3.351 \(\int \frac {x^3 \log (x)}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=286 \[ \frac {\left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {Li}_2\left (-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c^3}+\frac {\left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{2 c^3}+\frac {\log (x) \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c^3}+\frac {\log (x) \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c^3}+\frac {b x}{c^2}-\frac {b x \log (x)}{c^2}-\frac {x^2}{4 c}+\frac {x^2 \log (x)}{2 c} \]

[Out]

b*x/c^2-1/4*x^2/c-b*x*ln(x)/c^2+1/2*x^2*ln(x)/c+1/2*ln(x)*ln(1+2*c*x/(b-(-4*a*c+b^2)^(1/2)))*(b^2-a*c-b*(-3*a*
c+b^2)/(-4*a*c+b^2)^(1/2))/c^3+1/2*polylog(2,-2*c*x/(b-(-4*a*c+b^2)^(1/2)))*(b^2-a*c-b*(-3*a*c+b^2)/(-4*a*c+b^
2)^(1/2))/c^3+1/2*ln(x)*ln(1+2*c*x/(b+(-4*a*c+b^2)^(1/2)))*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^3+1/2
*polylog(2,-2*c*x/(b+(-4*a*c+b^2)^(1/2)))*(b^2-a*c+b*(-3*a*c+b^2)/(-4*a*c+b^2)^(1/2))/c^3

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Rubi [A]  time = 0.44, antiderivative size = 286, 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, 2295, 2304, 2317, 2391} \[ \frac {\left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {PolyLog}\left (2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )}{2 c^3}+\frac {\left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \text {PolyLog}\left (2,-\frac {2 c x}{\sqrt {b^2-4 a c}+b}\right )}{2 c^3}+\frac {\log (x) \left (-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \log \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )}{2 c^3}+\frac {\log (x) \left (\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}-a c+b^2\right ) \log \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )}{2 c^3}+\frac {b x}{c^2}-\frac {b x \log (x)}{c^2}-\frac {x^2}{4 c}+\frac {x^2 \log (x)}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*Log[x])/(a + b*x + c*x^2),x]

[Out]

(b*x)/c^2 - x^2/(4*c) - (b*x*Log[x])/c^2 + (x^2*Log[x])/(2*c) + ((b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a
*c])*Log[x]*Log[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(2*c^3) + ((b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a
*c])*Log[x]*Log[1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(2*c^3) + ((b^2 - a*c - (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a
*c])*PolyLog[2, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c])])/(2*c^3) + ((b^2 - a*c + (b*(b^2 - 3*a*c))/Sqrt[b^2 - 4*a*c]
)*PolyLog[2, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(2*c^3)

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 {x^3 \log (x)}{a+b x+c x^2} \, dx &=\int \left (-\frac {b \log (x)}{c^2}+\frac {x \log (x)}{c}+\frac {\left (a b+\left (b^2-a c\right ) x\right ) \log (x)}{c^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (a b+\left (b^2-a c\right ) x\right ) \log (x)}{a+b x+c x^2} \, dx}{c^2}-\frac {b \int \log (x) \, dx}{c^2}+\frac {\int x \log (x) \, dx}{c}\\ &=\frac {b x}{c^2}-\frac {x^2}{4 c}-\frac {b x \log (x)}{c^2}+\frac {x^2 \log (x)}{2 c}+\frac {\int \left (\frac {\left (b^2-a c+\frac {b \left (-b^2+3 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^2-a c-\frac {b \left (-b^2+3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \log (x)}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{c^2}\\ &=\frac {b x}{c^2}-\frac {x^2}{4 c}-\frac {b x \log (x)}{c^2}+\frac {x^2 \log (x)}{2 c}+\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log (x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{c^2}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {\log (x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{c^2}\\ &=\frac {b x}{c^2}-\frac {x^2}{4 c}-\frac {b x \log (x)}{c^2}+\frac {x^2 \log (x)}{2 c}+\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\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^3}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\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^3}-\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\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^3}-\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\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^3}\\ &=\frac {b x}{c^2}-\frac {x^2}{4 c}-\frac {b x \log (x)}{c^2}+\frac {x^2 \log (x)}{2 c}+\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\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^3}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\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^3}+\frac {\left (b^2-a c-\frac {b \left (b^2-3 a c\right )}{\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^3}+\frac {\left (b^2-a c+\frac {b \left (b^2-3 a c\right )}{\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^3}\\ \end {align*}

