\(\int \frac {x \log (x)}{a+b x+c x^2} \, dx\) [353]

Optimal result

Integrand size = 16, antiderivative size = 193 \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c} \]

[Out]

1/2*ln(x)*ln(1+2*c*x/(b-(-4*a*c+b^2)^(1/2)))*(1-b/(-4*a*c+b^2)^(1/2))/c+1/2*polylog(2,-2*c*x/(b-(-4*a*c+b^2)^(
1/2)))*(1-b/(-4*a*c+b^2)^(1/2))/c+1/2*ln(x)*ln(1+2*c*x/(b+(-4*a*c+b^2)^(1/2)))*(1+b/(-4*a*c+b^2)^(1/2))/c+1/2*
polylog(2,-2*c*x/(b+(-4*a*c+b^2)^(1/2)))*(1+b/(-4*a*c+b^2)^(1/2))/c

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2404, 2354, 2438} \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\log (x) \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c}+\frac {\log (x) \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c} \]

[In]

Int[(x*Log[x])/(a + b*x + c*x^2),x]

[Out]

((1 - b/Sqrt[b^2 - 4*a*c])*Log[x]*Log[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(2*c) + ((1 + b/Sqrt[b^2 - 4*a*c])
*Log[x]*Log[1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(2*c) + ((1 - b/Sqrt[b^2 - 4*a*c])*PolyLog[2, (-2*c*x)/(b -
Sqrt[b^2 - 4*a*c])])/(2*c) + ((1 + b/Sqrt[b^2 - 4*a*c])*PolyLog[2, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(2*c)

Rule 2354

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 2404

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 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]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x)}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x)}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx \\ & = \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log (x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx+\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log (x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx \\ & = \frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c}-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{x} \, dx}{2 c}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{x} \, dx}{2 c} \\ & = \frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.09 \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\frac {\log (x) \left (\left (-b+\sqrt {b^2-4 a c}\right ) \log \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )+\left (b+\sqrt {b^2-4 a c}\right ) \log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )+\left (-b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}} \]

[In]

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

[Out]

(Log[x]*((-b + Sqrt[b^2 - 4*a*c])*Log[(b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c])] + (b + Sqrt[b^2
- 4*a*c])*Log[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c])]) + (-b + Sqrt[b^2 - 4*a*c])*PolyLog[2,
(2*c*x)/(-b + Sqrt[b^2 - 4*a*c])] + (b + Sqrt[b^2 - 4*a*c])*PolyLog[2, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(2*
c*Sqrt[b^2 - 4*a*c])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(360\) vs. \(2(169)=338\).

Time = 1.05 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.87

method	result	size
risch	\(\frac {\ln \left (x \right ) \left (\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}-\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b +\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}+\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b \right )}{2 c \sqrt {-4 c a +b^{2}}}+\frac {\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )}{2 c}-\frac {\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b}{2 c \sqrt {-4 c a +b^{2}}}+\frac {\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )}{2 c}+\frac {\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b}{2 c \sqrt {-4 c a +b^{2}}}\)	\(361\)
default	\(\frac {\ln \left (x \right ) \left (\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}-\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b +\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}+\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b \right )}{2 c \sqrt {-4 c a +b^{2}}}+\frac {\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}-\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right ) b +\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) \sqrt {-4 c a +b^{2}}+\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right ) b}{2 c \sqrt {-4 c a +b^{2}}}\)	\(362\)

[In]

int(x*ln(x)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/2*ln(x)*(ln((-2*x*c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*(-4*a*c+b^2)^(1/2)-ln((-2*x*c+(-4*a*c+b^2
)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b+ln((2*x*c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*(-4*a*c+b^2)^(1/
2)+ln((2*x*c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b)/c/(-4*a*c+b^2)^(1/2)+1/2/c*dilog((-2*x*c+(-4*a*c
+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))-1/2/c/(-4*a*c+b^2)^(1/2)*dilog((-2*x*c+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a
*c+b^2)^(1/2)))*b+1/2/c*dilog((2*x*c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))+1/2/c/(-4*a*c+b^2)^(1/2)*di
log((2*x*c+(-4*a*c+b^2)^(1/2)+b)/(b+(-4*a*c+b^2)^(1/2)))*b

Fricas [F]

\[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\int { \frac {x \log \left (x\right )}{c x^{2} + b x + a} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\int \frac {x \log {\left (x \right )}}{a + b x + c x^{2}}\, dx \]

[In]

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

[Out]

Integral(x*log(x)/(a + b*x + c*x**2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

\[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\int { \frac {x \log \left (x\right )}{c x^{2} + b x + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\int \frac {x\,\ln \left (x\right )}{c\,x^2+b\,x+a} \,d x \]

[In]

int((x*log(x))/(a + b*x + c*x^2),x)

[Out]

int((x*log(x))/(a + b*x + c*x^2), x)