\(\int \frac {x \log (c+d x)}{a+b x} \, dx\) [267]

Optimal result

Integrand size = 15, antiderivative size = 81 \[ \int \frac {x \log (c+d x)}{a+b x} \, dx=-\frac {x}{b}+\frac {(c+d x) \log (c+d x)}{b d}-\frac {a \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{b^2}-\frac {a \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {45, 2463, 2436, 2332, 2441, 2440, 2438} \[ \int \frac {x \log (c+d x)}{a+b x} \, dx=-\frac {a \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{b^2}-\frac {a \log (c+d x) \log \left (-\frac {d (a+b x)}{b c-a d}\right )}{b^2}+\frac {(c+d x) \log (c+d x)}{b d}-\frac {x}{b} \]

[In]

Int[(x*Log[c + d*x])/(a + b*x),x]

[Out]

-(x/b) + ((c + d*x)*Log[c + d*x])/(b*d) - (a*Log[-((d*(a + b*x))/(b*c - a*d))]*Log[c + d*x])/b^2 - (a*PolyLog[
2, (b*(c + d*x))/(b*c - a*d)])/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 2441

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {x \log (c+d x)}{a+b x} \, dx=\frac {-b d x+\left (b c+b d x-a d \log \left (\frac {d (a+b x)}{-b c+a d}\right )\right ) \log (c+d x)-a d \operatorname {PolyLog}\left (2,\frac {b (c+d x)}{b c-a d}\right )}{b^2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.76 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {\frac {d \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )}{b}-\frac {a \,d^{2} \left (\frac {\operatorname {dilog}\left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b}+\frac {\ln \left (d x +c \right ) \ln \left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b}\right )}{b}}{d^{2}}\)	\(109\)
default	\(\frac {\frac {d \left (\left (d x +c \right ) \ln \left (d x +c \right )-d x -c \right )}{b}-\frac {a \,d^{2} \left (\frac {\operatorname {dilog}\left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b}+\frac {\ln \left (d x +c \right ) \ln \left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b}\right )}{b}}{d^{2}}\)	\(109\)
risch	\(\frac {x \ln \left (d x +c \right )}{b}+\frac {\ln \left (d x +c \right ) c}{d b}-\frac {x}{b}-\frac {c}{d b}-\frac {a \operatorname {dilog}\left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b^{2}}-\frac {a \ln \left (d x +c \right ) \ln \left (\frac {a d -c b +b \left (d x +c \right )}{a d -c b}\right )}{b^{2}}\)	\(114\)
parts	\(\frac {x \ln \left (d x +c \right )}{b}-\frac {\ln \left (d x +c \right ) a \ln \left (b x +a \right )}{b^{2}}-\frac {d \left (\frac {b x +a}{b d}-\frac {c \ln \left (a d -c b -d \left (b x +a \right )\right )}{d^{2}}+\frac {a \left (-\frac {\operatorname {dilog}\left (\frac {-a d +c b +d \left (b x +a \right )}{-a d +c b}\right )}{d}-\frac {\ln \left (b x +a \right ) \ln \left (\frac {-a d +c b +d \left (b x +a \right )}{-a d +c b}\right )}{d}\right )}{b}\right )}{b}\)	\(149\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x*log(d*x + c)/(b*x + a), x)

Sympy [F]

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

[In]

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

[Out]

Integral(x*log(c + d*x)/(a + b*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.37 \[ \int \frac {x \log (c+d x)}{a+b x} \, dx=d {\left (\frac {{\left (\log \left (b x + a\right ) \log \left (\frac {b d x + a d}{b c - a d} + 1\right ) + {\rm Li}_2\left (-\frac {b d x + a d}{b c - a d}\right )\right )} a}{b^{2} d} - \frac {x}{b d} + \frac {c \log \left (d x + c\right )}{b d^{2}}\right )} + {\left (\frac {x}{b} - \frac {a \log \left (b x + a\right )}{b^{2}}\right )} \log \left (d x + c\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(x*log(d*x + c)/(b*x + a), x)

Mupad [F(-1)]

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

[In]

int((x*log(c + d*x))/(a + b*x),x)

[Out]

int((x*log(c + d*x))/(a + b*x), x)