3.1490 \(\int \frac{x^9}{1+x^8} \, dx\)

Optimal. Leaf size=100 \[ \frac{x^2}{2}+\frac{\log \left (x^4-\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^4+\sqrt{2} x^2+1\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} x^2+1\right )}{4 \sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0645671, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.727, Rules used = {275, 321, 211, 1165, 628, 1162, 617, 204} \[ \frac{x^2}{2}+\frac{\log \left (x^4-\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^4+\sqrt{2} x^2+1\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} x^2+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(1 + x^8),x]

[Out]

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

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 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 211

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [A]  time = 0.075285, size = 191, normalized size = 1.91 \[ \frac{1}{16} \left (8 x^2-\sqrt{2} \log \left (x^2-2 x \sin \left (\frac{\pi }{8}\right )+1\right )-\sqrt{2} \log \left (x^2+2 x \sin \left (\frac{\pi }{8}\right )+1\right )+\sqrt{2} \log \left (x^2-2 x \cos \left (\frac{\pi }{8}\right )+1\right )+\sqrt{2} \log \left (x^2+2 x \cos \left (\frac{\pi }{8}\right )+1\right )-2 \sqrt{2} \tan ^{-1}\left (x \sec \left (\frac{\pi }{8}\right )-\tan \left (\frac{\pi }{8}\right )\right )+2 \sqrt{2} \tan ^{-1}\left (\csc \left (\frac{\pi }{8}\right ) \left (x+\cos \left (\frac{\pi }{8}\right )\right )\right )+2 \sqrt{2} \tan ^{-1}\left (\cot \left (\frac{\pi }{8}\right )-x \csc \left (\frac{\pi }{8}\right )\right )+2 \sqrt{2} \tan ^{-1}\left (\sec \left (\frac{\pi }{8}\right ) \left (x+\sin \left (\frac{\pi }{8}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(1 + x^8),x]

[Out]

(8*x^2 + 2*Sqrt[2]*ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]] + 2*Sqrt[2]*ArcTan[Cot[Pi/8] - x*Csc[Pi/8]] + 2*Sqrt[2]*A
rcTan[Sec[Pi/8]*(x + Sin[Pi/8])] - 2*Sqrt[2]*ArcTan[x*Sec[Pi/8] - Tan[Pi/8]] + Sqrt[2]*Log[1 + x^2 - 2*x*Cos[P
i/8]] + Sqrt[2]*Log[1 + x^2 + 2*x*Cos[Pi/8]] - Sqrt[2]*Log[1 + x^2 - 2*x*Sin[Pi/8]] - Sqrt[2]*Log[1 + x^2 + 2*
x*Sin[Pi/8]])/16

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Maple [A]  time = 0.002, size = 71, normalized size = 0.7 \begin{align*}{\frac{{x}^{2}}{2}}-{\frac{\arctan \left ( 1+{x}^{2}\sqrt{2} \right ) \sqrt{2}}{8}}-{\frac{\arctan \left ( -1+{x}^{2}\sqrt{2} \right ) \sqrt{2}}{8}}-{\frac{\sqrt{2}}{16}\ln \left ({\frac{1+{x}^{4}+{x}^{2}\sqrt{2}}{1+{x}^{4}-{x}^{2}\sqrt{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^8+1),x)

[Out]

1/2*x^2-1/8*arctan(1+x^2*2^(1/2))*2^(1/2)-1/8*arctan(-1+x^2*2^(1/2))*2^(1/2)-1/16*2^(1/2)*ln((1+x^4+x^2*2^(1/2
))/(1+x^4-x^2*2^(1/2)))

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Maxima [A]  time = 1.44987, size = 115, normalized size = 1.15 \begin{align*} \frac{1}{2} \, x^{2} - \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} + \sqrt{2}\right )}\right ) - \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} - \sqrt{2}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) \end{align*}

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 - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x^2 - sqrt(2))
) - 1/16*sqrt(2)*log(x^4 + sqrt(2)*x^2 + 1) + 1/16*sqrt(2)*log(x^4 - sqrt(2)*x^2 + 1)

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Fricas [A]  time = 1.37106, size = 333, normalized size = 3.33 \begin{align*} \frac{1}{2} \, x^{2} + \frac{1}{4} \, \sqrt{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} + \sqrt{2} x^{2} + 1} - 1\right ) + \frac{1}{4} \, \sqrt{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} - \sqrt{2} x^{2} + 1} + 1\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) \end{align*}

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*sqrt(2)*arctan(-sqrt(2)*x^2 + sqrt(2)*sqrt(x^4 + sqrt(2)*x^2 + 1) - 1) + 1/4*sqrt(2)*arctan(-sqr
t(2)*x^2 + sqrt(2)*sqrt(x^4 - sqrt(2)*x^2 + 1) + 1) - 1/16*sqrt(2)*log(x^4 + sqrt(2)*x^2 + 1) + 1/16*sqrt(2)*l
og(x^4 - sqrt(2)*x^2 + 1)

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Sympy [A]  time = 0.149146, size = 85, normalized size = 0.85 \begin{align*} \frac{x^{2}}{2} + \frac{\sqrt{2} \log{\left (x^{4} - \sqrt{2} x^{2} + 1 \right )}}{16} - \frac{\sqrt{2} \log{\left (x^{4} + \sqrt{2} x^{2} + 1 \right )}}{16} - \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x^{2} - 1 \right )}}{8} - \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x^{2} + 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 + sqrt(2)*log(x**4 - sqrt(2)*x**2 + 1)/16 - sqrt(2)*log(x**4 + sqrt(2)*x**2 + 1)/16 - sqrt(2)*atan(sqrt
(2)*x**2 - 1)/8 - sqrt(2)*atan(sqrt(2)*x**2 + 1)/8

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Giac [A]  time = 1.13899, size = 115, normalized size = 1.15 \begin{align*} \frac{1}{2} \, x^{2} - \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} + \sqrt{2}\right )}\right ) - \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} - \sqrt{2}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) \end{align*}

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 - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x^2 - sqrt(2))
) - 1/16*sqrt(2)*log(x^4 + sqrt(2)*x^2 + 1) + 1/16*sqrt(2)*log(x^4 - sqrt(2)*x^2 + 1)