\(\int \log (\frac {c x^2}{(b+a x)^2}) \, dx\) [99]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log \left (\frac {c x^2}{(b+a x)^2}\right )-\frac {2 b \log (b+a x)}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2536, 31} \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log \left (\frac {c x^2}{(a x+b)^2}\right )-\frac {2 b \log (a x+b)}{a} \]

[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 31

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

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]

Rubi steps \begin{align*} \text {integral}& = 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] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log \left (\frac {c x^2}{(b+a x)^2}\right )-\frac {2 b \log (b+a x)}{a} \]

[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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(x \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right )-\frac {2 b \ln \left (a x +b \right )}{a}\)	\(29\)
parts	\(x \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right )-\frac {2 b \ln \left (a x +b \right )}{a}\)	\(29\)
parallelrisch	\(-\frac {-2 \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right ) x a b +4 \ln \left (x \right ) b^{2}-2 b^{2} \ln \left (\frac {c \,x^{2}}{\left (a x +b \right )^{2}}\right )}{2 a b}\)	\(53\)
derivativedivides	\(-\frac {-\left (a x +b \right ) \ln \left (\frac {c \left (\frac {b}{a x +b}-1\right )^{2}}{a^{2}}\right )+2 b \left (-\ln \left (\frac {1}{a x +b}\right )+\ln \left (\frac {b}{a x +b}-1\right )\right )}{a}\)	\(59\)
default	\(-\frac {-\left (a x +b \right ) \ln \left (\frac {c \left (\frac {b}{a x +b}-1\right )^{2}}{a^{2}}\right )+2 b \left (-\ln \left (\frac {1}{a x +b}\right )+\ln \left (\frac {b}{a x +b}-1\right )\right )}{a}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=\frac {a x \log \left (\frac {c x^{2}}{a^{2} x^{2} + 2 \, a b x + b^{2}}\right ) - 2 \, b \log \left (a x + b\right )}{a} \]

[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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log {\left (\frac {c x^{2}}{\left (a x + b\right )^{2}} \right )} - \frac {2 b \log {\left (a x + b \right )}}{a} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log \left (\frac {c x^{2}}{{\left (a x + b\right )}^{2}}\right ) - \frac {2 \, b \log \left (a x + b\right )}{a} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x \log \left (\frac {c x^{2}}{{\left (a x + b\right )}^{2}}\right ) - \frac {2 \, b \log \left ({\left | a x + b \right |}\right )}{a} \]

[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

Mupad [B] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \log \left (\frac {c x^2}{(b+a x)^2}\right ) \, dx=x\,\ln \left (\frac {c\,x^2}{{\left (b+a\,x\right )}^2}\right )-\frac {2\,b\,\ln \left (b+a\,x\right )}{a} \]

[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