3.4.6 \(\int \frac {1}{x^4 (1-2 x^4+x^8)} \, dx\) [306]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, 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, 296, 331, 218, 212, 209} \begin {gather*} \frac {7 \text {ArcTan}(x)}{8}-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1-x^4\right )}+\frac {7}{8} \tanh ^{-1}(x) \end {gather*}

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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]

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*}

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.06 \begin {gather*} \frac {1}{48} \left (-\frac {16}{x^3}-\frac {12 x}{-1+x^4}+42 \tan ^{-1}(x)-21 \log (1-x)+21 \log (1+x)\right ) \end {gather*}

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 = 0.03, size = 47, normalized size = 1.31

method	result	size
risch	\(\frac {-\frac {7 x^{4}}{12}+\frac {1}{3}}{x^{3} \left (x^{4}-1\right )}-\frac {7 \ln \left (-1+x \right )}{16}+\frac {7 \arctan \left (x \right )}{8}+\frac {7 \ln \left (1+x \right )}{16}\)	\(36\)
default	\(-\frac {1}{16 \left (-1+x \right )}-\frac {7 \ln \left (-1+x \right )}{16}+\frac {x}{8 x^{2}+8}+\frac {7 \arctan \left (x \right )}{8}-\frac {1}{3 x^{3}}-\frac {1}{16 \left (1+x \right )}+\frac {7 \ln \left (1+x \right )}{16}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 37, normalized size = 1.03 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (26) = 52\).
time = 0.32, size = 63, normalized size = 1.75 \begin {gather*} -\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 {gather*}

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 = 0.09, size = 39, normalized size = 1.08 \begin {gather*} \frac {4 - 7 x^{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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4 - 7*x**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 = 3.28, size = 34, normalized size = 0.94 \begin {gather*} -\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 {gather*}

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))

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Mupad [B]
time = 1.29, size = 28, normalized size = 0.78 \begin {gather*} \frac {7\,\mathrm {atan}\left (x\right )}{8}+\frac {7\,\mathrm {atanh}\left (x\right )}{8}+\frac {\frac {7\,x^4}{12}-\frac {1}{3}}{x^3-x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^8 - 2*x^4 + 1)),x)

[Out]

(7*atan(x))/8 + (7*atanh(x))/8 + ((7*x^4)/12 - 1/3)/(x^3 - x^7)

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