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

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Rubi [A]  time = 0.0068001, antiderivative size = 28, normalized size of antiderivative = 1., 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 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]

Rule 31

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

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*}

Mathematica [A]  time = 0.0025771, size = 28, normalized size = 1. \[ x \log \left (\frac{c (a x+b)^2}{x^2}\right )+\frac{2 b \log (a x+b)}{a} \]

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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Maple [A]  time = 0.125, size = 40, normalized size = 1.4 \begin{align*} x\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{2} \right ) -2\,{\frac{b\ln \left ({x}^{-1} \right ) }{a}}+2\,{\frac{b}{a}\ln \left ( a+{\frac{b}{x}} \right ) } \end{align*}

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 = 1.19178, size = 38, normalized size = 1.36 \begin{align*} x \log \left (\frac{{\left (a x + b\right )}^{2} c}{x^{2}}\right ) + \frac{2 \, b \log \left (a x + b\right )}{a} \end{align*}

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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Fricas [A]  time = 1.87655, size = 93, normalized size = 3.32 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 0.311584, size = 26, normalized size = 0.93 \begin{align*} x \log{\left (\frac{c \left (a x + b\right )^{2}}{x^{2}} \right )} + \frac{2 b \log{\left (a x + b \right )}}{a} \end{align*}

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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Giac [A]  time = 1.14719, size = 39, normalized size = 1.39 \begin{align*} 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} \end{align*}

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