\(\int \frac {\log (x)}{a+b x} \, dx\) [616]

Optimal result

Integrand size = 10, antiderivative size = 29 \[ \int \frac {\log (x)}{a+b x} \, dx=\frac {\log (x) \log \left (1+\frac {b x}{a}\right )}{b}+\frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2354, 2438} \[ \int \frac {\log (x)}{a+b x} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b}+\frac {\log (x) \log \left (\frac {b x}{a}+1\right )}{b} \]

[In]

Int[Log[x]/(a + b*x),x]

[Out]

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

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 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}& = \frac {\log (x) \log \left (1+\frac {b x}{a}\right )}{b}-\frac {\int \frac {\log \left (1+\frac {b x}{a}\right )}{x} \, dx}{b} \\ & = \frac {\log (x) \log \left (1+\frac {b x}{a}\right )}{b}+\frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {\log (x)}{a+b x} \, dx=\frac {\log (x) \log \left (\frac {a+b x}{a}\right )}{b}+\frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b} \]

[In]

Integrate[Log[x]/(a + b*x),x]

[Out]

(Log[x]*Log[(a + b*x)/a])/b + PolyLog[2, -((b*x)/a)]/b

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {\operatorname {dilog}\left (\frac {b x +a}{a}\right )}{b}+\frac {\ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{b}\)	\(32\)
risch	\(\frac {\operatorname {dilog}\left (\frac {b x +a}{a}\right )}{b}+\frac {\ln \left (x \right ) \ln \left (\frac {b x +a}{a}\right )}{b}\)	\(32\)
parts	\(\frac {\ln \left (b x +a \right ) \ln \left (x \right )}{b}-\frac {\operatorname {dilog}\left (-\frac {b x}{a}\right )+\ln \left (b x +a \right ) \ln \left (-\frac {b x}{a}\right )}{b}\)	\(43\)

[In]

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

[Out]

dilog((b*x+a)/a)/b+ln(x)*ln((b*x+a)/a)/b

Fricas [F]

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

[In]

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

[Out]

integral(log(x)/(b*x + a), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.78 (sec) , antiderivative size = 177, normalized size of antiderivative = 6.10 \[ \int \frac {\log (x)}{a+b x} \, dx=\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {for}\: \frac {1}{\left |{\frac {a}{b} + x}\right |} < 1 \wedge \left |{\frac {a}{b} + x}\right | < 1 \\\frac {\log {\left (\frac {a}{b} \right )} \log {\left (\frac {a}{b} + x \right )}}{b} + \frac {i \pi \log {\left (\frac {a}{b} + x \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {for}\: \left |{\frac {a}{b} + x}\right | < 1 \\- \frac {\log {\left (\frac {a}{b} \right )} \log {\left (\frac {1}{\frac {a}{b} + x} \right )}}{b} - \frac {i \pi \log {\left (\frac {1}{\frac {a}{b} + x} \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {for}\: \frac {1}{\left |{\frac {a}{b} + x}\right |} < 1 \\- \frac {{G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {\frac {a}{b} + x} \right )} \log {\left (\frac {a}{b} \right )}}{b} - \frac {i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{b} + \frac {{G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )} \log {\left (\frac {a}{b} \right )}}{b} + \frac {i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-polylog(2, b*(a/b + x)/a)/b, (Abs(a/b + x) < 1) & (1/Abs(a/b + x) < 1)), (log(a/b)*log(a/b + x)/b
+ I*pi*log(a/b + x)/b - polylog(2, b*(a/b + x)/a)/b, Abs(a/b + x) < 1), (-log(a/b)*log(1/(a/b + x))/b - I*pi*l
og(1/(a/b + x))/b - polylog(2, b*(a/b + x)/a)/b, 1/Abs(a/b + x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), a/
b + x)*log(a/b)/b - I*pi*meijerg(((), (1, 1)), ((0, 0), ()), a/b + x)/b + meijerg(((1, 1), ()), ((), (0, 0)),
a/b + x)*log(a/b)/b + I*pi*meijerg(((1, 1), ()), ((), (0, 0)), a/b + x)/b - polylog(2, b*(a/b + x)/a)/b, True)
)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\log (x)}{a+b x} \, dx=\frac {\log \left (\frac {b x}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x}{a}\right )}{b} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(log(x)/(b*x + a), x)

Mupad [F(-1)]

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

[In]

int(log(x)/(a + b*x),x)

[Out]

int(log(x)/(a + b*x), x)