\(\int \frac {x}{(a+b x)^2} \, dx\) [173]

Optimal result

Integrand size = 9, antiderivative size = 23 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {a}{b^2 (a+b x)}+\frac {\log (a+b x)}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {45} \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {a}{b^2 (a+b x)}+\frac {\log (a+b x)}{b^2} \]

[In]

Int[x/(a + b*x)^2,x]

[Out]

a/(b^2*(a + b*x)) + Log[a + b*x]/b^2

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {\frac {a}{a+b x}+\log (a+b x)}{b^2} \]

[In]

Integrate[x/(a + b*x)^2,x]

[Out]

(a/(a + b*x) + Log[a + b*x])/b^2

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {a}{b^{2} \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b^{2}}\)	\(24\)
norman	\(\frac {a}{b^{2} \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b^{2}}\)	\(24\)
risch	\(\frac {a}{b^{2} \left (b x +a \right )}+\frac {\ln \left (b x +a \right )}{b^{2}}\)	\(24\)
parallelrisch	\(\frac {b \ln \left (b x +a \right ) x +a \ln \left (b x +a \right )+a}{b^{2} \left (b x +a \right )}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {{\left (b x + a\right )} \log \left (b x + a\right ) + a}{b^{3} x + a b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {a}{a b^{2} + b^{3} x} + \frac {\log {\left (a + b x \right )}}{b^{2}} \]

[In]

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

[Out]

a/(a*b**2 + b**3*x) + log(a + b*x)/b**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {a}{b^{3} x + a b^{2}} + \frac {\log \left (b x + a\right )}{b^{2}} \]

[In]

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

[Out]

a/(b^3*x + a*b^2) + log(b*x + a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {x}{(a+b x)^2} \, dx=-\frac {\frac {\log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {a}{{\left (b x + a\right )} b}}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(a+b x)^2} \, dx=\frac {\ln \left (a+b\,x\right )}{b^2}+\frac {a}{b^2\,\left (a+b\,x\right )} \]

[In]

int(x/(a + b*x)^2,x)

[Out]

log(a + b*x)/b^2 + a/(b^2*(a + b*x))