3.90.74 \(\int \frac {e^{-3+x}+10 x+e^{-3+x} x \log (x)}{5 x} \, dx\)

Optimal. Leaf size=30 \[ 1+\log (5)-\log \left (e^{8-2 e^3-2 x}\right )+\frac {1}{5} e^{-3+x} \log (x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 15, normalized size of antiderivative = 0.50, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {12, 14, 2288} \begin {gather*} 2 x+\frac {1}{5} e^{x-3} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-3 + x) + 10*x + E^(-3 + x)*x*Log[x])/(5*x),x]

[Out]

2*x + (E^(-3 + x)*Log[x])/5

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{-3+x}+10 x+e^{-3+x} x \log (x)}{x} \, dx\\ &=\frac {1}{5} \int \left (10+\frac {e^{-3+x} (1+x \log (x))}{x}\right ) \, dx\\ &=2 x+\frac {1}{5} \int \frac {e^{-3+x} (1+x \log (x))}{x} \, dx\\ &=2 x+\frac {1}{5} e^{-3+x} \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 16, normalized size = 0.53 \begin {gather*} \frac {1}{5} \left (10 x+e^{-3+x} \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + x) + 10*x + E^(-3 + x)*x*Log[x])/(5*x),x]

[Out]

(10*x + E^(-3 + x)*Log[x])/5

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 12, normalized size = 0.40 \begin {gather*} \frac {1}{5} \, e^{\left (x - 3\right )} \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x*exp(x-3)*log(x)+exp(x-3)+10*x)/x,x, algorithm="fricas")

[Out]

1/5*e^(x - 3)*log(x) + 2*x

________________________________________________________________________________________

giac [A]  time = 0.15, size = 15, normalized size = 0.50 \begin {gather*} \frac {1}{5} \, {\left (10 \, x e^{3} + e^{x} \log \relax (x)\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x*exp(x-3)*log(x)+exp(x-3)+10*x)/x,x, algorithm="giac")

[Out]

1/5*(10*x*e^3 + e^x*log(x))*e^(-3)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 13, normalized size = 0.43

method	result	size
default	\(2 x +\frac {\ln \relax (x ) {\mathrm e}^{x -3}}{5}\)	\(13\)
norman	\(2 x +\frac {\ln \relax (x ) {\mathrm e}^{x -3}}{5}\)	\(13\)
risch	\(2 x +\frac {\ln \relax (x ) {\mathrm e}^{x -3}}{5}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(x*exp(x-3)*ln(x)+exp(x-3)+10*x)/x,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 12, normalized size = 0.40 \begin {gather*} \frac {1}{5} \, e^{\left (x - 3\right )} \log \relax (x) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x*exp(x-3)*log(x)+exp(x-3)+10*x)/x,x, algorithm="maxima")

[Out]

1/5*e^(x - 3)*log(x) + 2*x

________________________________________________________________________________________

mupad [B]  time = 7.79, size = 12, normalized size = 0.40 \begin {gather*} 2\,x+\frac {{\mathrm {e}}^{-3}\,{\mathrm {e}}^x\,\ln \relax (x)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x + (exp(-3)*exp(x)*log(x))/5

________________________________________________________________________________________

sympy [A]  time = 0.29, size = 12, normalized size = 0.40 \begin {gather*} 2 x + \frac {e^{x - 3} \log {\relax (x )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(x*exp(x-3)*ln(x)+exp(x-3)+10*x)/x,x)

[Out]

2*x + exp(x - 3)*log(x)/5

________________________________________________________________________________________