\(\int \frac {1}{x^2 (a+b x)^3} \, dx\) [189]

Optimal result

Integrand size = 11, antiderivative size = 57 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=-\frac {1}{a^3 x}-\frac {b}{2 a^2 (a+b x)^2}-\frac {2 b}{a^3 (a+b x)}-\frac {3 b \log (x)}{a^4}+\frac {3 b \log (a+b x)}{a^4} \]

[Out]

-1/a^3/x-1/2*b/a^2/(b*x+a)^2-2*b/a^3/(b*x+a)-3*b*ln(x)/a^4+3*b*ln(b*x+a)/a^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, 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} \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=-\frac {3 b \log (x)}{a^4}+\frac {3 b \log (a+b x)}{a^4}-\frac {2 b}{a^3 (a+b x)}-\frac {1}{a^3 x}-\frac {b}{2 a^2 (a+b x)^2} \]

[In]

Int[1/(x^2*(a + b*x)^3),x]

[Out]

-(1/(a^3*x)) - b/(2*a^2*(a + b*x)^2) - (2*b)/(a^3*(a + b*x)) - (3*b*Log[x])/a^4 + (3*b*Log[a + b*x])/a^4

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^3 x^2}-\frac {3 b}{a^4 x}+\frac {b^2}{a^2 (a+b x)^3}+\frac {2 b^2}{a^3 (a+b x)^2}+\frac {3 b^2}{a^4 (a+b x)}\right ) \, dx \\ & = -\frac {1}{a^3 x}-\frac {b}{2 a^2 (a+b x)^2}-\frac {2 b}{a^3 (a+b x)}-\frac {3 b \log (x)}{a^4}+\frac {3 b \log (a+b x)}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=-\frac {\frac {a \left (2 a^2+9 a b x+6 b^2 x^2\right )}{x (a+b x)^2}+6 b \log (x)-6 b \log (a+b x)}{2 a^4} \]

[In]

Integrate[1/(x^2*(a + b*x)^3),x]

[Out]

-1/2*((a*(2*a^2 + 9*a*b*x + 6*b^2*x^2))/(x*(a + b*x)^2) + 6*b*Log[x] - 6*b*Log[a + b*x])/a^4

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {1}{a^{3} x}-\frac {b}{2 a^{2} \left (b x +a \right )^{2}}-\frac {2 b}{a^{3} \left (b x +a \right )}-\frac {3 b \ln \left (x \right )}{a^{4}}+\frac {3 b \ln \left (b x +a \right )}{a^{4}}\)	\(56\)
risch	\(\frac {-\frac {3 b^{2} x^{2}}{a^{3}}-\frac {9 b x}{2 a^{2}}-\frac {1}{a}}{x \left (b x +a \right )^{2}}+\frac {3 b \ln \left (-b x -a \right )}{a^{4}}-\frac {3 b \ln \left (x \right )}{a^{4}}\)	\(60\)
norman	\(\frac {-\frac {1}{a}+\frac {6 b^{2} x^{2}}{a^{3}}+\frac {9 b^{3} x^{3}}{2 a^{4}}}{x \left (b x +a \right )^{2}}-\frac {3 b \ln \left (x \right )}{a^{4}}+\frac {3 b \ln \left (b x +a \right )}{a^{4}}\)	\(61\)
parallelrisch	\(-\frac {6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+12 a \,b^{2} \ln \left (x \right ) x^{2}-12 \ln \left (b x +a \right ) x^{2} a \,b^{2}-9 b^{3} x^{3}+6 a^{2} b \ln \left (x \right ) x -6 \ln \left (b x +a \right ) x \,a^{2} b -12 a \,b^{2} x^{2}+2 a^{3}}{2 a^{4} x \left (b x +a \right )^{2}}\)	\(111\)

[In]

int(1/x^2/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/a^3/x-1/2*b/a^2/(b*x+a)^2-2*b/a^3/(b*x+a)-3*b*ln(x)/a^4+3*b*ln(b*x+a)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.91 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=-\frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (x\right )}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

[In]

integrate(1/x^2/(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=\frac {- 2 a^{2} - 9 a b x - 6 b^{2} x^{2}}{2 a^{5} x + 4 a^{4} b x^{2} + 2 a^{3} b^{2} x^{3}} + \frac {3 b \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{4}} \]

[In]

integrate(1/x**2/(b*x+a)**3,x)

[Out]

(-2*a**2 - 9*a*b*x - 6*b**2*x**2)/(2*a**5*x + 4*a**4*b*x**2 + 2*a**3*b**2*x**3) + 3*b*(-log(x) + log(a/b + x))
/a**4

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=-\frac {6 \, b^{2} x^{2} + 9 \, a b x + 2 \, a^{2}}{2 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} + \frac {3 \, b \log \left (b x + a\right )}{a^{4}} - \frac {3 \, b \log \left (x\right )}{a^{4}} \]

[In]

integrate(1/x^2/(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(6*b^2*x^2 + 9*a*b*x + 2*a^2)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x) + 3*b*log(b*x + a)/a^4 - 3*b*log(x)/a^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=\frac {3 \, b \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} a^{4} x} \]

[In]

integrate(1/x^2/(b*x+a)^3,x, algorithm="giac")

[Out]

3*b*log(abs(b*x + a))/a^4 - 3*b*log(abs(x))/a^4 - 1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/((b*x + a)^2*a^4*x)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^2 (a+b x)^3} \, dx=\frac {6\,b\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^4}-\frac {\frac {1}{a}+\frac {3\,b^2\,x^2}{a^3}+\frac {9\,b\,x}{2\,a^2}}{a^2\,x+2\,a\,b\,x^2+b^2\,x^3} \]

[In]

int(1/(x^2*(a + b*x)^3),x)

[Out]

(6*b*atanh((2*b*x)/a + 1))/a^4 - (1/a + (3*b^2*x^2)/a^3 + (9*b*x)/(2*a^2))/(a^2*x + b^2*x^3 + 2*a*b*x^2)