\(\int \frac {a+b \log (1+e x)}{x} \, dx\) [81]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {a+b \log (1+e x)}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-e x) \]

[Out]

a*ln(x)-b*polylog(2,-e*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2439, 2438} \[ \int \frac {a+b \log (1+e x)}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-e x) \]

[In]

Int[(a + b*Log[1 + e*x])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(e*x)]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + e*(x/d)]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rubi steps \begin{align*} \text {integral}& = a \log (x)+b \int \frac {\log (1+e x)}{x} \, dx \\ & = a \log (x)-b \text {Li}_2(-e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log (1+e x)}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-e x) \]

[In]

Integrate[(a + b*Log[1 + e*x])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(e*x)]

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
risch	\(\ln \left (x \right ) a -b \operatorname {dilog}\left (e x +1\right )\)	\(15\)
parts	\(\ln \left (x \right ) a -b \operatorname {dilog}\left (e x +1\right )\)	\(15\)
derivativedivides	\(a \ln \left (e x \right )-b \operatorname {dilog}\left (e x +1\right )\)	\(17\)
default	\(a \ln \left (e x \right )-b \operatorname {dilog}\left (e x +1\right )\)	\(17\)

[In]

int((a+b*ln(e*x+1))/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*a-b*dilog(e*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \log (1+e x)}{x} \, dx=-b {\rm Li}_2\left (-e x\right ) + a \log \left (x\right ) \]

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="fricas")

[Out]

-b*dilog(-e*x) + a*log(x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.84 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \log (1+e x)}{x} \, dx=a \log {\left (x \right )} - b \operatorname {Li}_{2}\left (e x e^{i \pi }\right ) \]

[In]

integrate((a+b*ln(e*x+1))/x,x)

[Out]

a*log(x) - b*polylog(2, e*x*exp_polar(I*pi))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86 \[ \int \frac {a+b \log (1+e x)}{x} \, dx={\left (\log \left (e x + 1\right ) \log \left (-e x\right ) + {\rm Li}_2\left (e x + 1\right )\right )} b + a \log \left (x\right ) \]

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="maxima")

[Out]

(log(e*x + 1)*log(-e*x) + dilog(e*x + 1))*b + a*log(x)

Giac [F]

\[ \int \frac {a+b \log (1+e x)}{x} \, dx=\int { \frac {b \log \left (e x + 1\right ) + a}{x} \,d x } \]

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="giac")

[Out]

integrate((b*log(e*x + 1) + a)/x, x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log (1+e x)}{x} \, dx=a\,\ln \left (x\right )-b\,\mathrm {polylog}\left (2,-e\,x\right ) \]

[In]

int((a + b*log(e*x + 1))/x,x)

[Out]

a*log(x) - b*polylog(2, -e*x)