\(\int (e^x+(1-2 x) \log (5)) \, dx\) [5072]

Optimal result

Integrand size = 12, antiderivative size = 17 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=-1+e^x-\left (2-x+x^2\right ) \log (5) \]

[Out]

exp(x)-1-ln(5)*(x^2-x+2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2225} \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=e^x-\frac {1}{4} (1-2 x)^2 \log (5) \]

[In]

Int[E^x + (1 - 2*x)*Log[5],x]

[Out]

E^x - ((1 - 2*x)^2*Log[5])/4

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}& = -\frac {1}{4} (1-2 x)^2 \log (5)+\int e^x \, dx \\ & = e^x-\frac {1}{4} (1-2 x)^2 \log (5) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=e^x+x \log (5)-x^2 \log (5) \]

[In]

Integrate[E^x + (1 - 2*x)*Log[5],x]

[Out]

E^x + x*Log[5] - x^2*Log[5]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
default	\(\left (-x^{2}+x \right ) \ln \left (5\right )+{\mathrm e}^{x}\)	\(14\)
norman	\(x \ln \left (5\right )-x^{2} \ln \left (5\right )+{\mathrm e}^{x}\)	\(15\)
risch	\(x \ln \left (5\right )-x^{2} \ln \left (5\right )+{\mathrm e}^{x}\)	\(15\)
parallelrisch	\(x \ln \left (5\right )-x^{2} \ln \left (5\right )+{\mathrm e}^{x}\)	\(15\)
parts	\(-\ln \left (5\right ) \left (x^{2}-x \right )+{\mathrm e}^{x}\)	\(15\)

[In]

int(exp(x)+(1-2*x)*ln(5),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=-{\left (x^{2} - x\right )} \log \left (5\right ) + e^{x} \]

[In]

integrate(exp(x)+(1-2*x)*log(5),x, algorithm="fricas")

[Out]

-(x^2 - x)*log(5) + e^x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=- x^{2} \log {\left (5 \right )} + x \log {\left (5 \right )} + e^{x} \]

[In]

integrate(exp(x)+(1-2*x)*ln(5),x)

[Out]

-x**2*log(5) + x*log(5) + exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=-{\left (x^{2} - x\right )} \log \left (5\right ) + e^{x} \]

[In]

integrate(exp(x)+(1-2*x)*log(5),x, algorithm="maxima")

[Out]

-(x^2 - x)*log(5) + e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx=-{\left (x^{2} - x\right )} \log \left (5\right ) + e^{x} \]

[In]

integrate(exp(x)+(1-2*x)*log(5),x, algorithm="giac")

[Out]

-(x^2 - x)*log(5) + e^x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \left (e^x+(1-2 x) \log (5)\right ) \, dx={\mathrm {e}}^x+x\,\ln \left (5\right )-x^2\,\ln \left (5\right ) \]

[In]

int(exp(x) - log(5)*(2*x - 1),x)

[Out]

exp(x) + x*log(5) - x^2*log(5)