\(\int (\frac {b}{x}+\frac {1}{x^2 (1+b x)}) \, dx\) [283]

Optimal result

Integrand size = 17, antiderivative size = 14 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=-\frac {1}{x}+b \log (1+b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {46} \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=b \log (b x+1)-\frac {1}{x} \]

[In]

Int[b/x + 1/(x^2*(1 + b*x)),x]

[Out]

-x^(-1) + b*Log[1 + b*x]

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=-\frac {1}{x}+b \log (1+b x) \]

[In]

Integrate[b/x + 1/(x^2*(1 + b*x)),x]

[Out]

-x^(-1) + b*Log[1 + b*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07

method	result	size
default	\(-\frac {1}{x}+b \ln \left (b x +1\right )\)	\(15\)
norman	\(-\frac {1}{x}+b \ln \left (b x +1\right )\)	\(15\)
risch	\(-\frac {1}{x}+b \ln \left (-b x -1\right )\)	\(16\)
parallelrisch	\(\frac {b \ln \left (b x +1\right ) x -1}{x}\)	\(16\)

[In]

int(b/x+1/x^2/(b*x+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=\frac {b x \log \left (b x + 1\right ) - 1}{x} \]

[In]

integrate(b/x+1/x^2/(b*x+1),x, algorithm="fricas")

[Out]

(b*x*log(b*x + 1) - 1)/x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=b \log {\left (b x + 1 \right )} - \frac {1}{x} \]

[In]

integrate(b/x+1/x**2/(b*x+1),x)

[Out]

b*log(b*x + 1) - 1/x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=b \log \left (b x + 1\right ) - \frac {1}{x} \]

[In]

integrate(b/x+1/x^2/(b*x+1),x, algorithm="maxima")

[Out]

b*log(b*x + 1) - 1/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.07 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=b \log \left ({\left | b x + 1 \right |}\right ) - \frac {1}{x} \]

[In]

integrate(b/x+1/x^2/(b*x+1),x, algorithm="giac")

[Out]

b*log(abs(b*x + 1)) - 1/x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \left (\frac {b}{x}+\frac {1}{x^2 (1+b x)}\right ) \, dx=b\,\ln \left (x\right )+2\,b\,\mathrm {atanh}\left (2\,b\,x+1\right )-\frac {1}{x} \]

[In]

int(1/(x^2*(b*x + 1)) + b/x,x)

[Out]

b*log(x) + 2*b*atanh(2*b*x + 1) - 1/x