\(\int \log ^3(\frac {c (b+a x)^2}{x^2}) \, dx\) [98]

Optimal result

Integrand size = 15, antiderivative size = 102 \[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {6 b \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \log \left (1-\frac {a x}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c (b+a x)^2}{x^2}\right ) \operatorname {PolyLog}\left (2,\frac {a x}{b+a x}\right )}{a}+\frac {48 b \operatorname {PolyLog}\left (3,\frac {a x}{b+a x}\right )}{a} \]

[Out]

x*ln(c*(a*x+b)^2/x^2)^3-6*b*ln(c*(a*x+b)^2/x^2)^2*ln(1-a*x/(a*x+b))/a+24*b*ln(c*(a*x+b)^2/x^2)*polylog(2,a*x/(
a*x+b))/a+48*b*polylog(3,a*x/(a*x+b))/a

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2536, 2550, 2379, 2421, 6724} \[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\frac {24 b \operatorname {PolyLog}\left (2,\frac {a x}{b+a x}\right ) \log \left (\frac {c (a x+b)^2}{x^2}\right )}{a}+x \log ^3\left (\frac {c (a x+b)^2}{x^2}\right )-\frac {6 b \log \left (1-\frac {a x}{a x+b}\right ) \log ^2\left (\frac {c (a x+b)^2}{x^2}\right )}{a}+\frac {48 b \operatorname {PolyLog}\left (3,\frac {a x}{b+a x}\right )}{a} \]

[In]

Int[Log[(c*(b + a*x)^2)/x^2]^3,x]

[Out]

x*Log[(c*(b + a*x)^2)/x^2]^3 - (6*b*Log[(c*(b + a*x)^2)/x^2]^2*Log[1 - (a*x)/(b + a*x)])/a + (24*b*Log[(c*(b +
a*x)^2)/x^2]*PolyLog[2, (a*x)/(b + a*x)])/a + (48*b*PolyLog[3, (a*x)/(b + a*x)])/a

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, 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 2550

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/b)^m, Subst[Int[x^m*((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[b*f - a*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*} \text {integral}& = x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )+(6 b) \int \frac {\log ^2\left (\frac {c (b+a x)^2}{x^2}\right )}{b+a x} \, dx \\ & = x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )+(6 b) \text {Subst}\left (\int \frac {\log ^2\left (c x^2\right )}{(a-x) x} \, dx,x,\frac {b+a x}{x}\right ) \\ & = x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {6 b \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \log \left (1-\frac {a x}{b+a x}\right )}{a}+\frac {(24 b) \text {Subst}\left (\int \frac {\log \left (1-\frac {a}{x}\right ) \log \left (c x^2\right )}{x} \, dx,x,\frac {b+a x}{x}\right )}{a} \\ & = x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {6 b \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \log \left (1-\frac {a x}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c (b+a x)^2}{x^2}\right ) \text {Li}_2\left (\frac {a x}{b+a x}\right )}{a}-\frac {(48 b) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {a}{x}\right )}{x} \, dx,x,\frac {b+a x}{x}\right )}{a} \\ & = x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {6 b \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \log \left (1-\frac {a x}{b+a x}\right )}{a}+\frac {24 b \log \left (\frac {c (b+a x)^2}{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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=-\frac {6 b \log \left (\frac {b}{b+a x}\right ) \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^3\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {24 b \log \left (\frac {c (b+a x)^2}{x^2}\right ) \operatorname {PolyLog}\left (2,\frac {a x}{b+a x}\right )}{a}+\frac {48 b \operatorname {PolyLog}\left (3,\frac {a x}{b+a x}\right )}{a} \]

[In]

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

[Out]

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

Maple [F]

\[\int \ln \left (\frac {c \left (a x +b \right )^{2}}{x^{2}}\right )^{3}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right )^{3} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=6 b \int \frac {\log {\left (a^{2} c + \frac {2 a b c}{x} + \frac {b^{2} c}{x^{2}} \right )}^{2}}{a x + b}\, dx + x \log {\left (\frac {c \left (a x + b\right )^{2}}{x^{2}} \right )}^{3} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right )^{3} \,d x } \]

[In]

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

Giac [F]

\[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\int { \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right )^{3} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \log ^3\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\int {\ln \left (\frac {c\,{\left (b+a\,x\right )}^2}{x^2}\right )}^3 \,d x \]

[In]

int(log((c*(b + a*x)^2)/x^2)^3,x)

[Out]

int(log((c*(b + a*x)^2)/x^2)^3, x)