\(\int \frac {1}{x^2 (1+b x)} \, dx\) [281]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int \frac {1}{x^2 (1+b x)} \, dx=-\frac {1}{x}-b \log (x)+b \log (1+b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^2 (1+b x)} \, dx=-b \log (x)+b \log (b x+1)-\frac {1}{x} \]

[In]

Int[1/(x^2*(1 + b*x)),x]

[Out]

-x^(-1) - b*Log[x] + 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}& = \int \left (\frac {1}{x^2}-\frac {b}{x}+\frac {b^2}{1+b x}\right ) \, dx \\ & = -\frac {1}{x}-b \log (x)+b \log (1+b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 (1+b x)} \, dx=-\frac {1}{x}-b \log (x)+b \log (1+b x) \]

[In]

Integrate[1/(x^2*(1 + b*x)),x]

[Out]

-x^(-1) - b*Log[x] + b*Log[1 + b*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05

method	result	size
default	\(-\frac {1}{x}-b \ln \left (x \right )+b \ln \left (b x +1\right )\)	\(20\)
norman	\(-\frac {1}{x}-b \ln \left (x \right )+b \ln \left (b x +1\right )\)	\(20\)
risch	\(-\frac {1}{x}-b \ln \left (x \right )+b \ln \left (-b x -1\right )\)	\(21\)
parallelrisch	\(-\frac {b \ln \left (x \right ) x -b \ln \left (b x +1\right ) x +1}{x}\)	\(23\)
meijerg	\(b \left (-\frac {1}{x b}-\ln \left (x \right )-\ln \left (b \right )+\ln \left (b x +1\right )\right )\)	\(26\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 (1+b x)} \, dx=\frac {b x \log \left (b x + 1\right ) - b x \log \left (x\right ) - 1}{x} \]

[In]

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

[Out]

(b*x*log(b*x + 1) - b*x*log(x) - 1)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^2 (1+b x)} \, dx=b \left (- \log {\left (x \right )} + \log {\left (x + \frac {1}{b} \right )}\right ) - \frac {1}{x} \]

[In]

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

[Out]

b*(-log(x) + log(x + 1/b)) - 1/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 (1+b x)} \, dx=b \log \left (b x + 1\right ) - b \log \left (x\right ) - \frac {1}{x} \]

[In]

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

[Out]

b*log(b*x + 1) - b*log(x) - 1/x

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 (1+b x)} \, dx=b \log \left ({\left | b x + 1 \right |}\right ) - b \log \left ({\left | x \right |}\right ) - \frac {1}{x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 (1+b x)} \, dx=2\,b\,\mathrm {atanh}\left (2\,b\,x+1\right )-\frac {1}{x} \]

[In]

int(1/(x^2*(b*x + 1)),x)

[Out]

2*b*atanh(2*b*x + 1) - 1/x