\(\int \frac {\log (x)}{a+b x+c x^2} \, dx\) [354]

Optimal result

Integrand size = 15, antiderivative size = 153 \[ \int \frac {\log (x)}{a+b x+c x^2} \, dx=\frac {\log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+\frac {\operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2404, 2354, 2438} \[ \int \frac {\log (x)}{a+b x+c x^2} \, dx=\frac {\operatorname {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+\frac {\log (x) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{\sqrt {b^2-4 a c}}-\frac {\log (x) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{\sqrt {b^2-4 a c}} \]

[In]

Int[Log[x]/(a + b*x + c*x^2),x]

[Out]

(Log[x]*Log[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[b^2 - 4*a*c] - (Log[x]*Log[1 + (2*c*x)/(b + Sqrt[b^2 -
4*a*c])])/Sqrt[b^2 - 4*a*c] + PolyLog[2, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])]/Sqrt[b^2 - 4*a*c] - PolyLog[2, (-2*
c*x)/(b + Sqrt[b^2 - 4*a*c])]/Sqrt[b^2 - 4*a*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 {2 c \log (x)}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}-\frac {2 c \log (x)}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right ) \, dx \\ & = \frac {(2 c) \int \frac {\log (x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {\log (x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}} \\ & = \frac {\log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\int \frac {\log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{x} \, dx}{\sqrt {b^2-4 a c}}+\frac {\int \frac {\log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{x} \, dx}{\sqrt {b^2-4 a c}} \\ & = \frac {\log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+\frac {\text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.94 \[ \int \frac {\log (x)}{a+b x+c x^2} \, dx=\frac {\log (x) \left (\log \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )-\log \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )+\operatorname {PolyLog}\left (2,\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )-\operatorname {PolyLog}\left (2,-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[In]

Integrate[Log[x]/(a + b*x + c*x^2),x]

[Out]

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

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.10

method	result	size
default	\(-\frac {\ln \left (x \right ) \left (\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )-\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )\right )}{\sqrt {-4 c a +b^{2}}}+\frac {\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )-\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}\)	\(169\)
risch	\(-\frac {\ln \left (x \right ) \left (\ln \left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )-\ln \left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )\right )}{\sqrt {-4 c a +b^{2}}}+\frac {\operatorname {dilog}\left (\frac {-2 x c +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}-\frac {\operatorname {dilog}\left (\frac {2 x c +\sqrt {-4 c a +b^{2}}+b}{b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}\)	\(178\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(log(x)/(c*x^2 + b*x + a), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(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]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(log(x)/(a + b*x + c*x^2),x)

[Out]

int(log(x)/(a + b*x + c*x^2), x)