3.269 \(\int \frac{x \log (1-a-b x)}{a+b x} \, dx\)

Optimal. Leaf size=43 \[ \frac{a \text{PolyLog}(2,a+b x)}{b^2}-\frac{(-a-b x+1) \log (-a-b x+1)}{b^2}-\frac{x}{b} \]

[Out]

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

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Rubi [A]  time = 0.0725418, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {43, 2416, 2389, 2295, 2393, 2391} \[ \frac{a \text{PolyLog}(2,a+b x)}{b^2}-\frac{(-a-b x+1) \log (-a-b x+1)}{b^2}-\frac{x}{b} \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[1 - a - b*x])/(a + b*x),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2389

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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{x \log (1-a-b x)}{a+b x} \, dx &=\int \left (\frac{\log (1-a-b x)}{b}-\frac{a \log (1-a-b x)}{b (a+b x)}\right ) \, dx\\ &=\frac{\int \log (1-a-b x) \, dx}{b}-\frac{a \int \frac{\log (1-a-b x)}{a+b x} \, dx}{b}\\ &=-\frac{\operatorname{Subst}(\int \log (x) \, dx,x,1-a-b x)}{b^2}-\frac{a \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,a+b x\right )}{b^2}\\ &=-\frac{x}{b}-\frac{(1-a-b x) \log (1-a-b x)}{b^2}+\frac{a \text{Li}_2(a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.0177899, size = 35, normalized size = 0.81 \[ \frac{a \text{PolyLog}(2,a+b x)+(a+b x-1) \log (-a-b x+1)-b x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 77, normalized size = 1.8 \begin{align*}{\frac{\ln \left ( -bx-a+1 \right ) x}{b}}+{\frac{a{\it dilog} \left ( -bx-a+1 \right ) }{{b}^{2}}}+{\frac{\ln \left ( -bx-a+1 \right ) a}{{b}^{2}}}-{\frac{x}{b}}-{\frac{\ln \left ( -bx-a+1 \right ) }{{b}^{2}}}-{\frac{a}{{b}^{2}}}+{b}^{-2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.03289, size = 111, normalized size = 2.58 \begin{align*} b{\left (\frac{{\left (\log \left (b x + a\right ) \log \left (-b x - a + 1\right ) +{\rm Li}_2\left (b x + a\right )\right )} a}{b^{3}} - \frac{x}{b^{2}} + \frac{{\left (a - 1\right )} \log \left (b x + a - 1\right )}{b^{3}}\right )} +{\left (\frac{x}{b} - \frac{a \log \left (b x + a\right )}{b^{2}}\right )} \log \left (-b x - a + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*((log(b*x + a)*log(-b*x - a + 1) + dilog(b*x + a))*a/b^3 - x/b^2 + (a - 1)*log(b*x + a - 1)/b^3) + (x/b - a*
log(b*x + a)/b^2)*log(-b*x - a + 1)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \log \left (-b x - a + 1\right )}{b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*log(-b*x - a + 1)/(b*x + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log{\left (- a - b x + 1 \right )}}{a + b x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*log(-a - b*x + 1)/(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log \left (-b x - a + 1\right )}{b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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