\(\int (1+e^x (3+x) \log (5)) \, dx\) [6367]

Optimal result

Integrand size = 11, antiderivative size = 12 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=-3+x+e^x (2+x) \log (5) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2207, 2225} \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x-e^x \log (5)+e^x (x+3) \log (5) \]

[In]

Int[1 + E^x*(3 + x)*Log[5],x]

[Out]

x - E^x*Log[5] + E^x*(3 + x)*Log[5]

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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 (3+x) \, dx \\ & = x+e^x (3+x) \log (5)-\log (5) \int e^x \, dx \\ & = x-e^x \log (5)+e^x (3+x) \log (5) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x+e^x (2+x) \log (5) \]

[In]

Integrate[1 + E^x*(3 + x)*Log[5],x]

[Out]

x + E^x*(2 + x)*Log[5]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
risch	\({\mathrm e}^{x} \left (2+x \right ) \ln \left (5\right )+x\)	\(11\)
default	\(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\)	\(15\)
norman	\(x +x \,{\mathrm e}^{x} \ln \left (5\right )+2 \,{\mathrm e}^{x} \ln \left (5\right )\)	\(15\)
parallelrisch	\(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\)	\(15\)
parts	\(x +\ln \left (5\right ) \left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right )\)	\(15\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left (x + 2\right )} e^{x} \log \left (5\right ) + x \]

[In]

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

[Out]

(x + 2)*e^x*log(5) + x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x + \left (x \log {\left (5 \right )} + 2 \log {\left (5 \right )}\right ) e^{x} \]

[In]

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

[Out]

x + (x*log(5) + 2*log(5))*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left ({\left (x - 1\right )} e^{x} + 3 \, e^{x}\right )} \log \left (5\right ) + x \]

[In]

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

[Out]

((x - 1)*e^x + 3*e^x)*log(5) + x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx={\left (x + 2\right )} e^{x} \log \left (5\right ) + x \]

[In]

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

[Out]

(x + 2)*e^x*log(5) + x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \left (1+e^x (3+x) \log (5)\right ) \, dx=x+2\,{\mathrm {e}}^x\,\ln \left (5\right )+x\,{\mathrm {e}}^x\,\ln \left (5\right ) \]

[In]

int(exp(x)*log(5)*(x + 3) + 1,x)

[Out]

x + 2*exp(x)*log(5) + x*exp(x)*log(5)