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

Optimal. Leaf size=102 \[ \frac{24 b \text{PolyLog}\left (2,\frac{a x}{a x+b}\right ) \log \left (\frac{c (a x+b)^2}{x^2}\right )}{a}+\frac{48 b \text{PolyLog}\left (3,\frac{a x}{a x+b}\right )}{a}+x \log ^3\left (\frac{c (a x+b)^2}{x^2}\right )-\frac{6 b \log \left (1-\frac{a x}{a x+b}\right ) \log ^2\left (\frac{c (a x+b)^2}{x^2}\right )}{a} \]

[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

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Rubi [A]  time = 0.131848, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2486, 2488, 2506, 6610} \[ \frac{24 b \text{PolyLog}\left (2,1-\frac{b}{a x+b}\right ) \log \left (\frac{c (a x+b)^2}{x^2}\right )}{a}+\frac{48 b \text{PolyLog}\left (3,1-\frac{b}{a x+b}\right )}{a}+x \log ^3\left (\frac{c (a x+b)^2}{x^2}\right )-\frac{6 b \log \left (\frac{b}{a x+b}\right ) \log ^2\left (\frac{c (a x+b)^2}{x^2}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[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, 1 - b/(b + a*x)])/a + (48*b*PolyLog[3, 1 - b/(b + a*x)])/a

Rule 2486

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p*
(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] &&
EqQ[p + q, 0] && IGtQ[s, 0]

Rule 2488

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_)),
x_Symbol] :> -Simp[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/h, x] + Dist[(p
*r*s*(b*c - a*d))/h, Int[(Log[-((b*c - a*d)/(d*(a + b*x)))]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a
+ b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q,
0] && EqQ[b*g - a*h, 0] && IGtQ[s, 0]

Rule 2506

Int[Log[v_]*Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_), x_Symbo
l] :> With[{g = Simplify[((v - 1)*(c + d*x))/(a + b*x)], h = Simplify[u*(a + b*x)*(c + d*x)]}, -Simp[(h*PolyLo
g[2, 1 - v]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(b*c - a*d), x] + Dist[h*p*r*s, Int[(PolyLog[2, 1 - v]*Log
[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{g, h}, x]] /; FreeQ[{a, b,
c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && EqQ[p + q, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \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 )+(6 b) \int \frac{\log ^2\left (\frac{c (b+a x)^2}{x^2}\right )}{b+a x} \, 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{\left (24 b^2\right ) \int \frac{\log \left (\frac{b}{b+a x}\right ) \log \left (\frac{c (b+a x)^2}{x^2}\right )}{x (b+a x)} \, dx}{a}\\ &=-\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 ) \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}\\ &=-\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 ) \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 = 0.0245463, size = 98, normalized size = 0.96 \[ \frac{24 b \text{PolyLog}\left (2,\frac{a x}{a x+b}\right ) \log \left (\frac{c (a x+b)^2}{x^2}\right )}{a}+\frac{48 b \text{PolyLog}\left (3,\frac{a x}{a x+b}\right )}{a}+x \log ^3\left (\frac{c (a x+b)^2}{x^2}\right )-\frac{6 b \log \left (\frac{b}{a x+b}\right ) \log ^2\left (\frac{c (a x+b)^2}{x^2}\right )}{a} \]

Antiderivative was successfully verified.

[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

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Maple [F]  time = 0.504, size = 0, normalized size = 0. \begin{align*} \int \left ( \ln \left ({\frac{c \left ( ax+b \right ) ^{2}}{{x}^{2}}} \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{4 \,{\left (2 \,{\left (a x + b\right )} \log \left (a x + b\right )^{3} + 3 \,{\left (a x \log \left (c\right ) - 2 \, a x \log \left (x\right )\right )} \log \left (a x + b\right )^{2}\right )}}{a} + \int \frac{a x \log \left (c\right )^{3} + b \log \left (c\right )^{3} - 8 \,{\left (a x + b\right )} \log \left (x\right )^{3} + 12 \,{\left (a x \log \left (c\right ) + b \log \left (c\right )\right )} \log \left (x\right )^{2} + 6 \,{\left ({\left (\log \left (c\right )^{2} - 4 \, \log \left (c\right )\right )} a x + b \log \left (c\right )^{2} + 4 \,{\left (a x + b\right )} \log \left (x\right )^{2} - 4 \,{\left (a x{\left (\log \left (c\right ) - 2\right )} + b \log \left (c\right )\right )} \log \left (x\right )\right )} \log \left (a x + b\right ) - 6 \,{\left (a x \log \left (c\right )^{2} + b \log \left (c\right )^{2}\right )} \log \left (x\right )}{a x + b}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\log \left (\frac{a^{2} c x^{2} + 2 \, a b c x + b^{2} c}{x^{2}}\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (\frac{{\left (a x + b\right )}^{2} c}{x^{2}}\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)