\(\int \frac {\log (a+b x)}{a+b x} \, dx\) [247]

Optimal result

Integrand size = 14, antiderivative size = 15 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log ^2(a+b x)}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2437, 2338} \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log ^2(a+b x)}{2 b} \]

[In]

Int[Log[a + b*x]/(a + b*x),x]

[Out]

Log[a + b*x]^2/(2*b)

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,a+b x\right )}{b} \\ & = \frac {\log ^2(a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log ^2(a+b x)}{2 b} \]

[In]

Integrate[Log[a + b*x]/(a + b*x),x]

[Out]

Log[a + b*x]^2/(2*b)

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)
default	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)
norman	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)
risch	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)
parallelrisch	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)
parts	\(\frac {\ln \left (b x +a \right )^{2}}{2 b}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*log(b*x + a)^2/b

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log {\left (a + b x \right )}^{2}}{2 b} \]

[In]

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

[Out]

log(a + b*x)**2/(2*b)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*log(b*x + a)^2/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

[In]

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

[Out]

1/2*log(b*x + a)^2/b

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {\log (a+b x)}{a+b x} \, dx=\frac {{\ln \left (a+b\,x\right )}^2}{2\,b} \]

[In]

int(log(a + b*x)/(a + b*x),x)

[Out]

log(a + b*x)^2/(2*b)