\(\int \frac {1}{x^3 (a+b x)^2} \, dx\) [177]

Optimal result

Integrand size = 11, antiderivative size = 58 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=-\frac {1}{2 a^2 x^2}+\frac {2 b}{a^3 x}+\frac {b^2}{a^3 (a+b x)}+\frac {3 b^2 \log (x)}{a^4}-\frac {3 b^2 \log (a+b x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, 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} \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {3 b^2 \log (x)}{a^4}-\frac {3 b^2 \log (a+b x)}{a^4}+\frac {b^2}{a^3 (a+b x)}+\frac {2 b}{a^3 x}-\frac {1}{2 a^2 x^2} \]

[In]

Int[1/(x^3*(a + b*x)^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {a \left (-\frac {a}{x^2}+\frac {4 b}{x}+\frac {2 b^2}{a+b x}\right )+6 b^2 \log (x)-6 b^2 \log (a+b x)}{2 a^4} \]

[In]

Integrate[1/(x^3*(a + b*x)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {1}{2 a^{2} x^{2}}+\frac {2 b}{a^{3} x}+\frac {b^{2}}{a^{3} \left (b x +a \right )}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}-\frac {3 b^{2} \ln \left (b x +a \right )}{a^{4}}\)	\(57\)
norman	\(\frac {-\frac {3 b^{3} x^{3}}{a^{4}}-\frac {1}{2 a}+\frac {3 b x}{2 a^{2}}}{x^{2} \left (b x +a \right )}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}-\frac {3 b^{2} \ln \left (b x +a \right )}{a^{4}}\)	\(61\)
risch	\(\frac {\frac {3 b^{2} x^{2}}{a^{3}}+\frac {3 b x}{2 a^{2}}-\frac {1}{2 a}}{x^{2} \left (b x +a \right )}+\frac {3 b^{2} \ln \left (-x \right )}{a^{4}}-\frac {3 b^{2} \ln \left (b x +a \right )}{a^{4}}\)	\(63\)
parallelrisch	\(\frac {6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+6 a \,b^{2} \ln \left (x \right ) x^{2}-6 \ln \left (b x +a \right ) x^{2} a \,b^{2}-6 b^{3} x^{3}+3 a^{2} b x -a^{3}}{2 a^{4} x^{2} \left (b x +a \right )}\)	\(87\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {6 \, a b^{2} x^{2} + 3 \, a^{2} b x - a^{3} - 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {- a^{2} + 3 a b x + 6 b^{2} x^{2}}{2 a^{4} x^{2} + 2 a^{3} b x^{3}} + \frac {3 b^{2} \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {6 \, b^{2} x^{2} + 3 \, a b x - a^{2}}{2 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}} - \frac {3 \, b^{2} \log \left (b x + a\right )}{a^{4}} + \frac {3 \, b^{2} \log \left (x\right )}{a^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {3 \, b^{2} \log \left ({\left | -\frac {a}{b x + a} + 1 \right |}\right )}{a^{4}} + \frac {b^{2}}{{\left (b x + a\right )} a^{3}} - \frac {\frac {6 \, a b^{2}}{b x + a} - 5 \, b^{2}}{2 \, a^{4} {\left (\frac {a}{b x + a} - 1\right )}^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^3 (a+b x)^2} \, dx=\frac {\frac {3\,b^2\,x^2}{a^3}-\frac {1}{2\,a}+\frac {3\,b\,x}{2\,a^2}}{b\,x^3+a\,x^2}-\frac {6\,b^2\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^4} \]

[In]

int(1/(x^3*(a + b*x)^2),x)

[Out]

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