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

Optimal. Leaf size=28 \[ x \log \left (\frac {c (a x+b)^2}{x^2}\right )+\frac {2 b \log (a x+b)}{a} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2486, 31} \[ x \log \left (\frac {c (a x+b)^2}{x^2}\right )+\frac {2 b \log (a x+b)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[(c*(b + a*x)^2)/x^2],x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

Rubi steps

\begin {align*} \int \log \left (\frac {c (b+a x)^2}{x^2}\right ) \, dx &=x \log \left (\frac {c (b+a x)^2}{x^2}\right )+(2 b) \int \frac {1}{b+a x} \, dx\\ &=\frac {2 b \log (b+a x)}{a}+x \log \left (\frac {c (b+a x)^2}{x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 28, normalized size = 1.00 \[ x \log \left (\frac {c (a x+b)^2}{x^2}\right )+\frac {2 b \log (a x+b)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(c*(b + a*x)^2)/x^2],x]

[Out]

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

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fricas [A]  time = 0.71, size = 42, normalized size = 1.50 \[ \frac {a x \log \left (\frac {a^{2} c x^{2} + 2 \, a b c x + b^{2} c}{x^{2}}\right ) + 2 \, b \log \left (a x + b\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 29, normalized size = 1.04 \[ x \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right ) + \frac {2 \, b \log \left ({\left | a x + b \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 40, normalized size = 1.43 \[ x \ln \left (\left (a +\frac {b}{x}\right )^{2} c \right )-\frac {2 b \ln \left (\frac {1}{x}\right )}{a}+\frac {2 b \ln \left (a +\frac {b}{x}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 28, normalized size = 1.00 \[ x \log \left (\frac {{\left (a x + b\right )}^{2} c}{x^{2}}\right ) + \frac {2 \, b \log \left (a x + b\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 28, normalized size = 1.00 \[ x\,\ln \left (\frac {c\,{\left (b+a\,x\right )}^2}{x^2}\right )+\frac {2\,b\,\ln \left (b+a\,x\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log((c*(b + a*x)^2)/x^2),x)

[Out]

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

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sympy [A]  time = 0.28, size = 26, normalized size = 0.93 \[ x \log {\left (\frac {c \left (a x + b\right )^{2}}{x^{2}} \right )} + \frac {2 b \log {\left (a x + b \right )}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*log(c*(a*x + b)**2/x**2) + 2*b*log(a*x + b)/a

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