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Mathematica [A]  time = 0.69, size = 464, normalized size = 1.62 \[ \frac {2 \left (b^2-a c\right ) \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {Li}_2\left (\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )+\frac {4 a b c \text {Li}_2\left (\frac {2 c x}{\sqrt {b^2-4 a c}-b}\right )}{\sqrt {b^2-4 a c}}+2 \left (b^2-a c\right ) \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )-\frac {4 a b c \text {Li}_2\left (-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+2 \log (x) \left (b^2-a c\right ) \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \log \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )+\frac {4 a b c \log (x) \log \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}+2 \log (x) \left (b^2-a c\right ) \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \log \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )-\frac {4 a b c \log (x) \log \left (\frac {\sqrt {b^2-4 a c}+b+2 c x}{\sqrt {b^2-4 a c}+b}\right )}{\sqrt {b^2-4 a c}}+4 b c x-4 b c x \log (x)-c^2 x^2+2 c^2 x^2 \log (x)}{4 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*Log[x])/(a + b*x + c*x^2),x]

[Out]

(4*b*c*x - c^2*x^2 - 4*b*c*x*Log[x] + 2*c^2*x^2*Log[x] + (4*a*b*c*Log[x]*Log[(b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(
b - Sqrt[b^2 - 4*a*c])])/Sqrt[b^2 - 4*a*c] + 2*(b^2 - a*c)*(1 - b/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])] - (4*a*b*c*Log[x]*Log[(b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2
- 4*a*c])])/Sqrt[b^2 - 4*a*c] + 2*(b^2 - a*c)*(1 + b/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])] + (4*a*b*c*PolyLog[2, (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])])/Sqrt[b^2 - 4*a*c] + 2*
(b^2 - a*c)*(1 - b/Sqrt[b^2 - 4*a*c])*PolyLog[2, (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])] - (4*a*b*c*PolyLog[2, (-2*c
*x)/(b + Sqrt[b^2 - 4*a*c])])/Sqrt[b^2 - 4*a*c] + 2*(b^2 - a*c)*(1 + b/Sqrt[b^2 - 4*a*c])*PolyLog[2, (-2*c*x)/
(b + Sqrt[b^2 - 4*a*c])])/(4*c^3)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \log \relax (x)}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(x)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \log \relax (x)}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*log(x)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 791, normalized size = 2.77 \[ -\frac {3 a 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 \sqrt {-4 a c +b^{2}}\, c^{2}}+\frac {3 a 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 \sqrt {-4 a c +b^{2}}\, c^{2}}+\frac {b^{3} \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}}\, c^{3}}-\frac {b^{3} \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}}\, c^{3}}+\frac {x^{2} \ln \relax (x )}{2 c}-\frac {3 a b \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}}\, c^{2}}+\frac {3 a b \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}}\, c^{2}}-\frac {a \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 c^{2}}-\frac {a \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 c^{2}}+\frac {b^{3} \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}}\, c^{3}}-\frac {b^{3} \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}}\, c^{3}}+\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 c^{3}}+\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 c^{3}}-\frac {b x \ln \relax (x )}{c^{2}}-\frac {x^{2}}{4 c}-\frac {a \dilog \left (\frac {2 c x +b +\sqrt {-4 a c +b^{2}}}{b +\sqrt {-4 a c +b^{2}}}\right )}{2 c^{2}}-\frac {a \dilog \left (\frac {-2 c x -b +\sqrt {-4 a c +b^{2}}}{-b +\sqrt {-4 a c +b^{2}}}\right )}{2 c^{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 c^{3}}+\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 c^{3}}+\frac {b x}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*ln(x)/(c*x^2+b*x+a),x)

[Out]

1/2*x^2*ln(x)/c-1/4/c*x^2-b*x*ln(x)/c^2+b*x/c^2-1/2/c^2*ln(x)*ln((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2
)^(1/2)))*a+1/2/c^3*ln(x)*ln((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b^2+3/2/c^2*ln(x)/(-4*a*c+
b^2)^(1/2)*ln((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*a*b-1/2/c^3*ln(x)/(-4*a*c+b^2)^(1/2)*ln((
-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b^3-1/2/c^2*ln(x)*ln((b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+(-4
*a*c+b^2)^(1/2)))*a+1/2/c^3*ln(x)*ln((b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))*b^2-3/2/c^2*ln(x)/(-
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)))*a*b+1/2/c^3*ln(x)/(-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)))*b^3-1/2/c^2*dilog((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-
4*a*c+b^2)^(1/2)))*a+1/2/c^3*dilog((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-b+(-4*a*c+b^2)^(1/2)))*b^2+3/2/c^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)))*a*b-1/2/c^3/(-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)))*b^3-1/2/c^2*dilog((b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+(-4*a*
c+b^2)^(1/2)))*a+1/2/c^3*dilog((b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+(-4*a*c+b^2)^(1/2)))*b^2-3/2/c^2/(-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)))*a*b+1/2/c^3/(-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)))*b^3

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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(x^3*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 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 \[ \int \frac {x^3\,\ln \relax (x)}{c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*log(x))/(a + b*x + c*x^2),x)

[Out]

int((x^3*log(x))/(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(x**3*ln(x)/(c*x**2+b*x+a),x)

[Out]

Timed out

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