\(\int \frac {x^5}{1-x^8} \, dx\) [1473]

Optimal result

Integrand size = 13, antiderivative size = 17 \[ \int \frac {x^5}{1-x^8} \, dx=-\frac {1}{4} \arctan \left (x^2\right )+\frac {\text {arctanh}\left (x^2\right )}{4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 304, 209, 212} \[ \int \frac {x^5}{1-x^8} \, dx=\frac {\text {arctanh}\left (x^2\right )}{4}-\frac {\arctan \left (x^2\right )}{4} \]

[In]

Int[x^5/(1 - x^8),x]

[Out]

-1/4*ArcTan[x^2] + ArcTanh[x^2]/4

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 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 304

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.82 \[ \int \frac {x^5}{1-x^8} \, dx=\frac {1}{4} \arctan \left (\frac {1}{x^2}\right )-\frac {1}{8} \log \left (1-x^2\right )+\frac {1}{8} \log \left (1+x^2\right ) \]

[In]

Integrate[x^5/(1 - x^8),x]

[Out]

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

Maple [A] (verified)

Time = 5.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41

method	result	size
risch	\(-\frac {\arctan \left (x^{2}\right )}{4}-\frac {\ln \left (x^{2}-1\right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}\)	\(24\)
default	\(-\frac {\ln \left (-1+x \right )}{8}-\frac {\ln \left (1+x \right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}-\frac {\arctan \left (x^{2}\right )}{4}\)	\(28\)
meijerg	\(-\frac {x^{6} \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (x^{8}\right )^{\frac {1}{4}}\right )\right )}{8 \left (x^{8}\right )^{\frac {3}{4}}}\)	\(40\)
parallelrisch	\(-\frac {\ln \left (1+x \right )}{8}-\frac {\ln \left (-1+x \right )}{8}+\frac {\ln \left (x -i\right )}{8}+\frac {\ln \left (x +i\right )}{8}+\frac {i \ln \left (x^{2}-i\right )}{8}-\frac {i \ln \left (x^{2}+i\right )}{8}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {x^5}{1-x^8} \, dx=-\frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {x^5}{1-x^8} \, dx=- \frac {\log {\left (x^{2} - 1 \right )}}{8} + \frac {\log {\left (x^{2} + 1 \right )}}{8} - \frac {\operatorname {atan}{\left (x^{2} \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {x^5}{1-x^8} \, dx=-\frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - 1\right ) \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 3.24 \[ \int \frac {x^5}{1-x^8} \, dx=\frac {1}{4} \, \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{4} \, \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {x^5}{1-x^8} \, dx=\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {\mathrm {atan}\left (x^2\right )}{4} \]

[In]

int(-x^5/(x^8 - 1),x)

[Out]

atanh(x^2)/4 - atan(x^2)/4