\(\int (-e^x+\log (5)) \, dx\) [5361]

Optimal result

Integrand size = 8, antiderivative size = 13 \[ \int \left (-e^x+\log (5)\right ) \, dx=-3-e^x+\log (2)+x \log (5) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2225} \[ \int \left (-e^x+\log (5)\right ) \, dx=x \log (5)-e^x \]

[In]

Int[-E^x + Log[5],x]

[Out]

-E^x + x*Log[5]

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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \left (-e^x+\log (5)\right ) \, dx=-e^x+x \log (5) \]

[In]

Integrate[-E^x + Log[5],x]

[Out]

-E^x + x*Log[5]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
default	\(-{\mathrm e}^{x}+x \ln \left (5\right )\)	\(10\)
norman	\(-{\mathrm e}^{x}+x \ln \left (5\right )\)	\(10\)
risch	\(-{\mathrm e}^{x}+x \ln \left (5\right )\)	\(10\)
parallelrisch	\(-{\mathrm e}^{x}+x \ln \left (5\right )\)	\(10\)
parts	\(-{\mathrm e}^{x}+x \ln \left (5\right )\)	\(10\)
derivativedivides	\(-{\mathrm e}^{x}+\ln \left (5\right ) \ln \left ({\mathrm e}^{x}\right )\)	\(12\)

[In]

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

[Out]

-exp(x)+x*ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \left (-e^x+\log (5)\right ) \, dx=x \log \left (5\right ) - e^{x} \]

[In]

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

[Out]

x*log(5) - e^x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.54 \[ \int \left (-e^x+\log (5)\right ) \, dx=x \log {\left (5 \right )} - e^{x} \]

[In]

integrate(-exp(x)+ln(5),x)

[Out]

x*log(5) - exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \left (-e^x+\log (5)\right ) \, dx=x \log \left (5\right ) - e^{x} \]

[In]

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

[Out]

x*log(5) - e^x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \left (-e^x+\log (5)\right ) \, dx=x \log \left (5\right ) - e^{x} \]

[In]

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

[Out]

x*log(5) - e^x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \left (-e^x+\log (5)\right ) \, dx=x\,\ln \left (5\right )-{\mathrm {e}}^x \]

[In]

int(log(5) - exp(x),x)

[Out]

x*log(5) - exp(x)