3.3.69 \(\int \frac {x \log (1-a-b x)}{a+b x} \, dx\) [269]

Optimal. Leaf size=43 \[ -\frac {x}{b}-\frac {(1-a-b x) \log (1-a-b x)}{b^2}+\frac {a \text {Li}_2(a+b x)}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 43, normalized size of antiderivative = 1.00, 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 = {45, 2463, 2436, 2332, 2440, 2438} \begin {gather*} \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} \end {gather*}

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 45

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 2332

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

Rule 2436

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 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]

Rule 2440

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 2463

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]

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 {\text {Subst}(\int \log (x) \, dx,x,1-a-b x)}{b^2}-\frac {a \text {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*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.81 \begin {gather*} \frac {-b x+(-1+a+b x) \log (1-a-b x)+a \text {Li}_2(a+b x)}{b^2} \end {gather*}

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 = 0.06, size = 47, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {-\left (-b x -a +1\right ) \ln \left (-b x -a +1\right )-b x -a +1+\dilog \left (-b x -a +1\right ) a}{b^{2}}\)	\(47\)
default	\(\frac {-\left (-b x -a +1\right ) \ln \left (-b x -a +1\right )-b x -a +1+\dilog \left (-b x -a +1\right ) a}{b^{2}}\)	\(47\)
risch	\(\frac {x \ln \left (-b x -a +1\right )}{b}+\frac {\dilog \left (-b x -a +1\right ) a}{b^{2}}+\frac {\ln \left (-b x -a +1\right ) a}{b^{2}}-\frac {x}{b}-\frac {\ln \left (-b x -a +1\right )}{b^{2}}-\frac {a}{b^{2}}+\frac {1}{b^{2}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (38) = 76\).
time = 0.31, size = 82, normalized size = 1.91 \begin {gather*} 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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \log {\left (- a - b x + 1 \right )}}{a + b x}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.32, size = 59, normalized size = 1.37 \begin {gather*} -\frac {\ln \left (1-b\,x-a\right )+b\,\left (x-x\,\ln \left (1-b\,x-a\right )\right )-a\,{\mathrm {Li}}_{\mathrm {2}}\left (1-b\,x-a\right )-a\,\ln \left (1-b\,x-a\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(1 - b*x - a))/(a + b*x),x)

[Out]

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

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