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

Optimal. Leaf size=251 \[ \frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 a^2}+\frac {\log (x) \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 a^2}+\frac {\log (x) \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 a^2}-\frac {b \log ^2(x)}{2 a^2}-\frac {1}{a x}-\frac {\log (x)}{a x} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2357, 2304, 2301, 2317, 2391} \[ \frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \text {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {PolyLog}\left (2,-\frac {2 c x}{\sqrt {b^2-4 a c}+b}\right )}{2 a^2}+\frac {\log (x) \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 a^2}+\frac {\log (x) \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 a^2}-\frac {b \log ^2(x)}{2 a^2}-\frac {1}{a x}-\frac {\log (x)}{a x} \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(x^2*(a + b*x + c*x^2)),x]

[Out]

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2317

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 2357

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 2391

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

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Mathematica [A]  time = 0.39, size = 255, normalized size = 1.02 \[ \frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \text {Li}_2\left (\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )+\left (\frac {2 a c-b^2}{\sqrt {b^2-4 a c}}+b\right ) \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )+\log (x) \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \log \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )+\log (x) \left (\frac {2 a c-b^2}{\sqrt {b^2-4 a c}}+b\right ) \log \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )-\frac {2 a}{x}-\frac {2 a \log (x)}{x}-b \log ^2(x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \relax (x)}{c x^{4} + b x^{3} + a x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(x)/(c*x^4 + b*x^3 + a*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \relax (x)}{{\left (c x^{2} + b x + a\right )} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 608, normalized size = 2.42 \[ \frac {c \ln \relax (x ) \ln \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{\sqrt {-4 a c +b^{2}}\, a}-\frac {c \ln \relax (x ) \ln \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{\sqrt {-4 a c +b^{2}}\, a}-\frac {b^{2} \ln \relax (x ) \ln \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{2 \sqrt {-4 a c +b^{2}}\, a^{2}}+\frac {b^{2} \ln \relax (x ) \ln \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{2 \sqrt {-4 a c +b^{2}}\, a^{2}}+\frac {c \dilog \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{\sqrt {-4 a c +b^{2}}\, a}-\frac {c \dilog \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{\sqrt {-4 a c +b^{2}}\, a}-\frac {b^{2} \dilog \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{2 \sqrt {-4 a c +b^{2}}\, a^{2}}+\frac {b^{2} \dilog \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{2 \sqrt {-4 a c +b^{2}}\, a^{2}}-\frac {b \ln \relax (x )^{2}}{2 a^{2}}+\frac {b \ln \relax (x ) \ln \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{2 a^{2}}+\frac {b \ln \relax (x ) \ln \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{2 a^{2}}+\frac {b \dilog \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{2 a^{2}}+\frac {b \dilog \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{2 a^{2}}-\frac {\ln \relax (x )}{a x}-\frac {1}{a x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(x)/x^2/(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 details)Is 4*a*c-b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \relax (x)}{x^2\,\left (c\,x^2+b\,x+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(x^2*(a + b*x + c*x^2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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