\(\int \frac {\log (a+b x)}{(a+b x)^2} \, dx\) [248]

Optimal result

Integrand size = 14, antiderivative size = 31 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {1}{b (a+b x)}-\frac {\log (a+b x)}{b (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, 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, 2341} \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {1}{b (a+b x)}-\frac {\log (a+b x)}{b (a+b x)} \]

[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 2341

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
]

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^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] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {1+\log (a+b x)}{a b+b^2 x} \]

[In]

Integrate[Log[a + b*x]/(a + b*x)^2,x]

[Out]

-((1 + Log[a + b*x])/(a*b + b^2*x))

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
norman	\(\frac {\frac {x}{a}-\frac {\ln \left (b x +a \right )}{b}}{b x +a}\)	\(26\)
parallelrisch	\(\frac {-\ln \left (b x +a \right ) b^{2}-b^{2}}{\left (b x +a \right ) b^{3}}\)	\(29\)
derivativedivides	\(\frac {-\frac {\ln \left (b x +a \right )}{b x +a}-\frac {1}{b x +a}}{b}\)	\(30\)
default	\(\frac {-\frac {\ln \left (b x +a \right )}{b x +a}-\frac {1}{b x +a}}{b}\)	\(30\)
risch	\(-\frac {1}{b \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{b \left (b x +a \right )}\)	\(32\)
parts	\(-\frac {1}{b \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{b \left (b x +a \right )}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {\log \left (b x + a\right ) + 1}{b^{2} x + a b} \]

[In]

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

[Out]

-(log(b*x + a) + 1)/(b^2*x + a*b)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=- \frac {\log {\left (a + b x \right )}}{a b + b^{2} x} - \frac {1}{a b + b^{2} x} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {\log \left (b x + a\right )}{{\left (b x + a\right )} b} - \frac {1}{{\left (b x + a\right )} b} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-b {\left (\frac {\log \left (b x + a\right )}{{\left (b x + a\right )} b^{2}} + \frac {1}{{\left (b x + a\right )} b^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {\log (a+b x)}{(a+b x)^2} \, dx=-\frac {a+a\,\ln \left (a+b\,x\right )}{a\,b\,\left (a+b\,x\right )} \]

[In]

int(log(a + b*x)/(a + b*x)^2,x)

[Out]

-(a + a*log(a + b*x))/(a*b*(a + b*x))