3.203 \(\int \frac {1}{x (b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=66 \[ -\frac {3 c^2 \log \left (b+c x^2\right )}{2 b^4}+\frac {3 c^2 \log (x)}{b^4}+\frac {c^2}{2 b^3 \left (b+c x^2\right )}+\frac {c}{b^3 x^2}-\frac {1}{4 b^2 x^4} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1584, 266, 44} \[ \frac {c^2}{2 b^3 \left (b+c x^2\right )}-\frac {3 c^2 \log \left (b+c x^2\right )}{2 b^4}+\frac {3 c^2 \log (x)}{b^4}+\frac {c}{b^3 x^2}-\frac {1}{4 b^2 x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(b*x^2 + c*x^4)^2),x]

[Out]

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

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 266

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]]

Rule 1584

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*} \int \frac {1}{x \left (b x^2+c x^4\right )^2} \, dx &=\int \frac {1}{x^5 \left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 (b+c x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{b^2 x^3}-\frac {2 c}{b^3 x^2}+\frac {3 c^2}{b^4 x}-\frac {c^3}{b^3 (b+c x)^2}-\frac {3 c^3}{b^4 (b+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{4 b^2 x^4}+\frac {c}{b^3 x^2}+\frac {c^2}{2 b^3 \left (b+c x^2\right )}+\frac {3 c^2 \log (x)}{b^4}-\frac {3 c^2 \log \left (b+c x^2\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 57, normalized size = 0.86 \[ \frac {-6 c^2 \log \left (b+c x^2\right )+b \left (\frac {2 c^2}{b+c x^2}-\frac {b}{x^4}+\frac {4 c}{x^2}\right )+12 c^2 \log (x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(b*x^2 + c*x^4)^2),x]

[Out]

(b*(-(b/x^4) + (4*c)/x^2 + (2*c^2)/(b + c*x^2)) + 12*c^2*Log[x] - 6*c^2*Log[b + c*x^2])/(4*b^4)

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fricas [A]  time = 0.65, size = 90, normalized size = 1.36 \[ \frac {6 \, b c^{2} x^{4} + 3 \, b^{2} c x^{2} - b^{3} - 6 \, {\left (c^{3} x^{6} + b c^{2} x^{4}\right )} \log \left (c x^{2} + b\right ) + 12 \, {\left (c^{3} x^{6} + b c^{2} x^{4}\right )} \log \relax (x)}{4 \, {\left (b^{4} c x^{6} + b^{5} x^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 86, normalized size = 1.30 \[ \frac {3 \, c^{2} \log \left (x^{2}\right )}{2 \, b^{4}} - \frac {3 \, c^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4}} + \frac {3 \, c^{3} x^{2} + 4 \, b c^{2}}{2 \, {\left (c x^{2} + b\right )} b^{4}} - \frac {9 \, c^{2} x^{4} - 4 \, b c x^{2} + b^{2}}{4 \, b^{4} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 61, normalized size = 0.92 \[ \frac {c^{2}}{2 \left (c \,x^{2}+b \right ) b^{3}}+\frac {3 c^{2} \ln \relax (x )}{b^{4}}-\frac {3 c^{2} \ln \left (c \,x^{2}+b \right )}{2 b^{4}}+\frac {c}{b^{3} x^{2}}-\frac {1}{4 b^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 70, normalized size = 1.06 \[ \frac {6 \, c^{2} x^{4} + 3 \, b c x^{2} - b^{2}}{4 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} - \frac {3 \, c^{2} \log \left (c x^{2} + b\right )}{2 \, b^{4}} + \frac {3 \, c^{2} \log \left (x^{2}\right )}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.17, size = 67, normalized size = 1.02 \[ \frac {\frac {3\,c\,x^2}{4\,b^2}-\frac {1}{4\,b}+\frac {3\,c^2\,x^4}{2\,b^3}}{c\,x^6+b\,x^4}-\frac {3\,c^2\,\ln \left (c\,x^2+b\right )}{2\,b^4}+\frac {3\,c^2\,\ln \relax (x)}{b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 + c*x^4)^2),x)

[Out]

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

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sympy [A]  time = 0.49, size = 68, normalized size = 1.03 \[ \frac {- b^{2} + 3 b c x^{2} + 6 c^{2} x^{4}}{4 b^{4} x^{4} + 4 b^{3} c x^{6}} + \frac {3 c^{2} \log {\relax (x )}}{b^{4}} - \frac {3 c^{2} \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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