3.291 \(\int \frac{x^9}{1-2 x^4+x^8} \, dx\)

Optimal. Leaf size=32 \[ \frac{x^6}{4 \left (1-x^4\right )}+\frac{3 x^2}{4}-\frac{3}{4} \tanh ^{-1}\left (x^2\right ) \]

[Out]

(3*x^2)/4 + x^6/(4*(1 - x^4)) - (3*ArcTanh[x^2])/4

________________________________________________________________________________________

Rubi [A]  time = 0.0130546, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {28, 275, 288, 321, 207} \[ \frac{x^6}{4 \left (1-x^4\right )}+\frac{3 x^2}{4}-\frac{3}{4} \tanh ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^9/(1 - 2*x^4 + x^8),x]

[Out]

(3*x^2)/4 + x^6/(4*(1 - x^4)) - (3*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 275

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 288

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]

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{x^9}{1-2 x^4+x^8} \, dx &=\int \frac{x^9}{\left (-1+x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (-1+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{x^6}{4 \left (1-x^4\right )}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,x^2\right )\\ &=\frac{3 x^2}{4}+\frac{x^6}{4 \left (1-x^4\right )}+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,x^2\right )\\ &=\frac{3 x^2}{4}+\frac{x^6}{4 \left (1-x^4\right )}-\frac{3}{4} \tanh ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0236068, size = 39, normalized size = 1.22 \[ \frac{1}{8} \left (2 \left (\frac{1}{1-x^4}+2\right ) x^2+3 \log \left (1-x^2\right )-3 \log \left (x^2+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(1 - 2*x^4 + x^8),x]

[Out]

(2*x^2*(2 + (1 - x^4)^(-1)) + 3*Log[1 - x^2] - 3*Log[1 + x^2])/8

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 41, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}}{2}}-{\frac{1}{8\,{x}^{2}+8}}-{\frac{3\,\ln \left ({x}^{2}+1 \right ) }{8}}-{\frac{1}{8\,{x}^{2}-8}}+{\frac{3\,\ln \left ({x}^{2}-1 \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^8-2*x^4+1),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01735, size = 46, normalized size = 1.44 \begin{align*} \frac{1}{2} \, x^{2} - \frac{x^{2}}{4 \,{\left (x^{4} - 1\right )}} - \frac{3}{8} \, \log \left (x^{2} + 1\right ) + \frac{3}{8} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

1/2*x^2 - 1/4*x^2/(x^4 - 1) - 3/8*log(x^2 + 1) + 3/8*log(x^2 - 1)

________________________________________________________________________________________

Fricas [A]  time = 1.45559, size = 115, normalized size = 3.59 \begin{align*} \frac{4 \, x^{6} - 6 \, x^{2} - 3 \,{\left (x^{4} - 1\right )} \log \left (x^{2} + 1\right ) + 3 \,{\left (x^{4} - 1\right )} \log \left (x^{2} - 1\right )}{8 \,{\left (x^{4} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

1/8*(4*x^6 - 6*x^2 - 3*(x^4 - 1)*log(x^2 + 1) + 3*(x^4 - 1)*log(x^2 - 1))/(x^4 - 1)

________________________________________________________________________________________

Sympy [A]  time = 0.126685, size = 34, normalized size = 1.06 \begin{align*} \frac{x^{2}}{2} - \frac{x^{2}}{4 x^{4} - 4} + \frac{3 \log{\left (x^{2} - 1 \right )}}{8} - \frac{3 \log{\left (x^{2} + 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(x**8-2*x**4+1),x)

[Out]

x**2/2 - x**2/(4*x**4 - 4) + 3*log(x**2 - 1)/8 - 3*log(x**2 + 1)/8

________________________________________________________________________________________

Giac [A]  time = 1.1067, size = 47, normalized size = 1.47 \begin{align*} \frac{1}{2} \, x^{2} - \frac{x^{2}}{4 \,{\left (x^{4} - 1\right )}} - \frac{3}{8} \, \log \left (x^{2} + 1\right ) + \frac{3}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

1/2*x^2 - 1/4*x^2/(x^4 - 1) - 3/8*log(x^2 + 1) + 3/8*log(abs(x^2 - 1))