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

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

[Out]

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

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Rubi [A]  time = 0.0066862, 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} \[ x \log \left (\frac{c x^2}{(a x+b)^2}\right )-\frac{2 b \log (a x+b)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[(c*x^2)/(b + a*x)^2],x]

[Out]

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

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

Mathematica [A]  time = 0.0028988, size = 28, normalized size = 1. \[ x \log \left (\frac{c x^2}{(a x+b)^2}\right )-\frac{2 b \log (a x+b)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[(c*x^2)/(b + a*x)^2],x]

[Out]

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

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Maple [B]  time = 0.224, size = 79, normalized size = 2.8 \begin{align*} \ln \left ({\frac{c}{{a}^{2}} \left ({\frac{b}{ax+b}}-1 \right ) ^{2}} \right ) x+2\,{\frac{b\ln \left ( \left ( ax+b \right ) ^{-1} \right ) }{a}}-2\,{\frac{b}{a}\ln \left ({\frac{b}{ax+b}}-1 \right ) }+{\frac{b}{a}\ln \left ({\frac{c}{{a}^{2}} \left ({\frac{b}{ax+b}}-1 \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.24105, size = 38, normalized size = 1.36 \begin{align*} x \log \left (\frac{c x^{2}}{{\left (a x + b\right )}^{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*x^2/(a*x+b)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.96246, size = 88, normalized size = 3.14 \begin{align*} \frac{a x \log \left (\frac{c x^{2}}{a^{2} x^{2} + 2 \, a b x + b^{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*x^2/(a*x+b)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.363537, size = 26, normalized size = 0.93 \begin{align*} x \log{\left (\frac{c x^{2}}{\left (a x + b\right )^{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*x**2/(a*x+b)**2),x)

[Out]

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

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Giac [A]  time = 1.26297, size = 39, normalized size = 1.39 \begin{align*} x \log \left (\frac{c x^{2}}{{\left (a x + b\right )}^{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*x^2/(a*x+b)^2),x, algorithm="giac")

[Out]

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