3.15.71 \(\int \frac {x^9}{1-x^8} \, dx\) [1471]

Optimal. Leaf size=24 \[ -\frac {x^2}{2}+\frac {1}{4} \tan ^{-1}\left (x^2\right )+\frac {1}{4} \tanh ^{-1}\left (x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {281, 327, 218, 212, 209} \begin {gather*} \frac {\text {ArcTan}\left (x^2\right )}{4}-\frac {x^2}{2}+\frac {1}{4} \tanh ^{-1}\left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(1 - x^8),x]

[Out]

-1/2*x^2 + ArcTan[x^2]/4 + 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 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 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 327

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.58 \begin {gather*} -\frac {x^2}{2}-\frac {1}{4} \tan ^{-1}\left (\frac {1}{x^2}\right )-\frac {1}{8} \log \left (1-x^2\right )+\frac {1}{8} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(1 - x^8),x]

[Out]

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

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Maple [A]
time = 0.19, size = 33, normalized size = 1.38

method	result	size
risch	\(-\frac {x^{2}}{2}+\frac {\ln \left (x^{2}+1\right )}{8}+\frac {\arctan \left (x^{2}\right )}{4}-\frac {\ln \left (x^{2}-1\right )}{8}\)	\(29\)
default	\(-\frac {x^{2}}{2}-\frac {\ln \left (x -1\right )}{8}-\frac {\ln \left (x +1\right )}{8}+\frac {\ln \left (x^{2}+1\right )}{8}+\frac {\arctan \left (x^{2}\right )}{4}\)	\(33\)
meijerg	\(\frac {\left (-1\right )^{\frac {3}{4}} \left (4 x^{2} \left (-1\right )^{\frac {1}{4}}+\frac {x^{2} \left (-1\right )^{\frac {1}{4}} \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (x^{8}\right )^{\frac {1}{4}}\right )\right )}{\left (x^{8}\right )^{\frac {1}{4}}}\right )}{8}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(-x^8+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 28, normalized size = 1.17 \begin {gather*} -\frac {1}{2} \, x^{2} + \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 28, normalized size = 1.17 \begin {gather*} -\frac {1}{2} \, x^{2} + \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.06, size = 27, normalized size = 1.12 \begin {gather*} - \frac {x^{2}}{2} - \frac {\log {\left (x^{2} - 1 \right )}}{8} + \frac {\log {\left (x^{2} + 1 \right )}}{8} + \frac {\operatorname {atan}{\left (x^{2} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.06, size = 29, normalized size = 1.21 \begin {gather*} -\frac {1}{2} \, x^{2} + \frac {1}{4} \, \arctan \left (x^{2}\right ) + \frac {1}{8} \, \log \left (x^{2} + 1\right ) - \frac {1}{8} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.06, size = 18, normalized size = 0.75 \begin {gather*} \frac {\mathrm {atan}\left (x^2\right )}{4}+\frac {\mathrm {atanh}\left (x^2\right )}{4}-\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^9/(x^8 - 1),x)

[Out]

atan(x^2)/4 + atanh(x^2)/4 - x^2/2

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