\(\int \log (e^{a+b x}) \, dx\) [282]

Optimal result

Integrand size = 8, antiderivative size = 17 \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {\log ^2\left (e^{a+b x}\right )}{2 b} \]

[Out]

1/2*ln(exp(b*x+a))^2/b

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 = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2188, 30} \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {\log ^2\left (e^{a+b x}\right )}{2 b} \]

[In]

Int[Log[E^(a + b*x)],x]

[Out]

Log[E^(a + b*x)]^2/(2*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {\log ^2\left (e^{a+b x}\right )}{2 b} \]

[In]

Integrate[Log[E^(a + b*x)],x]

[Out]

Log[E^(a + b*x)]^2/(2*b)

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\ln \left ({\mathrm e}^{b x +a}\right )^{2}}{2 b}\)	\(15\)
default	\(\frac {\ln \left ({\mathrm e}^{b x +a}\right )^{2}}{2 b}\)	\(15\)
norman	\(\frac {\ln \left ({\mathrm e}^{b x +a}\right )^{2}}{2 b}\)	\(15\)
risch	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {b \,x^{2}}{2}\)	\(17\)
parallelrisch	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {b \,x^{2}}{2}\)	\(17\)
parts	\(x \ln \left ({\mathrm e}^{b x +a}\right )-\frac {b \,x^{2}}{2}\)	\(17\)

[In]

int(ln(exp(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(exp(b*x+a))^2/b

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {1}{2} \, b x^{2} + a x \]

[In]

integrate(log(exp(b*x+a)),x, algorithm="fricas")

[Out]

1/2*b*x^2 + a*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \log \left (e^{a+b x}\right ) \, dx=a x + \frac {b x^{2}}{2} \]

[In]

integrate(ln(exp(b*x+a)),x)

[Out]

a*x + b*x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {1}{2} \, b x^{2} + a x \]

[In]

integrate(log(exp(b*x+a)),x, algorithm="maxima")

[Out]

1/2*b*x^2 + a*x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \log \left (e^{a+b x}\right ) \, dx=\frac {1}{2} \, b x^{2} + a x \]

[In]

integrate(log(exp(b*x+a)),x, algorithm="giac")

[Out]

1/2*b*x^2 + a*x

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \log \left (e^{a+b x}\right ) \, dx=x\,\ln \left ({\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\right )-\frac {b\,x^2}{2} \]

[In]

int(log(exp(a + b*x)),x)

[Out]

x*log(exp(b*x)*exp(a)) - (b*x^2)/2