3.616 \(\int \frac {\log (x)}{a+b x} \, dx\)

Optimal. Leaf size=29 \[ \frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b}+\frac {\log (x) \log \left (\frac {b x}{a}+1\right )}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2317, 2391} \[ \frac {\text {PolyLog}\left (2,-\frac {b x}{a}\right )}{b}+\frac {\log (x) \log \left (\frac {b x}{a}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(a + b*x),x]

[Out]

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

Rule 2317

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 2391

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*} \int \frac {\log (x)}{a+b x} \, dx &=\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 {\text {Li}_2\left (-\frac {b x}{a}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 30, normalized size = 1.03 \[ \frac {\operatorname {PolyLog}\left (2,-\frac {b x}{a}\right )}{b}+\frac {\log (x) \log \left (\frac {a+b x}{a}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(a + b*x),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log (x)}{a+b x} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Log[x]/(a + b*x),x]

[Out]

Could not integrate

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \relax (x)}{b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(x)/(b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \relax (x)}{b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(x)/(b*x + a), x)

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maple [A]  time = 0.31, size = 32, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.43, size = 25, normalized size = 0.86 \[ \frac {\log \left (\frac {b x}{a} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x}{a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\ln \relax (x)}{a+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(a + b*x),x)

[Out]

int(log(x)/(a + b*x), x)

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sympy [C]  time = 6.44, size = 151, normalized size = 5.21 \[ \begin {cases} \frac {\log {\left (\frac {a}{b} \right )} \log {\left (\frac {a}{b} + x \right )}}{b} + \frac {i \pi \log {\left (\frac {a}{b} + x \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {for}\: \left |{\frac {a}{b} + x}\right | < 1 \\- \frac {\log {\left (\frac {a}{b} \right )} \log {\left (\frac {1}{\frac {a}{b} + x} \right )}}{b} - \frac {i \pi \log {\left (\frac {1}{\frac {a}{b} + x} \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {for}\: \frac {1}{\left |{\frac {a}{b} + x}\right |} < 1 \\- \frac {{G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {\frac {a}{b} + x} \right )} \log {\left (\frac {a}{b} \right )}}{b} - \frac {i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{b} + \frac {{G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )} \log {\left (\frac {a}{b} \right )}}{b} + \frac {i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {\frac {a}{b} + x} \right )}}{b} - \frac {\operatorname {Li}_{2}\left (\frac {b \left (\frac {a}{b} + x\right )}{a}\right )}{b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((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), (-l
og(a/b)*log(1/(a/b + x))/b - I*pi*log(1/(a/b + x))/b - polylog(2, b*(a/b + x)/a)/b, 1/Abs(a/b + x) < 1), (-mei
jerg(((), (1, 1)), ((0, 0), ()), a/b + x)*log(a/b)/b - I*pi*meijerg(((), (1, 1)), ((0, 0), ()), a/b + x)/b + m
eijerg(((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))

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