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

Optimal. Leaf size=20 \[ \frac {\log ^3\left (\frac {c x}{a+b x}\right )}{3 a} \]

[Out]

1/3*ln(c*x/(b*x+a))^3/a

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Rubi [A]  time = 0.06, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2505} \[ \frac {\log ^3\left (\frac {c x}{a+b x}\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[(c*x)/(a + b*x)]^3/(3*a)

Rule 2505

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

Rubi steps

\begin {align*} \int \frac {\log ^2\left (\frac {c x}{a+b x}\right )}{x (a+b x)} \, dx &=\frac {\log ^3\left (\frac {c x}{a+b x}\right )}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 20, normalized size = 1.00 \[ \frac {\log ^3\left (\frac {c x}{a+b x}\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[(c*x)/(a + b*x)]^3/(3*a)

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fricas [A]  time = 0.68, size = 18, normalized size = 0.90 \[ \frac {\log \left (\frac {c x}{b x + a}\right )^{3}}{3 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*log(c*x/(b*x + a))^3/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 29, normalized size = 1.45 \[ \frac {\ln \left (-\frac {a c}{\left (b x +a \right ) b}+\frac {c}{b}\right )^{3}}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/a*ln(-1/(b*x+a)*a/b*c+1/b*c)^3

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maxima [B]  time = 0.61, size = 141, normalized size = 7.05 \[ -{\left (\frac {\log \left (b x + a\right )}{a} - \frac {\log \relax (x)}{a}\right )} \log \left (\frac {c x}{b x + a}\right )^{2} - \frac {{\left (c \log \left (b x + a\right )^{2} - 2 \, c \log \left (b x + a\right ) \log \relax (x) + c \log \relax (x)^{2}\right )} \log \left (\frac {c x}{b x + a}\right )}{a c} - \frac {c^{2} \log \left (b x + a\right )^{3} - 3 \, c^{2} \log \left (b x + a\right )^{2} \log \relax (x) + 3 \, c^{2} \log \left (b x + a\right ) \log \relax (x)^{2} - c^{2} \log \relax (x)^{3}}{3 \, a c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 18, normalized size = 0.90 \[ \frac {{\ln \left (\frac {c\,x}{a+b\,x}\right )}^3}{3\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((c*x)/(a + b*x))^3/(3*a)

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sympy [A]  time = 0.29, size = 14, normalized size = 0.70 \[ \frac {\log {\left (\frac {c x}{a + b x} \right )}^{3}}{3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x/(a + b*x))**3/(3*a)

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