3.2.1 \(\int \log ^3(\frac {c x^2}{(b+a x)^2}) \, dx\) [101]

Optimal. Leaf size=98 \[ x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {6 b \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \text {Li}_2\left (\frac {a x}{b+a x}\right )}{a}-\frac {48 b \text {Li}_3\left (\frac {a x}{b+a x}\right )}{a} \]

[Out]

x*ln(c*x^2/(a*x+b)^2)^3+6*b*ln(c*x^2/(a*x+b)^2)^2*ln(b/(a*x+b))/a+24*b*ln(c*x^2/(a*x+b)^2)*polylog(2,a*x/(a*x+
b))/a-48*b*polylog(3,a*x/(a*x+b))/a

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2536, 2552, 2354, 2421, 6724} \begin {gather*} \frac {24 b \text {PolyLog}\left (2,\frac {a x}{a x+b}\right ) \log \left (\frac {c x^2}{(a x+b)^2}\right )}{a}-\frac {48 b \text {PolyLog}\left (3,\frac {a x}{a x+b}\right )}{a}+x \log ^3\left (\frac {c x^2}{(a x+b)^2}\right )+\frac {6 b \log \left (\frac {b}{a x+b}\right ) \log ^2\left (\frac {c x^2}{(a x+b)^2}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[(c*x^2)/(b + a*x)^2]^3,x]

[Out]

x*Log[(c*x^2)/(b + a*x)^2]^3 + (6*b*Log[(c*x^2)/(b + a*x)^2]^2*Log[b/(b + a*x)])/a + (24*b*Log[(c*x^2)/(b + a*
x)^2]*PolyLog[2, (a*x)/(b + a*x)])/a - (48*b*PolyLog[3, (a*x)/(b + a*x)])/a

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 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-PolyLog[2, (-d)*f*x^m])*((a + b*Log[c*x^n])^p/m), x] + Dist[b*n*(p/m), Int[PolyLog[2, (-d)*f*x^m]*((a + b*L
og[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2536

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.), x_Symbol] :> Simp[
(a + b*x)*((A + B*Log[e*((a + b*x)^n/(c + d*x)^n)])^p/b), x] - Dist[B*n*p*((b*c - a*d)/b), Int[(A + B*Log[e*((
a + b*x)^n/(c + d*x)^n)])^(p - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x] && EqQ[n + mn, 0] &&
NeQ[b*c - a*d, 0] && IGtQ[p, 0]

Rule 2552

Int[((A_.) + Log[(e_.)*((a_.) + (b_.)*(x_))^(n_.)*((c_.) + (d_.)*(x_))^(mn_)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_)
)^(m_.), x_Symbol] :> Dist[(b*c - a*d)^(m + 1)*(g/d)^m, Subst[Int[(A + B*Log[e*x^n])^p/(b - d*x)^(m + 2), x],
x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, n}, x] && EqQ[n + mn, 0] && IGtQ[n, 0] && NeQ
[b*c - a*d, 0] && IntegersQ[m, p] && EqQ[d*f - c*g, 0] && (GtQ[p, 0] || LtQ[m, -1])

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \log ^3\left (\frac {c x^2}{(b+a x)^2}\right ) \, dx &=x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )-(6 b) \int \frac {\log ^2\left (\frac {c x^2}{(b+a x)^2}\right )}{b+a x} \, dx\\ &=x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {6 b \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}-\frac {\left (24 b^2\right ) \int \frac {\log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{x (b+a x)} \, dx}{a}\\ &=x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {6 b \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \text {Li}_2\left (1-\frac {b}{b+a x}\right )}{a}-\frac {\left (48 b^2\right ) \int \frac {\text {Li}_2\left (1-\frac {b}{b+a x}\right )}{x (b+a x)} \, dx}{a}\\ &=x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {6 b \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \text {Li}_2\left (1-\frac {b}{b+a x}\right )}{a}-\frac {48 b \text {Li}_3\left (1-\frac {b}{b+a x}\right )}{a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 98, normalized size = 1.00 \begin {gather*} x \log ^3\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {6 b \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \text {Li}_2\left (\frac {a x}{b+a x}\right )}{a}-\frac {48 b \text {Li}_3\left (\frac {a x}{b+a x}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*Log[(c*x^2)/(b + a*x)^2]^3 + (6*b*Log[(c*x^2)/(b + a*x)^2]^2*Log[b/(b + a*x)])/a + (24*b*Log[(c*x^2)/(b + a*
x)^2]*PolyLog[2, (a*x)/(b + a*x)])/a - (48*b*PolyLog[3, (a*x)/(b + a*x)])/a

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(2*(a*x + b)*log(a*x + b)^3 - 3*(a*x*log(c) + 2*a*x*log(x))*log(a*x + b)^2)/a - integrate(-(a*x*log(c)^3 +
b*log(c)^3 + 8*(a*x + b)*log(x)^3 + 12*(a*x*log(c) + b*log(c))*log(x)^2 - 6*((log(c)^2 + 4*log(c))*a*x + b*log
(c)^2 + 4*(a*x + b)*log(x)^2 + 4*(a*x*(log(c) + 2) + b*log(c))*log(x))*log(a*x + b) + 6*(a*x*log(c)^2 + b*log(
c)^2)*log(x))/(a*x + b), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - 6 b \int \frac {\log {\left (\frac {c x^{2}}{a^{2} x^{2} + 2 a b x + b^{2}} \right )}^{2}}{a x + b}\, dx + x \log {\left (\frac {c x^{2}}{\left (a x + b\right )^{2}} \right )}^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*b*Integral(log(c*x**2/(a**2*x**2 + 2*a*b*x + b**2))**2/(a*x + b), x) + x*log(c*x**2/(a*x + b)**2)**3

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\ln \left (\frac {c\,x^2}{{\left (b+a\,x\right )}^2}\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*x^2)/(b + a*x)^2)^3,x)

[Out]

int(log((c*x^2)/(b + a*x)^2)^3, x)

________________________________________________________________________________________