3.15.90 \(\int \frac {x^9}{1+x^8} \, dx\) [1490]

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

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

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Rubi [A]
time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 = {281, 327, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\text {ArcTan}\left (1-\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\text {ArcTan}\left (\sqrt {2} x^2+1\right )}{4 \sqrt {2}}+\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}} \end {gather*}

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 210

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

Rule 217

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 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]

Rule 631

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 642

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 1176

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 1179

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]

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-x^2}{1+x^4} \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,x^2\right )\\ &=\frac {x^2}{2}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,x^2\right )+\frac {\text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,x^2\right )}{8 \sqrt {2}}+\frac {\text {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 {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x^2\right )}{4 \sqrt {2}}+\frac {\text {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*}

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Mathematica [A]
time = 0.06, size = 191, normalized size = 1.91 \begin {gather*} \frac {1}{16} \left (8 x^2+2 \sqrt {2} \tan ^{-1}\left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\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 )-2 \sqrt {2} \tan ^{-1}\left (x \sec \left (\frac {\pi }{8}\right )-\tan \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right )+\sqrt {2} \log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right )-\sqrt {2} \log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right )\right ) \end {gather*}

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 = 0.19, size = 66, normalized size = 0.66

method	result	size
risch	\(\frac {x^{2}}{2}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x^{2}-\textit {\_R} \right )\right )}{8}\)	\(28\)
default	\(\frac {x^{2}}{2}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{4}+x^{2} \sqrt {2}}{1+x^{4}-x^{2} \sqrt {2}}\right )+2 \arctan \left (1+x^{2} \sqrt {2}\right )+2 \arctan \left (-1+x^{2} \sqrt {2}\right )\right )}{16}\)	\(66\)
meijerg	\(\frac {x^{2}}{2}-\frac {x^{2} \left (-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{2 \left (x^{8}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{\left (x^{8}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{2 \left (x^{8}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{\left (x^{8}\right )^{\frac {1}{4}}}\right )}{8}\)	\(136\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 85, normalized size = 0.85 \begin {gather*} \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 {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 - 1/8*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 = 0.37, size = 117, normalized size = 1.17 \begin {gather*} \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 (4 \, x^{4} + 4 \, \sqrt {2} x^{2} + 4\right ) + \frac {1}{16} \, \sqrt {2} \log \left (4 \, x^{4} - 4 \, \sqrt {2} x^{2} + 4\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*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(4*x^4 + 4*sqrt(2)*x^2 + 4) + 1/16*sqrt(
2)*log(4*x^4 - 4*sqrt(2)*x^2 + 4)

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Sympy [A]
time = 0.06, size = 85, normalized size = 0.85 \begin {gather*} \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 {gather*}

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 = 0.49, size = 85, normalized size = 0.85 \begin {gather*} \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 {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 - 1/8*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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Mupad [B]
time = 0.07, size = 42, normalized size = 0.42 \begin {gather*} \frac {x^2}{2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^8 + 1),x)

[Out]

x^2/2 - 2^(1/2)*atan(2^(1/2)*x^2*(1/2 + 1i/2))*(1/8 - 1i/8) - 2^(1/2)*atan(2^(1/2)*x^2*(1/2 - 1i/2))*(1/8 + 1i
/8)

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