3.247 \(\int \frac {\log (a+b x)}{a+b x} \, dx\)

Optimal. Leaf size=15 \[ \frac {\log ^2(a+b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2390, 2301} \[ \frac {\log ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Log[a + b*x]/(a + b*x),x]

[Out]

Log[a + b*x]^2/(2*b)

Rule 2301

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 2390

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {\log ^2(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a + b*x]/(a + b*x),x]

[Out]

Log[a + b*x]^2/(2*b)

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fricas [A]  time = 0.52, size = 13, normalized size = 0.87 \[ \frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x + a)^2/b

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giac [A]  time = 0.16, size = 13, normalized size = 0.87 \[ \frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x + a)^2/b

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maple [A]  time = 0.06, size = 14, normalized size = 0.93 \[ \frac {\ln \left (b x +a \right )^{2}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.70, size = 13, normalized size = 0.87 \[ \frac {\log \left (b x + a\right )^{2}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(b*x + a)^2/b

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mupad [B]  time = 0.55, size = 13, normalized size = 0.87 \[ \frac {{\ln \left (a+b\,x\right )}^2}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a + b*x)/(a + b*x),x)

[Out]

log(a + b*x)^2/(2*b)

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sympy [A]  time = 0.26, size = 10, normalized size = 0.67 \[ \frac {\log {\left (a + b x \right )}^{2}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a + b*x)**2/(2*b)

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