### 3.270 $$\int \frac{(b+2 c x) \log (x)}{x (b+c x)} \, dx$$

Optimal. Leaf size=30 $\text{PolyLog}\left (2,-\frac{c x}{b}\right )+\log (x) \log \left (\frac{c x}{b}+1\right )+\frac{\log ^2(x)}{2}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0988166, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {2357, 2301, 2317, 2391} $\text{PolyLog}\left (2,-\frac{c x}{b}\right )+\log (x) \log \left (\frac{c x}{b}+1\right )+\frac{\log ^2(x)}{2}$

Antiderivative was successfully veriﬁed.

[In]

Int[((b + 2*c*x)*Log[x])/(x*(b + c*x)),x]

[Out]

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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{(b+2 c x) \log (x)}{x (b+c x)} \, dx &=\int \left (\frac{\log (x)}{x}+\frac{c \log (x)}{b+c x}\right ) \, dx\\ &=c \int \frac{\log (x)}{b+c x} \, dx+\int \frac{\log (x)}{x} \, dx\\ &=\frac{\log ^2(x)}{2}+\log (x) \log \left (1+\frac{c x}{b}\right )-\int \frac{\log \left (1+\frac{c x}{b}\right )}{x} \, dx\\ &=\frac{\log ^2(x)}{2}+\log (x) \log \left (1+\frac{c x}{b}\right )+\text{Li}_2\left (-\frac{c x}{b}\right )\\ \end{align*}

Mathematica [A]  time = 0.0102847, size = 31, normalized size = 1.03 $\text{PolyLog}\left (2,-\frac{c x}{b}\right )+\log (x) \log \left (\frac{b+c x}{b}\right )+\frac{\log ^2(x)}{2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((b + 2*c*x)*Log[x])/(x*(b + c*x)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 31, normalized size = 1. \begin{align*}{\frac{ \left ( \ln \left ( x \right ) \right ) ^{2}}{2}}+\ln \left ( x \right ) \ln \left ({\frac{cx+b}{b}} \right ) +{\it dilog} \left ({\frac{cx+b}{b}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*ln(x)/x/(c*x+b),x)

[Out]

1/2*ln(x)^2+ln(x)*ln((c*x+b)/b)+dilog((c*x+b)/b)

________________________________________________________________________________________

Maxima [A]  time = 1.02865, size = 66, normalized size = 2.2 \begin{align*}{\left (\log \left (c x + b\right ) + \log \left (x\right )\right )} \log \left (x\right ) - \log \left (c x + b\right ) \log \left (x\right ) + \log \left (\frac{c x}{b} + 1\right ) \log \left (x\right ) - \frac{1}{2} \, \log \left (x\right )^{2} +{\rm Li}_2\left (-\frac{c x}{b}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*log(x)/x/(c*x+b),x, algorithm="maxima")

[Out]

(log(c*x + b) + log(x))*log(x) - log(c*x + b)*log(x) + log(c*x/b + 1)*log(x) - 1/2*log(x)^2 + dilog(-c*x/b)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (2 \, c x + b\right )} \log \left (x\right )}{c x^{2} + b x}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*log(x)/x/(c*x+b),x, algorithm="fricas")

[Out]

integral((2*c*x + b)*log(x)/(c*x^2 + b*x), x)

________________________________________________________________________________________

Sympy [C]  time = 83.6595, size = 192, normalized size = 6.4 \begin{align*} b \left (\begin{cases} - \frac{1}{c x} & \text{for}\: b = 0 \\\frac{\begin{cases} \log{\left (c \right )} \log{\left (x \right )} + \operatorname{Li}_{2}\left (\frac{b e^{i \pi }}{c x}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (c \right )} \log{\left (\frac{1}{x} \right )} + \operatorname{Li}_{2}\left (\frac{b e^{i \pi }}{c x}\right ) & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (c \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (c \right )} + \operatorname{Li}_{2}\left (\frac{b e^{i \pi }}{c x}\right ) & \text{otherwise} \end{cases}}{b} & \text{otherwise} \end{cases}\right ) - b \left (\begin{cases} \frac{1}{c x} & \text{for}\: b = 0 \\\frac{\log{\left (\frac{b}{x} + c \right )}}{b} & \text{otherwise} \end{cases}\right ) \log{\left (x \right )} - 2 c \left (\begin{cases} \frac{x}{b} & \text{for}\: c = 0 \\\frac{\begin{cases} \log{\left (b \right )} \log{\left (x \right )} - \operatorname{Li}_{2}\left (\frac{c x e^{i \pi }}{b}\right ) & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (b \right )} \log{\left (\frac{1}{x} \right )} - \operatorname{Li}_{2}\left (\frac{c x e^{i \pi }}{b}\right ) & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (b \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (b \right )} - \operatorname{Li}_{2}\left (\frac{c x e^{i \pi }}{b}\right ) & \text{otherwise} \end{cases}}{c} & \text{otherwise} \end{cases}\right ) + 2 c \left (\begin{cases} \frac{x}{b} & \text{for}\: c = 0 \\\frac{\log{\left (b + c x \right )}}{c} & \text{otherwise} \end{cases}\right ) \log{\left (x \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*ln(x)/x/(c*x+b),x)

[Out]

b*Piecewise((-1/(c*x), Eq(b, 0)), (Piecewise((log(c)*log(x) + polylog(2, b*exp_polar(I*pi)/(c*x)), Abs(x) < 1)
, (-log(c)*log(1/x) + polylog(2, b*exp_polar(I*pi)/(c*x)), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ())
, x)*log(c) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(c) + polylog(2, b*exp_polar(I*pi)/(c*x)), True))/b, T
rue)) - b*Piecewise((1/(c*x), Eq(b, 0)), (log(b/x + c)/b, True))*log(x) - 2*c*Piecewise((x/b, Eq(c, 0)), (Piec
ewise((log(b)*log(x) - polylog(2, c*x*exp_polar(I*pi)/b), Abs(x) < 1), (-log(b)*log(1/x) - polylog(2, c*x*exp_
polar(I*pi)/b), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(b) + meijerg(((1, 1), ()), ((), (0
, 0)), x)*log(b) - polylog(2, c*x*exp_polar(I*pi)/b), True))/c, True)) + 2*c*Piecewise((x/b, Eq(c, 0)), (log(b
+ c*x)/c, True))*log(x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (2 \, c x + b\right )} \log \left (x\right )}{{\left (c x + b\right )} x}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*log(x)/x/(c*x+b),x, algorithm="giac")

[Out]

integrate((2*c*x + b)*log(x)/((c*x + b)*x), x)