\(\int \frac {1}{x^2 (a x^2+b x^3)} \, dx\) [221]

Optimal result

Integrand size = 17, antiderivative size = 56 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=-\frac {1}{3 a x^3}+\frac {b}{2 a^2 x^2}-\frac {b^2}{a^3 x}-\frac {b^3 \log (x)}{a^4}+\frac {b^3 \log (a+b x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 46} \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=-\frac {b^3 \log (x)}{a^4}+\frac {b^3 \log (a+b x)}{a^4}-\frac {b^2}{a^3 x}+\frac {b}{2 a^2 x^2}-\frac {1}{3 a x^3} \]

[In]

Int[1/(x^2*(a*x^2 + b*x^3)),x]

[Out]

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

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^4 (a+b x)} \, dx \\ & = \int \left (\frac {1}{a x^4}-\frac {b}{a^2 x^3}+\frac {b^2}{a^3 x^2}-\frac {b^3}{a^4 x}+\frac {b^4}{a^4 (a+b x)}\right ) \, dx \\ & = -\frac {1}{3 a x^3}+\frac {b}{2 a^2 x^2}-\frac {b^2}{a^3 x}-\frac {b^3 \log (x)}{a^4}+\frac {b^3 \log (a+b x)}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=-\frac {1}{3 a x^3}+\frac {b}{2 a^2 x^2}-\frac {b^2}{a^3 x}-\frac {b^3 \log (x)}{a^4}+\frac {b^3 \log (a+b x)}{a^4} \]

[In]

Integrate[1/(x^2*(a*x^2 + b*x^3)),x]

[Out]

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

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {1}{3 a \,x^{3}}+\frac {b}{2 a^{2} x^{2}}-\frac {b^{2}}{a^{3} x}-\frac {b^{3} \ln \left (x \right )}{a^{4}}+\frac {b^{3} \ln \left (b x +a \right )}{a^{4}}\)	\(53\)
norman	\(\frac {-\frac {1}{3 a}+\frac {b x}{2 a^{2}}-\frac {b^{2} x^{2}}{a^{3}}}{x^{3}}+\frac {b^{3} \ln \left (b x +a \right )}{a^{4}}-\frac {b^{3} \ln \left (x \right )}{a^{4}}\)	\(53\)
parallelrisch	\(-\frac {6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+6 a \,b^{2} x^{2}-3 a^{2} b x +2 a^{3}}{6 a^{4} x^{3}}\)	\(55\)
risch	\(\frac {-\frac {1}{3 a}+\frac {b x}{2 a^{2}}-\frac {b^{2} x^{2}}{a^{3}}}{x^{3}}-\frac {b^{3} \ln \left (x \right )}{a^{4}}+\frac {b^{3} \ln \left (-b x -a \right )}{a^{4}}\)	\(56\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=\frac {6 \, b^{3} x^{3} \log \left (b x + a\right ) - 6 \, b^{3} x^{3} \log \left (x\right ) - 6 \, a b^{2} x^{2} + 3 \, a^{2} b x - 2 \, a^{3}}{6 \, a^{4} x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=\frac {- 2 a^{2} + 3 a b x - 6 b^{2} x^{2}}{6 a^{3} x^{3}} + \frac {b^{3} \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=\frac {b^{3} \log \left (b x + a\right )}{a^{4}} - \frac {b^{3} \log \left (x\right )}{a^{4}} - \frac {6 \, b^{2} x^{2} - 3 \, a b x + 2 \, a^{2}}{6 \, a^{3} x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=\frac {b^{3} \log \left ({\left | b x + a \right |}\right )}{a^{4}} - \frac {b^{3} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {6 \, a b^{2} x^{2} - 3 \, a^{2} b x + 2 \, a^{3}}{6 \, a^{4} x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^2 \left (a x^2+b x^3\right )} \, dx=\frac {2\,b^3\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^4}-\frac {\frac {a^3}{3}-\frac {a^2\,b\,x}{2}+a\,b^2\,x^2}{a^4\,x^3} \]

[In]

int(1/(x^2*(a*x^2 + b*x^3)),x)

[Out]

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