\(\int \log (\frac {c (b+a x)^2}{x^2}) \, dx\) [96]

Optimal result

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

[Out]

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

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

[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 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 (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] (verified)

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

[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]

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {2 b \ln \left (a x +b \right )}{a}+x \ln \left (\frac {c \left (a x +b \right )^{2}}{x^{2}}\right )\)	\(29\)
parts	\(\frac {2 b \ln \left (a x +b \right )}{a}+x \ln \left (\frac {c \left (a x +b \right )^{2}}{x^{2}}\right )\)	\(29\)
derivativedivides	\(x \ln \left (c \left (a +\frac {b}{x}\right )^{2}\right )-2 b \left (\frac {\ln \left (\frac {1}{x}\right )}{a}-\frac {\ln \left (a +\frac {b}{x}\right )}{a}\right )\)	\(41\)
default	\(x \ln \left (c \left (a +\frac {b}{x}\right )^{2}\right )-2 b \left (\frac {\ln \left (\frac {1}{x}\right )}{a}-\frac {\ln \left (a +\frac {b}{x}\right )}{a}\right )\)	\(41\)
parallelrisch	\(-\frac {-2 \ln \left (\frac {c \left (a x +b \right )^{2}}{x^{2}}\right ) x a -4 \ln \left (x \right ) b -2 b \ln \left (\frac {c \left (a x +b \right )^{2}}{x^{2}}\right )}{2 a}\)	\(45\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \log \left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=\frac {a x \log \left (\frac {a^{2} c x^{2} + 2 \, a b c x + b^{2} c}{x^{2}}\right ) + 2 \, b \log \left (a x + b\right )}{a} \]

[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

Sympy [A] (verification not implemented)

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

[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

Maxima [A] (verification not implemented)

none

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

[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

Giac [A] (verification not implemented)

none

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

[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

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \log \left (\frac {c (b+a x)^2}{x^2}\right ) \, dx=x\,\ln \left (\frac {c\,{\left (b+a\,x\right )}^2}{x^2}\right )+\frac {2\,b\,\ln \left (b+a\,x\right )}{a} \]

[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