\(\int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx\) [7828]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-159-e^{5+x}-\log ^2(x) \]

[Out]

-159-exp(5+x)-ln(x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {14, 2225, 2338} \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-e^{x+5}-\log ^2(x) \]

[In]

Int[(-(E^(5 + x)*x) - 2*Log[x])/x,x]

[Out]

-E^(5 + x) - Log[x]^2

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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{5+x}-\frac {2 \log (x)}{x}\right ) \, dx \\ & = -\left (2 \int \frac {\log (x)}{x} \, dx\right )-\int e^{5+x} \, dx \\ & = -e^{5+x}-\log ^2(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-e^{5+x}-\log ^2(x) \]

[In]

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

[Out]

-E^(5 + x) - Log[x]^2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(-\ln \left (x \right )^{2}-{\mathrm e}^{5+x}\)	\(14\)
norman	\(-\ln \left (x \right )^{2}-{\mathrm e}^{5+x}\)	\(14\)
risch	\(-\ln \left (x \right )^{2}-{\mathrm e}^{5+x}\)	\(14\)
parallelrisch	\(-\ln \left (x \right )^{2}-{\mathrm e}^{5+x}\)	\(14\)
parts	\(-\ln \left (x \right )^{2}-{\mathrm e}^{5+x}\)	\(14\)

[In]

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

[Out]

-ln(x)^2-exp(5+x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-\log \left (x\right )^{2} - e^{\left (x + 5\right )} \]

[In]

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

[Out]

-log(x)^2 - e^(x + 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=- e^{x + 5} - \log {\left (x \right )}^{2} \]

[In]

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

[Out]

-exp(x + 5) - log(x)**2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-\log \left (x\right )^{2} - e^{\left (x + 5\right )} \]

[In]

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

[Out]

-log(x)^2 - e^(x + 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-\log \left (x\right )^{2} - e^{\left (x + 5\right )} \]

[In]

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

[Out]

-log(x)^2 - e^(x + 5)

Mupad [B] (verification not implemented)

Time = 15.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {-e^{5+x} x-2 \log (x)}{x} \, dx=-{\ln \left (x\right )}^2-{\mathrm {e}}^{x+5} \]

[In]

int(-(2*log(x) + x*exp(x + 5))/x,x)

[Out]

- exp(x + 5) - log(x)^2