\(\int \log ^2(\frac {c (b+a x)^2}{x^2}) \, dx\) [97]

Optimal result

Integrand size = 15, antiderivative size = 67 \[ \int \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=-\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {8 b \operatorname {PolyLog}\left (2,1-\frac {b}{b+a x}\right )}{a} \]

[Out]

-4*b*ln(b/(a*x+b))*ln(c*(a*x+b)^2/x^2)/a+x*ln(c*(a*x+b)^2/x^2)^2+8*b*polylog(2,1-b/(a*x+b))/a

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2536, 2542, 2458, 2378, 2370, 2352} \[ \int \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=x \log ^2\left (\frac {c (a x+b)^2}{x^2}\right )-\frac {4 b \log \left (\frac {b}{a x+b}\right ) \log \left (\frac {c (a x+b)^2}{x^2}\right )}{a}+\frac {8 b \operatorname {PolyLog}\left (2,1-\frac {b}{b+a x}\right )}{a} \]

[In]

Int[Log[(c*(b + a*x)^2)/x^2]^2,x]

[Out]

(-4*b*Log[b/(b + a*x)]*Log[(c*(b + a*x)^2)/x^2])/a + x*Log[(c*(b + a*x)^2)/x^2]^2 + (8*b*PolyLog[2, 1 - b/(b +
a*x)])/a

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2370

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

Rule 2378

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

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 2542

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

Rubi steps \begin{align*} \text {integral}& = x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+(4 b) \int \frac {\log \left (\frac {c (b+a x)^2}{x^2}\right )}{b+a x} \, dx \\ & = -\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {\left (8 b^2\right ) \int \frac {\log \left (\frac {b}{b+a x}\right )}{x (b+a x)} \, dx}{a} \\ & = -\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )-\frac {\left (8 b^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {b}{x}\right )}{x \left (-\frac {b}{a}+\frac {x}{a}\right )} \, dx,x,b+a x\right )}{a^2} \\ & = -\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {\left (8 b^2\right ) \text {Subst}\left (\int \frac {\log (b x)}{\left (-\frac {b}{a}+\frac {1}{a x}\right ) x} \, dx,x,\frac {1}{b+a x}\right )}{a^2} \\ & = -\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {\left (8 b^2\right ) \text {Subst}\left (\int \frac {\log (b x)}{\frac {1}{a}-\frac {b x}{a}} \, dx,x,\frac {1}{b+a x}\right )}{a^2} \\ & = -\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {8 b \text {Li}_2\left (\frac {a x}{b+a x}\right )}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.58 \[ \int \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=-\frac {8 b \log \left (-\frac {a x}{b}\right ) \log \left (\frac {b}{b+a x}\right )}{a}-\frac {4 b \log ^2\left (\frac {b}{b+a x}\right )}{a}-\frac {4 b \log \left (\frac {b}{b+a x}\right ) \log \left (\frac {c (b+a x)^2}{x^2}\right )}{a}+x \log ^2\left (\frac {c (b+a x)^2}{x^2}\right )+\frac {8 b \operatorname {PolyLog}\left (2,\frac {b+a x}{b}\right )}{a} \]

[In]

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

[Out]

(-8*b*Log[-((a*x)/b)]*Log[b/(b + a*x)])/a - (4*b*Log[b/(b + a*x)]^2)/a - (4*b*Log[b/(b + a*x)]*Log[(c*(b + a*x
)^2)/x^2])/a + x*Log[(c*(b + a*x)^2)/x^2]^2 + (8*b*PolyLog[2, (b + a*x)/b])/a

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

\[ \int \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=4 b \int \frac {\log {\left (a^{2} c + \frac {2 a b c}{x} + \frac {b^{2} c}{x^{2}} \right )}}{a x + b}\, dx + x \log {\left (\frac {c \left (a x + b\right )^{2}}{x^{2}} \right )}^{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.76 \[ \int \log ^2\left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=x \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right )^{2} + \frac {4 \, b \log \left (a x + b\right ) \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right )}{a} + \frac {4 \, {\left ({\left (\frac {c \log \left (a x + b\right )^{2}}{a} - \frac {2 \, {\left (\log \left (\frac {a x}{b} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {a x}{b}\right )\right )} c}{a}\right )} b - \frac {2 \, {\left (c \log \left (a x + b\right ) - c \log \left (x\right )\right )} b \log \left (a x + b\right )}{a}\right )}}{c} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(log((c*(b + a*x)^2)/x^2)^2,x)

[Out]

int(log((c*(b + a*x)^2)/x^2)^2, x)