\(\int \frac {\log (x)}{(a+b x)^2} \, dx\) [617]

Optimal result

Integrand size = 10, antiderivative size = 29 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=\frac {x \log (x)}{a (a+b x)}-\frac {\log (a+b x)}{a b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2351, 31} \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=\frac {x \log (x)}{a (a+b x)}-\frac {\log (a+b x)}{a b} \]

[In]

Int[Log[x]/(a + b*x)^2,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2351

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

Rubi steps \begin{align*} \text {integral}& = \frac {x \log (x)}{a (a+b x)}-\frac {\int \frac {1}{a+b x} \, dx}{a} \\ & = \frac {x \log (x)}{a (a+b x)}-\frac {\log (a+b x)}{a b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=\frac {\frac {x \log (x)}{a+b x}-\frac {\log (a+b x)}{b}}{a} \]

[In]

Integrate[Log[x]/(a + b*x)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {x \ln \left (x \right )}{a \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}\)	\(30\)
norman	\(\frac {x \ln \left (x \right )}{a \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}\)	\(30\)
parts	\(-\frac {\ln \left (x \right )}{b \left (b x +a \right )}+\frac {\frac {\ln \left (x \right )}{a}-\frac {\ln \left (b x +a \right )}{a}}{b}\)	\(38\)
parallelrisch	\(\frac {-b \ln \left (b x +a \right ) x +b x \ln \left (x \right )-\ln \left (b x +a \right ) a}{a b \left (b x +a \right )}\)	\(40\)
risch	\(-\frac {\ln \left (x \right )}{b \left (b x +a \right )}-\frac {\ln \left (b x +a \right )}{a b}+\frac {\ln \left (-x \right )}{a b}\)	\(41\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=\frac {b x \log \left (x\right ) - {\left (b x + a\right )} \log \left (b x + a\right )}{a b^{2} x + a^{2} b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=- \frac {\log {\left (x \right )}}{a b + b^{2} x} + \frac {\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}}{a b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=-\frac {\frac {\log \left (b x + a\right )}{a} - \frac {\log \left (x\right )}{a}}{b} - \frac {\log \left (x\right )}{{\left (b x + a\right )} b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (29) = 58\).

Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 4.76 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=b^{2} {\left (\frac {\log \left (\frac {{\left (b x + a\right )}^{2} {\left | b \right |} {\left | \frac {a}{b x + a} - 1 \right |}}{b^{2} {\left | b x + a \right |}}\right )}{a b^{3}} + \frac {\log \left (-\frac {a + \frac {{\left (b x + a\right )} b {\left (\frac {a}{b x + a} - 1\right )} - a b}{b}}{b}\right )}{{\left ({\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a\right )} b^{3}} - \frac {\log \left ({\left | -{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} + a \right |}\right )}{a b^{3}}\right )} \]

[In]

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

[Out]

b^2*(log((b*x + a)^2*abs(b)*abs(a/(b*x + a) - 1)/(b^2*abs(b*x + a)))/(a*b^3) + log(-(a + ((b*x + a)*b*(a/(b*x
+ a) - 1) - a*b)/b)/b)/(((b*x + a)*(a/(b*x + a) - 1) - a)*b^3) - log(abs(-(b*x + a)*(a/(b*x + a) - 1) + a))/(a
*b^3))

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\log (x)}{(a+b x)^2} \, dx=\frac {x^2\,\ln \left (x\right )}{a\,\left (b\,x^2+a\,x\right )}-\frac {\ln \left (a+b\,x\right )}{a\,b} \]

[In]

int(log(x)/(a + b*x)^2,x)

[Out]

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