\(\int \frac {x^5}{1-2 x^4+x^8} \, dx\) [293]

Optimal result

Integrand size = 16, antiderivative size = 25 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {x^2}{4 \left (1-x^4\right )}-\frac {\text {arctanh}\left (x^2\right )}{4} \]

[Out]

1/4*x^2/(-x^4+1)-1/4*arctanh(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {28, 281, 294, 213} \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {x^2}{4 \left (1-x^4\right )}-\frac {\text {arctanh}\left (x^2\right )}{4} \]

[In]

Int[x^5/(1 - 2*x^4 + x^8),x]

[Out]

x^2/(4*(1 - x^4)) - 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 213

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

Rule 281

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 294

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^5}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{4 \left (1-x^4\right )}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{4 \left (1-x^4\right )}-\frac {1}{4} \tanh ^{-1}\left (x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=\frac {1}{8} \left (-\frac {2 x^2}{-1+x^4}+\log \left (1-x^2\right )-\log \left (1+x^2\right )\right ) \]

[In]

Integrate[x^5/(1 - 2*x^4 + x^8),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20

method	result	size
risch	\(-\frac {x^{2}}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x^{2}-1\right )}{8}-\frac {\ln \left (x^{2}+1\right )}{8}\)	\(30\)
norman	\(-\frac {x^{2}}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}-\frac {\ln \left (x^{2}+1\right )}{8}\)	\(34\)
default	\(-\frac {1}{8 \left (x^{2}+1\right )}-\frac {\ln \left (x^{2}+1\right )}{8}-\frac {1}{8 \left (x^{2}-1\right )}+\frac {\ln \left (x^{2}-1\right )}{8}\)	\(36\)
parallelrisch	\(\frac {\ln \left (x -1\right ) x^{4}+\ln \left (x +1\right ) x^{4}-\ln \left (x^{2}+1\right ) x^{4}-2 x^{2}-\ln \left (x -1\right )-\ln \left (x +1\right )+\ln \left (x^{2}+1\right )}{8 x^{4}-8}\)	\(61\)

[In]

int(x^5/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (19) = 38\).

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {2 \, x^{2} + {\left (x^{4} - 1\right )} \log \left (x^{2} + 1\right ) - {\left (x^{4} - 1\right )} \log \left (x^{2} - 1\right )}{8 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(x^5/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=- \frac {x^{2}}{4 x^{4} - 4} + \frac {\log {\left (x^{2} - 1 \right )}}{8} - \frac {\log {\left (x^{2} + 1 \right )}}{8} \]

[In]

integrate(x**5/(x**8-2*x**4+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]

[In]

integrate(x^5/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {x^{2}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \log \left (x^{2} + 1\right ) + \frac {1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \]

[In]

integrate(x^5/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.44 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {x^5}{1-2 x^4+x^8} \, dx=-\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {x^2}{4\,\left (x^4-1\right )} \]

[In]

int(x^5/(x^8 - 2*x^4 + 1),x)

[Out]

- atanh(x^2)/4 - x^2/(4*(x^4 - 1))