\(\int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx\) [2226]

Optimal result

Integrand size = 32, antiderivative size = 24 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=1+x+\frac {1}{2} \left (-22-x-\frac {1}{2} e^{\frac {1}{x}} x \log (x)\right ) \]

[Out]

-10+1/2*x-1/4*x*exp(1/x)*ln(x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 14, 2326} \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\frac {x}{2}-\frac {1}{4} e^{\frac {1}{x}} x \log (x) \]

[In]

Int[(2*x - E^x^(-1)*x + E^x^(-1)*(1 - x)*Log[x])/(4*x),x]

[Out]

x/2 - (E^x^(-1)*x*Log[x])/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2326

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{x} \, dx \\ & = \frac {1}{4} \int \left (2-\frac {e^{\frac {1}{x}} (x-\log (x)+x \log (x))}{x}\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{4} \int \frac {e^{\frac {1}{x}} (x-\log (x)+x \log (x))}{x} \, dx \\ & = \frac {x}{2}-\frac {1}{4} e^{\frac {1}{x}} x \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\frac {1}{4} \left (2 x-e^{\frac {1}{x}} x \log (x)\right ) \]

[In]

Integrate[(2*x - E^x^(-1)*x + E^x^(-1)*(1 - x)*Log[x])/(4*x),x]

[Out]

(2*x - E^x^(-1)*x*Log[x])/4

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58

method	result	size
norman	\(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\)	\(14\)
risch	\(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\)	\(14\)
parallelrisch	\(\frac {x}{2}-\frac {x \,{\mathrm e}^{\frac {1}{x}} \ln \left (x \right )}{4}\)	\(14\)
derivativedivides	\(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\)	\(39\)
default	\(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\)	\(39\)
parts	\(\frac {x}{2}-\frac {\left (\frac {\left (\ln \left (x \right )+\ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{\frac {1}{x}}}{x}-\frac {{\mathrm e}^{\frac {1}{x}} \ln \left (\frac {1}{x}\right )}{x}\right ) x^{2}}{4}\)	\(39\)

[In]

int(1/4*((1-x)*exp(1/x)*ln(x)-x*exp(1/x)+2*x)/x,x,method=_RETURNVERBOSE)

[Out]

1/2*x-1/4*x*exp(1/x)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {1}{4} \, x e^{\frac {1}{x}} \log \left (x\right ) + \frac {1}{2} \, x \]

[In]

integrate(1/4*((1-x)*exp(1/x)*log(x)-x*exp(1/x)+2*x)/x,x, algorithm="fricas")

[Out]

-1/4*x*e^(1/x)*log(x) + 1/2*x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=- \frac {x e^{\frac {1}{x}} \log {\left (x \right )}}{4} + \frac {x}{2} \]

[In]

integrate(1/4*((1-x)*exp(1/x)*ln(x)-x*exp(1/x)+2*x)/x,x)

[Out]

-x*exp(1/x)*log(x)/4 + x/2

Maxima [F]

\[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=\int { -\frac {{\left (x - 1\right )} e^{\frac {1}{x}} \log \left (x\right ) + x e^{\frac {1}{x}} - 2 \, x}{4 \, x} \,d x } \]

[In]

integrate(1/4*((1-x)*exp(1/x)*log(x)-x*exp(1/x)+2*x)/x,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {1}{4} \, x e^{\frac {1}{x}} \log \left (x\right ) + \frac {1}{2} \, x \]

[In]

integrate(1/4*((1-x)*exp(1/x)*log(x)-x*exp(1/x)+2*x)/x,x, algorithm="giac")

[Out]

-1/4*x*e^(1/x)*log(x) + 1/2*x

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.50 \[ \int \frac {2 x-e^{\frac {1}{x}} x+e^{\frac {1}{x}} (1-x) \log (x)}{4 x} \, dx=-\frac {x\,\left ({\mathrm {e}}^{1/x}\,\ln \left (x\right )-2\right )}{4} \]

[In]

int(-((x*exp(1/x))/4 - x/2 + (exp(1/x)*log(x)*(x - 1))/4)/x,x)

[Out]

-(x*(exp(1/x)*log(x) - 2))/4