\(\int \frac {1}{x^2 (a+b x)^2} \, dx\) [30]

Optimal result

Integrand size = 11, antiderivative size = 42 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {1}{a^2 x}-\frac {b}{a^2 (a+b x)}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log (a+b x)}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, 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} \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {2 b \log (x)}{a^3}+\frac {2 b \log (a+b x)}{a^3}-\frac {b}{a^2 (a+b x)}-\frac {1}{a^2 x} \]

[In]

Int[1/(x^2*(a + b*x)^2),x]

[Out]

-(1/(a^2*x)) - b/(a^2*(a + b*x)) - (2*b*Log[x])/a^3 + (2*b*Log[a + b*x])/a^3

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}{a^2 x^2}-\frac {2 b}{a^3 x}+\frac {b^2}{a^2 (a+b x)^2}+\frac {2 b^2}{a^3 (a+b x)}\right ) \, dx \\ & = -\frac {1}{a^2 x}-\frac {b}{a^2 (a+b x)}-\frac {2 b \log (x)}{a^3}+\frac {2 b \log (a+b x)}{a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {a \left (\frac {1}{x}+\frac {b}{a+b x}\right )+2 b \log (x)-2 b \log (a+b x)}{a^3} \]

[In]

Integrate[1/(x^2*(a + b*x)^2),x]

[Out]

-((a*(x^(-1) + b/(a + b*x)) + 2*b*Log[x] - 2*b*Log[a + b*x])/a^3)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {1}{a^{2} x}-\frac {b}{a^{2} \left (b x +a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (b x +a \right )}{a^{3}}\)	\(43\)
risch	\(\frac {-\frac {2 b x}{a^{2}}-\frac {1}{a}}{x \left (b x +a \right )}+\frac {2 b \ln \left (-b x -a \right )}{a^{3}}-\frac {2 b \ln \left (x \right )}{a^{3}}\)	\(49\)
norman	\(\frac {\frac {2 b^{2} x^{2}}{a^{3}}-\frac {1}{a}}{x \left (b x +a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {2 b \ln \left (b x +a \right )}{a^{3}}\)	\(50\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) x^{2} b^{2}-2 \ln \left (b x +a \right ) x^{2} b^{2}+2 \ln \left (x \right ) x a b -2 \ln \left (b x +a \right ) x a b -2 x^{2} b^{2}+a^{2}}{a^{3} x \left (b x +a \right )}\)	\(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {2 \, a b x + a^{2} - 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (b x + a\right ) + 2 \, {\left (b^{2} x^{2} + a b x\right )} \log \left (x\right )}{a^{3} b x^{2} + a^{4} x} \]

[In]

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

[Out]

-(2*a*b*x + a^2 - 2*(b^2*x^2 + a*b*x)*log(b*x + a) + 2*(b^2*x^2 + a*b*x)*log(x))/(a^3*b*x^2 + a^4*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=\frac {- a - 2 b x}{a^{3} x + a^{2} b x^{2}} + \frac {2 b \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {2 \, b x + a}{a^{2} b x^{2} + a^{3} x} + \frac {2 \, b \log \left (b x + a\right )}{a^{3}} - \frac {2 \, b \log \left (x\right )}{a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=-\frac {2 \, b \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{3}} - \frac {b}{{\left (b x + a\right )} a^{2}} + \frac {b}{a^{3} {\left (\frac {a}{b x + a} - 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^2 (a+b x)^2} \, dx=\frac {2\,b\,\ln \left (\frac {a+b\,x}{x}\right )}{a^3}-\frac {1}{a\,x\,\left (a+b\,x\right )}-\frac {2\,b}{a^2\,\left (a+b\,x\right )} \]

[In]

int(1/(x^2*(a + b*x)^2),x)

[Out]

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