^{-{\mathrm e}^{8}-{\mathrm e}^{5}-1}-\frac {\left (-2 \ln \relax (3) {\mathrm e}^{{\mathrm e}^{8}+{\mathrm e}^{5}+1}+\frac {2 \ln \relax (3) {\mathrm e}^{2 x}}{x}\right ) x \,{\mathrm e}^{-{\mathrm e}^{8}-{\mathrm e}^{5}-1}}{2}\)	\(76\)
default	\(\ln \relax (x ) {\mathrm e}^{-{\mathrm e}^{8}-{\mathrm e}^{5}+2 x -1}-x \,{\mathrm e}^{-{\mathrm e}^{8}-{\mathrm e}^{5}+2 x -1}-\left (\ln \left (3 \,{\mathrm e}^{x -\ln \relax (x )}\right )-x +\ln \relax (x )\right ) {\mathrm e}^{-{\mathrm e}^{8}-{\mathrm e}^{5}+2 x -1}+\ln \left (3 \,{\mathrm e}^{x -\ln \relax (x )}\right ) x\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(-exp(8)-exp(5)+2*x-1)+x)*ln(3*exp(x-ln(x)))+(1-x)*exp(-exp(8)-exp(5)+2*x-1)+x^2-x)/x,x,method=_
RETURNVERBOSE)

[Out]

ln(exp(x)/x)*(exp(exp(8)+exp(5)+1)-exp(2*x)/x)*x*exp(-exp(8)-exp(5)-1)-1/2*(-2*ln(3)*exp(exp(8)+exp(5)+1)+2*ln
(3)/x*exp(2*x))*x*exp(-exp(8)-exp(5)-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\rm Ei}\left (2 \, x\right ) e^{\left (-e^{8} - e^{5} - 1\right )} - e^{\left (-e^{8} - e^{5} - 1\right )} \int \frac {{\left (2 \, x^{2} + 2 \, x \log \relax (3) + 1\right )} e^{\left (2 \, x\right )}}{x}\,{d x} + e^{\left (2 \, x - e^{8} - e^{5} - 1\right )} \log \relax (x) + x \log \left (\frac {3 \, e^{x}}{x}\right ) - \frac {1}{2} \, e^{\left (2 \, x - e^{8} - e^{5} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(-exp(8)-exp(5)+2*x-1)+x)*log(3*exp(x-log(x)))+(-x+1)*exp(-exp(8)-exp(5)+2*x-1)+x^2-x)/x,x
, algorithm="maxima")

[Out]

Ei(2*x)*e^(-e^8 - e^5 - 1) - e^(-e^8 - e^5 - 1)*integrate((2*x^2 + 2*x*log(3) + 1)*e^(2*x)/x, x) + e^(2*x - e^
8 - e^5 - 1)*log(x) + x*log(3*e^x/x) - 1/2*e^(2*x - e^8 - e^5 - 1)

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mupad [B]  time = 3.74, size = 36, normalized size = 1.16 \begin {gather*} -{\mathrm {e}}^{-{\mathrm {e}}^5-{\mathrm {e}}^8-1}\,\left ({\mathrm {e}}^{2\,x}-x\,{\mathrm {e}}^{{\mathrm {e}}^5+{\mathrm {e}}^8+1}\right )\,\left (x+\ln \left (\frac {3}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + exp(2*x - exp(5) - exp(8) - 1)*(x - 1) - log(3*exp(x - log(x)))*(x - 2*x*exp(2*x - exp(5) - exp(8) -
1)) - x^2)/x,x)

[Out]

-exp(- exp(5) - exp(8) - 1)*(exp(2*x) - x*exp(exp(5) + exp(8) + 1))*(x + log(3/x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(-exp(8)-exp(5)+2*x-1)+x)*ln(3*exp(x-ln(x)))+(-x+1)*exp(-exp(8)-exp(5)+2*x-1)+x**2-x)/x,x)

[Out]

Timed out

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