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

Optimal. Leaf size=36 \[ \frac{1}{4 x^3 \left (1-x^4\right )}-\frac{7}{12 x^3}+\frac{7}{8} \tan ^{-1}(x)+\frac{7}{8} \tanh ^{-1}(x) \]

[Out]

-7/(12*x^3) + 1/(4*x^3*(1 - x^4)) + (7*ArcTan[x])/8 + (7*ArcTanh[x])/8

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Rubi [A]  time = 0.0088647, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {28, 290, 325, 212, 206, 203} \[ \frac{1}{4 x^3 \left (1-x^4\right )}-\frac{7}{12 x^3}+\frac{7}{8} \tan ^{-1}(x)+\frac{7}{8} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(1 - 2*x^4 + x^8)),x]

[Out]

-7/(12*x^3) + 1/(4*x^3*(1 - x^4)) + (7*ArcTan[x])/8 + (7*ArcTanh[x])/8

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 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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (1-2 x^4+x^8\right )} \, dx &=\int \frac{1}{x^4 \left (-1+x^4\right )^2} \, dx\\ &=\frac{1}{4 x^3 \left (1-x^4\right )}-\frac{7}{4} \int \frac{1}{x^4 \left (-1+x^4\right )} \, dx\\ &=-\frac{7}{12 x^3}+\frac{1}{4 x^3 \left (1-x^4\right )}-\frac{7}{4} \int \frac{1}{-1+x^4} \, dx\\ &=-\frac{7}{12 x^3}+\frac{1}{4 x^3 \left (1-x^4\right )}+\frac{7}{8} \int \frac{1}{1-x^2} \, dx+\frac{7}{8} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{7}{12 x^3}+\frac{1}{4 x^3 \left (1-x^4\right )}+\frac{7}{8} \tan ^{-1}(x)+\frac{7}{8} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0186844, size = 38, normalized size = 1.06 \[ \frac{1}{48} \left (-\frac{12 x}{x^4-1}-\frac{16}{x^3}-21 \log (1-x)+21 \log (x+1)+42 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16/x^3 - (12*x)/(-1 + x^4) + 42*ArcTan[x] - 21*Log[1 - x] + 21*Log[1 + x])/48

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Maple [A]  time = 0.015, size = 47, normalized size = 1.3 \begin{align*}{\frac{x}{8\,{x}^{2}+8}}+{\frac{7\,\arctan \left ( x \right ) }{8}}-{\frac{1}{3\,{x}^{3}}}-{\frac{1}{16+16\,x}}+{\frac{7\,\ln \left ( 1+x \right ) }{16}}-{\frac{1}{16\,x-16}}-{\frac{7\,\ln \left ( x-1 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x/(x^2+1)+7/8*arctan(x)-1/3/x^3-1/16/(1+x)+7/16*ln(1+x)-1/16/(x-1)-7/16*ln(x-1)

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Maxima [A]  time = 1.60184, size = 50, normalized size = 1.39 \begin{align*} -\frac{7 \, x^{4} - 4}{12 \,{\left (x^{7} - x^{3}\right )}} + \frac{7}{8} \, \arctan \left (x\right ) + \frac{7}{16} \, \log \left (x + 1\right ) - \frac{7}{16} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(7*x^4 - 4)/(x^7 - x^3) + 7/8*arctan(x) + 7/16*log(x + 1) - 7/16*log(x - 1)

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Fricas [B]  time = 1.46941, size = 157, normalized size = 4.36 \begin{align*} -\frac{28 \, x^{4} - 42 \,{\left (x^{7} - x^{3}\right )} \arctan \left (x\right ) - 21 \,{\left (x^{7} - x^{3}\right )} \log \left (x + 1\right ) + 21 \,{\left (x^{7} - x^{3}\right )} \log \left (x - 1\right ) - 16}{48 \,{\left (x^{7} - x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(28*x^4 - 42*(x^7 - x^3)*arctan(x) - 21*(x^7 - x^3)*log(x + 1) + 21*(x^7 - x^3)*log(x - 1) - 16)/(x^7 -
x^3)

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Sympy [A]  time = 0.187099, size = 39, normalized size = 1.08 \begin{align*} - \frac{7 x^{4} - 4}{12 x^{7} - 12 x^{3}} - \frac{7 \log{\left (x - 1 \right )}}{16} + \frac{7 \log{\left (x + 1 \right )}}{16} + \frac{7 \operatorname{atan}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(7*x**4 - 4)/(12*x**7 - 12*x**3) - 7*log(x - 1)/16 + 7*log(x + 1)/16 + 7*atan(x)/8

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Giac [A]  time = 1.11398, size = 46, normalized size = 1.28 \begin{align*} -\frac{x}{4 \,{\left (x^{4} - 1\right )}} - \frac{1}{3 \, x^{3}} + \frac{7}{8} \, \arctan \left (x\right ) + \frac{7}{16} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{7}{16} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x/(x^4 - 1) - 1/3/x^3 + 7/8*arctan(x) + 7/16*log(abs(x + 1)) - 7/16*log(abs(x - 1))