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

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.09 \[ \int \frac {x \log (x)}{a+b x+c x^2} \, dx=\frac {\log (x) \left (\left (-b+\sqrt {b^2-4 a c}\right ) \log \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )+\left (b+\sqrt {b^2-4 a c}\right ) \log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )+\left (-b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+\left (b+\sqrt {b^2-4 a c}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}} \] Input:

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

Output:

(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])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2804, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2804

\(\displaystyle \int \left (\frac {\log (x) \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right )}{-\sqrt {b^2-4 a c}+b+2 c x}+\frac {\log (x) \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right )}{\sqrt {b^2-4 a c}+b+2 c x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c}+\frac {\log (x) \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c}+\frac {\log (x) \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c}\)

Input:

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

Output:

((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 - Sq 
rt[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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2804

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]
 

Maple [B] (verified)

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

Time = 1.86 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.87

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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