3.297 \(\int \frac{1}{x^3 (1-2 x^4+x^8)} \, dx\)

Optimal. Leaf size=32 \[ \frac{1}{4 x^2 \left (1-x^4\right )}-\frac{3}{4 x^2}+\frac{3}{4} \tanh ^{-1}\left (x^2\right ) \]

[Out]

-3/(4*x^2) + 1/(4*x^2*(1 - x^4)) + (3*ArcTanh[x^2])/4

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Rubi [A]  time = 0.0126115, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {28, 275, 290, 325, 207} \[ \frac{1}{4 x^2 \left (1-x^4\right )}-\frac{3}{4 x^2}+\frac{3}{4} \tanh ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 - 2*x^4 + x^8)),x]

[Out]

-3/(4*x^2) + 1/(4*x^2*(1 - x^4)) + (3*ArcTanh[x^2])/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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1-2 x^4+x^8\right )} \, dx &=\int \frac{1}{x^3 \left (-1+x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (-1+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{1}{4 x^2 \left (1-x^4\right )}-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (-1+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{3}{4 x^2}+\frac{1}{4 x^2 \left (1-x^4\right )}-\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,x^2\right )\\ &=-\frac{3}{4 x^2}+\frac{1}{4 x^2 \left (1-x^4\right )}+\frac{3}{4} \tanh ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.017206, size = 41, normalized size = 1.28 \[ \frac{1}{8} \left (\frac{4-6 x^4}{x^2 \left (x^4-1\right )}-3 \log \left (1-x^2\right )+3 \log \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 - 2*x^4 + x^8)),x]

[Out]

((4 - 6*x^4)/(x^2*(-1 + x^4)) - 3*Log[1 - x^2] + 3*Log[1 + x^2])/8

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Maple [A]  time = 0.017, size = 50, normalized size = 1.6 \begin{align*} -{\frac{1}{8\,{x}^{2}+8}}+{\frac{3\,\ln \left ({x}^{2}+1 \right ) }{8}}-{\frac{1}{2\,{x}^{2}}}+{\frac{1}{16+16\,x}}-{\frac{3\,\ln \left ( 1+x \right ) }{8}}-{\frac{1}{16\,x-16}}-{\frac{3\,\ln \left ( x-1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^8-2*x^4+1),x)

[Out]

-1/8/(x^2+1)+3/8*ln(x^2+1)-1/2/x^2+1/16/(1+x)-3/8*ln(1+x)-1/16/(x-1)-3/8*ln(x-1)

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Maxima [A]  time = 1.01926, size = 50, normalized size = 1.56 \begin{align*} -\frac{3 \, x^{4} - 2}{4 \,{\left (x^{6} - x^{2}\right )}} + \frac{3}{8} \, \log \left (x^{2} + 1\right ) - \frac{3}{8} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/4*(3*x^4 - 2)/(x^6 - x^2) + 3/8*log(x^2 + 1) - 3/8*log(x^2 - 1)

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Fricas [B]  time = 1.42239, size = 119, normalized size = 3.72 \begin{align*} -\frac{6 \, x^{4} - 3 \,{\left (x^{6} - x^{2}\right )} \log \left (x^{2} + 1\right ) + 3 \,{\left (x^{6} - x^{2}\right )} \log \left (x^{2} - 1\right ) - 4}{8 \,{\left (x^{6} - x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/8*(6*x^4 - 3*(x^6 - x^2)*log(x^2 + 1) + 3*(x^6 - x^2)*log(x^2 - 1) - 4)/(x^6 - x^2)

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Sympy [A]  time = 0.155546, size = 36, normalized size = 1.12 \begin{align*} - \frac{3 x^{4} - 2}{4 x^{6} - 4 x^{2}} - \frac{3 \log{\left (x^{2} - 1 \right )}}{8} + \frac{3 \log{\left (x^{2} + 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(x**8-2*x**4+1),x)

[Out]

-(3*x**4 - 2)/(4*x**6 - 4*x**2) - 3*log(x**2 - 1)/8 + 3*log(x**2 + 1)/8

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Giac [A]  time = 1.12248, size = 51, normalized size = 1.59 \begin{align*} -\frac{3 \, x^{4} - 2}{4 \,{\left (x^{6} - x^{2}\right )}} + \frac{3}{8} \, \log \left (x^{2} + 1\right ) - \frac{3}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4*(3*x^4 - 2)/(x^6 - x^2) + 3/8*log(x^2 + 1) - 3/8*log(abs(x^2 - 1))