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

Optimal. Leaf size=153 \[ \frac {\log (x) \log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\log (x) \log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+\frac {\text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[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)

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Rubi [A]
time = 0.10, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2404, 2354, 2438} \begin {gather*} \frac {\text {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}-\frac {\text {PolyLog}\left (2,-\frac {2 c x}{\sqrt {b^2-4 a c}+b}\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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\log (x)}{a+b x+c x^2} \, dx &=\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*}

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Mathematica [A]
time = 0.02, size = 116, normalized size = 0.76 \begin {gather*} \frac {\log (x) \left (\log \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )-\log \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )\right )+\text {Li}_2\left (\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )-\text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[x]/(a + b*x + c*x^2),x]

[Out]

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

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Maple [A]
time = 0.75, size = 168, normalized size = 1.10

method	result	size
default	\(\frac {\ln \left (x \right ) \left (\ln \left (\frac {-2 c x +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )-\ln \left (\frac {b +2 c x +\sqrt {-4 c a +b^{2}}}{b +\sqrt {-4 c a +b^{2}}}\right )\right )}{\sqrt {-4 c a +b^{2}}}+\frac {\dilog \left (\frac {-2 c x +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )-\dilog \left (\frac {b +2 c x +\sqrt {-4 c a +b^{2}}}{b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}\)	\(168\)
risch	\(\frac {\ln \left (x \right ) \left (\ln \left (\frac {-2 c x +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )-\ln \left (\frac {b +2 c x +\sqrt {-4 c a +b^{2}}}{b +\sqrt {-4 c a +b^{2}}}\right )\right )}{\sqrt {-4 c a +b^{2}}}+\frac {\dilog \left (\frac {-2 c x +\sqrt {-4 c a +b^{2}}-b}{-b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}-\frac {\dilog \left (\frac {b +2 c x +\sqrt {-4 c a +b^{2}}}{b +\sqrt {-4 c a +b^{2}}}\right )}{\sqrt {-4 c a +b^{2}}}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

integral(log(x)/(c*x^2 + b*x + a), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (x \right )}}{a + b x + c x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (x\right )}{c\,x^2+b\,x+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(a + b*x + c*x^2),x)

[Out]

int(log(x)/(a + b*x + c*x^2), x)

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