\(\int \frac {a+b \log (c x)}{(d+\frac {e}{x}) x} \, dx\) [341]

Optimal result

Integrand size = 21, antiderivative size = 36 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x} \, dx=\frac {(a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )}{d}+\frac {b \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d} \]

[Out]

(a+b*ln(c*x))*ln(1+d*x/e)/d+b*polylog(2,-d*x/e)/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2370, 2354, 2438} \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x} \, dx=\frac {\log \left (\frac {d x}{e}+1\right ) (a+b \log (c x))}{d}+\frac {b \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d} \]

[In]

Int[(a + b*Log[c*x])/((d + e/x)*x),x]

[Out]

((a + b*Log[c*x])*Log[1 + (d*x)/e])/d + (b*PolyLog[2, -((d*x)/e)])/d

Rule 2354

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

Rule 2370

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x} \, dx=\frac {(a+b \log (c x)) \log \left (1+\frac {d x}{e}\right )+b \operatorname {PolyLog}\left (2,-\frac {d x}{e}\right )}{d} \]

[In]

Integrate[(a + b*Log[c*x])/((d + e/x)*x),x]

[Out]

((a + b*Log[c*x])*Log[1 + (d*x)/e] + b*PolyLog[2, -((d*x)/e)])/d

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.64

method	result	size
risch	\(\frac {a \ln \left (d x +e \right )}{d}+\frac {b \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {b \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\)	\(59\)
parts	\(\frac {a \ln \left (d x +e \right )}{d}+\frac {b \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {b \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\)	\(59\)
derivativedivides	\(\frac {a \ln \left (c d x +c e \right )}{d}+\frac {b \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {b \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\)	\(62\)
default	\(\frac {a \ln \left (c d x +c e \right )}{d}+\frac {b \operatorname {dilog}\left (\frac {c d x +c e}{e c}\right )}{d}+\frac {b \ln \left (x c \right ) \ln \left (\frac {c d x +c e}{e c}\right )}{d}\)	\(62\)

[In]

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

[Out]

a*ln(d*x+e)/d+b*dilog((c*d*x+c*e)/e/c)/d+b*ln(x*c)*ln((c*d*x+c*e)/e/c)/d

Fricas [F]

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

[In]

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

[Out]

integral((b*log(c*x) + a)/(d*x + e), x)

Sympy [F]

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

[In]

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

[Out]

Integral((a + b*log(c*x))/(d*x + e), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \log (c x)}{\left (d+\frac {e}{x}\right ) x} \, dx=\frac {{\left (\log \left (\frac {d x}{e} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {d x}{e}\right )\right )} b}{d} + \frac {{\left (b \log \left (c\right ) + a\right )} \log \left (d x + e\right )}{d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*log(c*x))/(x*(d + e/x)),x)

[Out]

int((a + b*log(c*x))/(x*(d + e/x)), x)