3.5.99 \(\int \frac {1}{x (a^2+2 a b x^2+b^2 x^4)^2} \, dx\) [499]

Optimal. Leaf size=70 \[ \frac {1}{6 a \left (a+b x^2\right )^3}+\frac {1}{4 a^2 \left (a+b x^2\right )^2}+\frac {1}{2 a^3 \left (a+b x^2\right )}+\frac {\log (x)}{a^4}-\frac {\log \left (a+b x^2\right )}{2 a^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 272, 46} \begin {gather*} -\frac {\log \left (a+b x^2\right )}{2 a^4}+\frac {\log (x)}{a^4}+\frac {1}{2 a^3 \left (a+b x^2\right )}+\frac {1}{4 a^2 \left (a+b x^2\right )^2}+\frac {1}{6 a \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*a*(a + b*x^2)^3) + 1/(4*a^2*(a + b*x^2)^2) + 1/(2*a^3*(a + b*x^2)) + Log[x]/a^4 - Log[a + b*x^2]/(2*a^4)

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*} \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {1}{x \left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac {1}{2} b^4 \text {Subst}\left (\int \frac {1}{x \left (a b+b^2 x\right )^4} \, dx,x,x^2\right )\\ &=\frac {1}{2} b^4 \text {Subst}\left (\int \left (\frac {1}{a^4 b^4 x}-\frac {1}{a b^3 (a+b x)^4}-\frac {1}{a^2 b^3 (a+b x)^3}-\frac {1}{a^3 b^3 (a+b x)^2}-\frac {1}{a^4 b^3 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{6 a \left (a+b x^2\right )^3}+\frac {1}{4 a^2 \left (a+b x^2\right )^2}+\frac {1}{2 a^3 \left (a+b x^2\right )}+\frac {\log (x)}{a^4}-\frac {\log \left (a+b x^2\right )}{2 a^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 54, normalized size = 0.77 \begin {gather*} \frac {\frac {a \left (11 a^2+15 a b x^2+6 b^2 x^4\right )}{\left (a+b x^2\right )^3}+12 \log (x)-6 \log \left (a+b x^2\right )}{12 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(11*a^2 + 15*a*b*x^2 + 6*b^2*x^4))/(a + b*x^2)^3 + 12*Log[x] - 6*Log[a + b*x^2])/(12*a^4)

________________________________________________________________________________________

Maple [A]
time = 0.04, size = 76, normalized size = 1.09

method	result	size
norman	\(\frac {-\frac {3 b \,x^{2}}{2 a^{2}}-\frac {9 b^{2} x^{4}}{4 a^{3}}-\frac {11 b^{3} x^{6}}{12 a^{4}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\ln \left (x \right )}{a^{4}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{4}}\)	\(63\)
default	\(-\frac {b \left (-\frac {a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}-\frac {a}{b \left (b \,x^{2}+a \right )}-\frac {a^{3}}{3 b \left (b \,x^{2}+a \right )^{3}}+\frac {\ln \left (b \,x^{2}+a \right )}{b}\right )}{2 a^{4}}+\frac {\ln \left (x \right )}{a^{4}}\)	\(76\)
risch	\(\frac {\frac {b^{2} x^{4}}{2 a^{3}}+\frac {5 b \,x^{2}}{4 a^{2}}+\frac {11}{12 a}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {\ln \left (x \right )}{a^{4}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{4}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 82, normalized size = 1.17 \begin {gather*} \frac {6 \, b^{2} x^{4} + 15 \, a b x^{2} + 11 \, a^{2}}{12 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {\log \left (x^{2}\right )}{2 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*b^2*x^4 + 15*a*b*x^2 + 11*a^2)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6) - 1/2*log(b*x^2 + a)/
a^4 + 1/2*log(x^2)/a^4

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (62) = 124\).
time = 0.35, size = 134, normalized size = 1.91 \begin {gather*} \frac {6 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 11 \, a^{3} - 6 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right ) + 12 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (x\right )}{12 \, {\left (a^{4} b^{3} x^{6} + 3 \, a^{5} b^{2} x^{4} + 3 \, a^{6} b x^{2} + a^{7}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a*b^2*x^4 + 15*a^2*b*x^2 + 11*a^3 - 6*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(b*x^2 + a) + 12*
(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(x))/(a^4*b^3*x^6 + 3*a^5*b^2*x^4 + 3*a^6*b*x^2 + a^7)

________________________________________________________________________________________

Sympy [A]
time = 0.25, size = 80, normalized size = 1.14 \begin {gather*} \frac {11 a^{2} + 15 a b x^{2} + 6 b^{2} x^{4}}{12 a^{6} + 36 a^{5} b x^{2} + 36 a^{4} b^{2} x^{4} + 12 a^{3} b^{3} x^{6}} + \frac {\log {\left (x \right )}}{a^{4}} - \frac {\log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*a**2 + 15*a*b*x**2 + 6*b**2*x**4)/(12*a**6 + 36*a**5*b*x**2 + 36*a**4*b**2*x**4 + 12*a**3*b**3*x**6) + log
(x)/a**4 - log(a/b + x**2)/(2*a**4)

________________________________________________________________________________________

Giac [A]
time = 2.79, size = 70, normalized size = 1.00 \begin {gather*} \frac {\log \left (x^{2}\right )}{2 \, a^{4}} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4}} + \frac {11 \, b^{3} x^{6} + 39 \, a b^{2} x^{4} + 48 \, a^{2} b x^{2} + 22 \, a^{3}}{12 \, {\left (b x^{2} + a\right )}^{3} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(x^2)/a^4 - 1/2*log(abs(b*x^2 + a))/a^4 + 1/12*(11*b^3*x^6 + 39*a*b^2*x^4 + 48*a^2*b*x^2 + 22*a^3)/((b*
x^2 + a)^3*a^4)

________________________________________________________________________________________

Mupad [B]
time = 4.47, size = 78, normalized size = 1.11 \begin {gather*} \frac {\ln \left (x\right )}{a^4}+\frac {\frac {11}{12\,a}+\frac {5\,b\,x^2}{4\,a^2}+\frac {b^2\,x^4}{2\,a^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}-\frac {\ln \left (b\,x^2+a\right )}{2\,a^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/a^4 + (11/(12*a) + (5*b*x^2)/(4*a^2) + (b^2*x^4)/(2*a^3))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) -
log(a + b*x^2)/(2*a^4)

________________________________________________________________________________________