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

Optimal. Leaf size=97 \[ -\frac {6 b \text {Li}_2\left (\frac {b}{a x}+1\right ) \log \left (c \left (a+\frac {b}{x}\right )\right )}{a}+\frac {(a x+b) \log ^3\left (a c+\frac {b c}{x}\right )}{a}-\frac {3 b \log \left (-\frac {b}{a x}\right ) \log ^2\left (c \left (a+\frac {b}{x}\right )\right )}{a}+\frac {6 b \text {Li}_3\left (\frac {b}{a x}+1\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2453, 2449, 2454, 2396, 2433, 2374, 6589} \[ -\frac {6 b \text {PolyLog}\left (2,\frac {b}{a x}+1\right ) \log \left (c \left (a+\frac {b}{x}\right )\right )}{a}+\frac {6 b \text {PolyLog}\left (3,\frac {b}{a x}+1\right )}{a}+\frac {(a x+b) \log ^3\left (a c+\frac {b c}{x}\right )}{a}-\frac {3 b \log \left (-\frac {b}{a x}\right ) \log ^2\left (c \left (a+\frac {b}{x}\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[(c*(b + a*x))/x]^3,x]

[Out]

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

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(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*Log
[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 2396

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

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2449

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

Rule 2453

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.), x_Symbol] :> Int[(a + b*Log[c*ExpandToSum[v, x]^p])^q, x] /;
FreeQ[{a, b, c, p, q}, x] && BinomialQ[v, x] &&  !BinomialMatchQ[v, x]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 6589

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

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Mathematica [A]  time = 0.03, size = 91, normalized size = 0.94 \[ \frac {-6 b \text {Li}_2\left (\frac {b}{a x}+1\right ) \log \left (\frac {c (a x+b)}{x}\right )+\left ((a x+b) \log \left (\frac {c (a x+b)}{x}\right )-3 b \log \left (-\frac {b}{a x}\right )\right ) \log ^2\left (\frac {c (a x+b)}{x}\right )+6 b \text {Li}_3\left (\frac {b}{a x}+1\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(c*(b + a*x))/x]^3,x]

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\log \left (\frac {a c x + b c}{x}\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((a*c*x + b*c)/x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((a*x + b)*c/x)^3, x)

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maple [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int \ln \left (\frac {\left (a x +b \right ) c}{x}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (a x + b\right )} \log \left (a x + b\right )^{3} + 3 \, {\left (a x \log \relax (c) - a x \log \relax (x)\right )} \log \left (a x + b\right )^{2}}{a} + \int \frac {a x \log \relax (c)^{3} + b \log \relax (c)^{3} - {\left (a x + b\right )} \log \relax (x)^{3} + 3 \, {\left (a x \log \relax (c) + b \log \relax (c)\right )} \log \relax (x)^{2} + 3 \, {\left ({\left (\log \relax (c)^{2} - 2 \, \log \relax (c)\right )} a x + b \log \relax (c)^{2} + {\left (a x + b\right )} \log \relax (x)^{2} - 2 \, {\left (a x {\left (\log \relax (c) - 1\right )} + b \log \relax (c)\right )} \log \relax (x)\right )} \log \left (a x + b\right ) - 3 \, {\left (a x \log \relax (c)^{2} + b \log \relax (c)^{2}\right )} \log \relax (x)}{a x + b}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*x + b)*log(a*x + b)^3 + 3*(a*x*log(c) - a*x*log(x))*log(a*x + b)^2)/a + integrate((a*x*log(c)^3 + b*log(c)
^3 - (a*x + b)*log(x)^3 + 3*(a*x*log(c) + b*log(c))*log(x)^2 + 3*((log(c)^2 - 2*log(c))*a*x + b*log(c)^2 + (a*
x + b)*log(x)^2 - 2*(a*x*(log(c) - 1) + b*log(c))*log(x))*log(a*x + b) - 3*(a*x*log(c)^2 + b*log(c)^2)*log(x))
/(a*x + b), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*(b + a*x))/x)^3,x)

[Out]

int(log((c*(b + a*x))/x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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