\(\int \frac {1}{x^3 (a^2+2 a b x^2+b^2 x^4)} \, dx\) [482]

Optimal result

Integrand size = 24, antiderivative size = 49 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {1}{2 a^2 x^2}-\frac {b}{2 a^2 \left (a+b x^2\right )}-\frac {2 b \log (x)}{a^3}+\frac {b \log \left (a+b x^2\right )}{a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 46} \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {b \log \left (a+b x^2\right )}{a^3}-\frac {2 b \log (x)}{a^3}-\frac {b}{2 a^2 \left (a+b x^2\right )}-\frac {1}{2 a^2 x^2} \]

[In]

Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {a \left (\frac {1}{x^2}+\frac {b}{a+b x^2}\right )+4 b \log (x)-2 b \log \left (a+b x^2\right )}{2 a^3} \]

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04

method	result	size
norman	\(\frac {-\frac {b \,x^{2}}{a^{2}}-\frac {1}{2 a}}{x^{2} \left (b \,x^{2}+a \right )}+\frac {b \ln \left (b \,x^{2}+a \right )}{a^{3}}-\frac {2 b \ln \left (x \right )}{a^{3}}\)	\(51\)
risch	\(\frac {-\frac {b \,x^{2}}{a^{2}}-\frac {1}{2 a}}{x^{2} \left (b \,x^{2}+a \right )}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {b \ln \left (-b \,x^{2}-a \right )}{a^{3}}\)	\(54\)
default	\(-\frac {1}{2 a^{2} x^{2}}-\frac {2 b \ln \left (x \right )}{a^{3}}+\frac {b^{2} \left (\frac {2 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{3}}\)	\(55\)
parallelrisch	\(-\frac {4 b^{2} \ln \left (x \right ) x^{4}-2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{2}-2 b^{2} x^{4}+4 a b \ln \left (x \right ) x^{2}-2 \ln \left (b \,x^{2}+a \right ) x^{2} a b +a^{2}}{2 a^{3} x^{2} \left (b \,x^{2}+a \right )}\)	\(80\)

[In]

int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.49 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {2 \, a b x^{2} + a^{2} - 2 \, {\left (b^{2} x^{4} + a b x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{2} x^{4} + a b x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {- a - 2 b x^{2}}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} - \frac {2 b \log {\left (x \right )}}{a^{3}} + \frac {b \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{3}} \]

[In]

integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {2 \, b x^{2} + a}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac {b \log \left (b x^{2} + a\right )}{a^{3}} - \frac {b \log \left (x^{2}\right )}{a^{3}} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {b \log \left (x^{2}\right )}{a^{3}} + \frac {b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{3}} - \frac {2 \, b x^{2} + a}{2 \, {\left (b x^{4} + a x^{2}\right )} a^{2}} \]

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {b\,\ln \left (b\,x^2+a\right )}{a^3}-\frac {\frac {1}{2\,a}+\frac {b\,x^2}{a^2}}{b\,x^4+a\,x^2}-\frac {2\,b\,\ln \left (x\right )}{a^3} \]

[In]

int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

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