\(\int \frac {a+b \log (e x)}{x} \, dx\) [82]

Optimal result

Integrand size = 12, antiderivative size = 17 \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {(a+b \log (e x))^2}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2338} \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {(a+b \log (e x))^2}{2 b} \]

[In]

Int[(a + b*Log[e*x])/x,x]

[Out]

(a + b*Log[e*x])^2/(2*b)

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \log (e x)}{x} \, dx=a \log (x)+\frac {1}{2} b \log ^2(e x) \]

[In]

Integrate[(a + b*Log[e*x])/x,x]

[Out]

a*Log[x] + (b*Log[e*x]^2)/2

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {1}{2} \, b \log \left (e x\right )^{2} + a \log \left (e x\right ) \]

[In]

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

[Out]

1/2*b*log(e*x)^2 + a*log(e*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log (e x)}{x} \, dx=a \log {\left (x \right )} + \frac {b \log {\left (e x \right )}^{2}}{2} \]

[In]

integrate((a+b*ln(e*x))/x,x)

[Out]

a*log(x) + b*log(e*x)**2/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {{\left (b \log \left (e x\right ) + a\right )}^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*(b*log(e*x) + a)^2/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {1}{2} \, b \log \left (e x\right )^{2} + a \log \left (x\right ) \]

[In]

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

[Out]

1/2*b*log(e*x)^2 + a*log(x)

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log (e x)}{x} \, dx=\frac {b\,{\ln \left (e\,x\right )}^2}{2}+a\,\ln \left (x\right ) \]

[In]

int((a + b*log(e*x))/x,x)

[Out]

a*log(x) + (b*log(e*x)^2)/2