\(\int \frac {x^7}{1-2 x^4+x^8} \, dx\) [292]

Optimal result

Integrand size = 16, antiderivative size = 26 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=\frac {1}{4 \left (1-x^4\right )}+\frac {1}{4} \log \left (1-x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {28, 272, 45} \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=\frac {1}{4 \left (1-x^4\right )}+\frac {1}{4} \log \left (1-x^4\right ) \]

[In]

Int[x^7/(1 - 2*x^4 + x^8),x]

[Out]

1/(4*(1 - x^4)) + Log[1 - x^4]/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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^7}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {x}{(-1+x)^2} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {1}{(-1+x)^2}+\frac {1}{-1+x}\right ) \, dx,x,x^4\right ) \\ & = \frac {1}{4 \left (1-x^4\right )}+\frac {1}{4} \log \left (1-x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \left (-1+x^4\right )}+\frac {1}{4} \log \left (-1+x^4\right ) \]

[In]

Integrate[x^7/(1 - 2*x^4 + x^8),x]

[Out]

-1/4*1/(-1 + x^4) + Log[-1 + x^4]/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73

method	result	size
default	\(-\frac {1}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x^{4}-1\right )}{4}\)	\(19\)
risch	\(-\frac {1}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x^{4}-1\right )}{4}\)	\(19\)
norman	\(-\frac {1}{4 \left (x^{4}-1\right )}+\frac {\ln \left (x -1\right )}{4}+\frac {\ln \left (x +1\right )}{4}+\frac {\ln \left (x^{2}+1\right )}{4}\)	\(31\)
parallelrisch	\(\frac {\ln \left (x -1\right ) x^{4}+\ln \left (x +1\right ) x^{4}+\ln \left (x^{2}+1\right ) x^{4}-1-\ln \left (x -1\right )-\ln \left (x +1\right )-\ln \left (x^{2}+1\right )}{4 x^{4}-4}\)	\(58\)

[In]

int(x^7/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=\frac {{\left (x^{4} - 1\right )} \log \left (x^{4} - 1\right ) - 1}{4 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(x^7/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

1/4*((x^4 - 1)*log(x^4 - 1) - 1)/(x^4 - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=\frac {\log {\left (x^{4} - 1 \right )}}{4} - \frac {1}{4 x^{4} - 4} \]

[In]

integrate(x**7/(x**8-2*x**4+1),x)

[Out]

log(x**4 - 1)/4 - 1/(4*x**4 - 4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} + \frac {1}{4} \, \log \left (x^{4} - 1\right ) \]

[In]

integrate(x^7/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/4/(x^4 - 1) + 1/4*log(x^4 - 1)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} + \frac {1}{4} \, \log \left ({\left | x^{4} - 1 \right |}\right ) \]

[In]

integrate(x^7/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {x^7}{1-2 x^4+x^8} \, dx=\frac {\ln \left (x^4-1\right )}{4}-\frac {1}{4\,\left (x^4-1\right )} \]

[In]

int(x^7/(x^8 - 2*x^4 + 1),x)

[Out]

log(x^4 - 1)/4 - 1/(4*(x^4 - 1))