3.1.100 \(\int \log ^2(\frac {c x^2}{(b+a x)^2}) \, dx\) [100]

Optimal. Leaf size=67 \[ x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {8 b \text {Li}_2\left (1-\frac {b}{b+a x}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2536, 2542, 2458, 2378, 2370, 2352} \begin {gather*} \frac {8 b \text {PolyLog}\left (2,1-\frac {b}{a x+b}\right )}{a}+x \log ^2\left (\frac {c x^2}{(a x+b)^2}\right )+\frac {4 b \log \left (\frac {b}{a x+b}\right ) \log \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]^2,x]

[Out]

x*Log[(c*x^2)/(b + a*x)^2]^2 + (4*b*Log[(c*x^2)/(b + a*x)^2]*Log[b/(b + a*x)])/a + (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*} \int \log ^2\left (\frac {c x^2}{(b+a x)^2}\right ) \, dx &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )-(4 b) \int \frac {\log \left (\frac {c x^2}{(b+a x)^2}\right )}{b+a x} \, dx\\ &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}-\frac {\left (8 b^2\right ) \int \frac {\log \left (\frac {b}{b+a x}\right )}{x (b+a x)} \, dx}{a}\\ &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}-\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}\\ &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\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}\\ &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\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}\\ &=x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {8 b \text {Li}_2\left (\frac {a x}{b+a x}\right )}{a}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 106, normalized size = 1.58 \begin {gather*} x \log ^2\left (\frac {c x^2}{(b+a x)^2}\right )-\frac {8 b \log \left (-\frac {a x}{b}\right ) \log \left (\frac {b}{b+a x}\right )}{a}+\frac {4 b \log \left (\frac {c x^2}{(b+a x)^2}\right ) \log \left (\frac {b}{b+a x}\right )}{a}-\frac {4 b \log ^2\left (\frac {b}{b+a x}\right )}{a}+\frac {8 b \text {Li}_2\left (\frac {b+a x}{b}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 118, normalized size = 1.76 \begin {gather*} x \log \left (\frac {c x^{2}}{{\left (a x + b\right )}^{2}}\right )^{2} - \frac {4 \, b \log \left (a x + b\right ) \log \left (\frac {c x^{2}}{{\left (a x + b\right )}^{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - 4 b \int \frac {\log {\left (\frac {c x^{2}}{a^{2} x^{2} + 2 a b x + b^{2}} \right )}}{a x + b}\, dx + x \log {\left (\frac {c x^{2}}{\left (a x + b\right )^{2}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="giac")

[Out]

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

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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 )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*x^2)/(b + a*x)^2)^2,x)

[Out]

int(log((c*x^2)/(b + a*x)^2)^2, x)

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