\(\int \frac {a+b \log (c (\frac {1}{c}+e x))}{x} \, dx\) [72]

Optimal result

Integrand size = 18, antiderivative size = 15 \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-c e x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2439, 2438} \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-c e x) \]

[In]

Int[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*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+c e x)}{x} \, dx \\ & = a \log (x)-b \text {Li}_2(-c e x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=a \log (x)-b \operatorname {PolyLog}(2,-c e x) \]

[In]

Integrate[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*e*x)]

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07

method	result	size
risch	\(\ln \left (x \right ) a -b \operatorname {dilog}\left (c e x +1\right )\)	\(16\)
parts	\(\ln \left (x \right ) a -b \operatorname {dilog}\left (c e x +1\right )\)	\(16\)
derivativedivides	\(a \ln \left (c e x \right )-b \operatorname {dilog}\left (c e x +1\right )\)	\(19\)
default	\(a \ln \left (c e x \right )-b \operatorname {dilog}\left (c e x +1\right )\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=-b {\rm Li}_2\left (-c e x\right ) + a \log \left (x\right ) \]

[In]

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

[Out]

-b*dilog(-c*e*x) + a*log(x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=a \log {\left (x \right )} - b \operatorname {Li}_{2}\left (c e x e^{i \pi }\right ) \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx=a\,\ln \left (x\right )-b\,\mathrm {polylog}\left (2,-c\,e\,x\right ) \]

[In]

int((a + b*log(c*(e*x + 1/c)))/x,x)

[Out]

a*log(x) - b*polylog(2, -c*e*x)