### 3.248 $$\int \frac{\log (a+b x)}{(a+b x)^2} \, dx$$

Optimal. Leaf size=31 $-\frac{1}{b (a+b x)}-\frac{\log (a+b x)}{b (a+b x)}$

[Out]

-(1/(b*(a + b*x))) - Log[a + b*x]/(b*(a + b*x))

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Rubi [A]  time = 0.0218602, antiderivative size = 31, normalized size of antiderivative = 1., 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, 2304} $-\frac{1}{b (a+b x)}-\frac{\log (a+b x)}{b (a+b x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[Log[a + b*x]/(a + b*x)^2,x]

[Out]

-(1/(b*(a + b*x))) - Log[a + b*x]/(b*(a + b*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]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

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

Mathematica [A]  time = 0.0060834, size = 21, normalized size = 0.68 $-\frac{\log (a+b x)+1}{a b+b^2 x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Log[a + b*x]/(a + b*x)^2,x]

[Out]

-((1 + Log[a + b*x])/(a*b + b^2*x))

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Maple [A]  time = 0.003, size = 32, normalized size = 1. \begin{align*} -{\frac{1}{b \left ( bx+a \right ) }}-{\frac{\ln \left ( bx+a \right ) }{b \left ( bx+a \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b/(b*x+a)-ln(b*x+a)/b/(b*x+a)

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Maxima [A]  time = 0.998021, size = 42, normalized size = 1.35 \begin{align*} -\frac{\log \left (b x + a\right )}{{\left (b x + a\right )} b} - \frac{1}{{\left (b x + a\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)^2,x, algorithm="maxima")

[Out]

-log(b*x + a)/((b*x + a)*b) - 1/((b*x + a)*b)

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Fricas [A]  time = 1.82127, size = 47, normalized size = 1.52 \begin{align*} -\frac{\log \left (b x + a\right ) + 1}{b^{2} x + a b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)^2,x, algorithm="fricas")

[Out]

-(log(b*x + a) + 1)/(b^2*x + a*b)

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Sympy [A]  time = 0.345217, size = 26, normalized size = 0.84 \begin{align*} - \frac{\log{\left (a + b x \right )}}{a b + b^{2} x} - \frac{1}{a b + b^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(b*x+a)/(b*x+a)**2,x)

[Out]

-log(a + b*x)/(a*b + b**2*x) - 1/(a*b + b**2*x)

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Giac [A]  time = 1.31106, size = 42, normalized size = 1.35 \begin{align*} -\frac{\log \left (b x + a\right )}{{\left (b x + a\right )} b} - \frac{1}{{\left (b x + a\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(b*x+a)/(b*x+a)^2,x, algorithm="giac")

[Out]

-log(b*x + a)/((b*x + a)*b) - 1/((b*x + a)*b)