\(\int (-125+125 x+125 \log (\frac {e^{10+x}}{x})) \, dx\) [7468]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=5+125 x \log \left (\frac {e^{10+x}}{x}\right ) \]

[Out]

125*ln(exp(2*x+9)/x/exp(-1+x))*x+5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2628} \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 x \log \left (\frac {e^{x+10}}{x}\right ) \]

[In]

Int[-125 + 125*x + 125*Log[E^(10 + x)/x],x]

[Out]

125*x*Log[E^(10 + x)/x]

Rule 2628

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{align*} \text {integral}& = -125 x+\frac {125 x^2}{2}+125 \int \log \left (\frac {e^{10+x}}{x}\right ) \, dx \\ & = -125 x+\frac {125 x^2}{2}+125 x \log \left (\frac {e^{10+x}}{x}\right )-125 \int (-1+x) \, dx \\ & = 125 x \log \left (\frac {e^{10+x}}{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 x \log \left (\frac {e^{10+x}}{x}\right ) \]

[In]

Integrate[-125 + 125*x + 125*Log[E^(10 + x)/x],x]

[Out]

125*x*Log[E^(10 + x)/x]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
norman	\(125 x \ln \left (\frac {{\mathrm e}^{11} {\mathrm e}^{-1+x}}{x}\right )\)	\(15\)
default	\(125 \ln \left (\frac {{\mathrm e}^{2 x +9} {\mathrm e}^{1-x}}{x}\right ) x\)	\(21\)
parallelrisch	\(125 \ln \left (\frac {{\mathrm e}^{2 x +9} {\mathrm e}^{1-x}}{x}\right ) x\)	\(21\)
parts	\(125 \ln \left (\frac {{\mathrm e}^{2 x +9} {\mathrm e}^{1-x}}{x}\right ) x\)	\(21\)
risch	\(125 x \ln \left ({\mathrm e}^{-1+x}\right )+\frac {125 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-1+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-1+x}}{x}\right )^{2}}{2}-\frac {125 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{-1+x}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{-1+x}}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {125 i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-1+x}}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}-\frac {125 i \pi x \operatorname {csgn}\left (\frac {i {\mathrm e}^{-1+x}}{x}\right )^{3}}{2}-125 x \ln \left (x \right )+1375 x\)	\(118\)

[In]

int(125*ln(exp(2*x+9)/x/exp(-1+x))+125*x-125,x,method=_RETURNVERBOSE)

[Out]

125*x*ln(exp(11)*exp(-1+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 \, x \log \left (\frac {e^{\left (x + 10\right )}}{x}\right ) \]

[In]

integrate(125*log(exp(2*x+9)/x/exp(-1+x))+125*x-125,x, algorithm="fricas")

[Out]

125*x*log(e^(x + 10)/x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 x \log {\left (\frac {e^{11} e^{x - 1}}{x} \right )} \]

[In]

integrate(125*ln(exp(2*x+9)/x/exp(-1+x))+125*x-125,x)

[Out]

125*x*log(exp(11)*exp(x - 1)/x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 \, x \log \left (\frac {e^{\left (x + 10\right )}}{x}\right ) \]

[In]

integrate(125*log(exp(2*x+9)/x/exp(-1+x))+125*x-125,x, algorithm="maxima")

[Out]

125*x*log(e^(x + 10)/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125 \, x^{2} - 125 \, x \log \left (x\right ) + 1250 \, x \]

[In]

integrate(125*log(exp(2*x+9)/x/exp(-1+x))+125*x-125,x, algorithm="giac")

[Out]

125*x^2 - 125*x*log(x) + 1250*x

Mupad [B] (verification not implemented)

Time = 12.36 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \left (-125+125 x+125 \log \left (\frac {e^{10+x}}{x}\right )\right ) \, dx=125\,x\,\left (x+\ln \left (\frac {1}{x}\right )+10\right ) \]

[In]

int(125*x + 125*log((exp(1 - x)*exp(2*x + 9))/x) - 125,x)

[Out]

125*x*(x + log(1/x) + 10)