3.1498 \(\int \frac {1}{x^7 (1+x^8)} \, dx\)

Optimal. Leaf size=100 \[ -\frac {1}{6 x^6}+\frac {\tan ^{-1}\left (1-\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x^2+1\right )}{4 \sqrt {2}}+\frac {\log \left (x^4-\sqrt {2} x^2+1\right )}{8 \sqrt {2}}-\frac {\log \left (x^4+\sqrt {2} x^2+1\right )}{8 \sqrt {2}} \]

[Out]

-1/6/x^6-1/8*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 = 0.07, 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 = {275, 325, 211, 1165, 628, 1162, 617, 204} \[ -\frac {1}{6 x^6}+\frac {\log \left (x^4-\sqrt {2} x^2+1\right )}{8 \sqrt {2}}-\frac {\log \left (x^4+\sqrt {2} x^2+1\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x^2+1\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + x^8)),x]

[Out]

-1/(6*x^6) + 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 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])

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (1+x^8\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}-\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 {1}{6 x^6}-\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 {1}{6 x^6}+\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 {1}{6 x^6}+\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 = 0.08, size = 193, normalized size = 1.93 \[ \frac {1}{48} \left (-\frac {8}{x^6}-3 \sqrt {2} \log \left (x^2-2 x \sin \left (\frac {\pi }{8}\right )+1\right )-3 \sqrt {2} \log \left (x^2+2 x \sin \left (\frac {\pi }{8}\right )+1\right )+3 \sqrt {2} \log \left (x^2-2 x \cos \left (\frac {\pi }{8}\right )+1\right )+3 \sqrt {2} \log \left (x^2+2 x \cos \left (\frac {\pi }{8}\right )+1\right )-6 \sqrt {2} \tan ^{-1}\left (x \sec \left (\frac {\pi }{8}\right )-\tan \left (\frac {\pi }{8}\right )\right )+6 \sqrt {2} \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x+\cos \left (\frac {\pi }{8}\right )\right )\right )+6 \sqrt {2} \tan ^{-1}\left (\cot \left (\frac {\pi }{8}\right )-x \csc \left (\frac {\pi }{8}\right )\right )+6 \sqrt {2} \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + x^8)),x]

[Out]

(-8/x^6 + 6*Sqrt[2]*ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]] + 6*Sqrt[2]*ArcTan[Cot[Pi/8] - x*Csc[Pi/8]] + 6*Sqrt[2]*
ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])] - 6*Sqrt[2]*ArcTan[x*Sec[Pi/8] - Tan[Pi/8]] + 3*Sqrt[2]*Log[1 + x^2 - 2*x*Co
s[Pi/8]] + 3*Sqrt[2]*Log[1 + x^2 + 2*x*Cos[Pi/8]] - 3*Sqrt[2]*Log[1 + x^2 - 2*x*Sin[Pi/8]] - 3*Sqrt[2]*Log[1 +
x^2 + 2*x*Sin[Pi/8]])/48

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fricas [A]  time = 0.91, size = 125, normalized size = 1.25 \[ \frac {12 \, \sqrt {2} x^{6} \arctan \left (-\sqrt {2} x^{2} + \sqrt {2} \sqrt {x^{4} + \sqrt {2} x^{2} + 1} - 1\right ) + 12 \, \sqrt {2} x^{6} \arctan \left (-\sqrt {2} x^{2} + \sqrt {2} \sqrt {x^{4} - \sqrt {2} x^{2} + 1} + 1\right ) - 3 \, \sqrt {2} x^{6} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) + 3 \, \sqrt {2} x^{6} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) - 8}{48 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+1),x, algorithm="fricas")

[Out]

1/48*(12*sqrt(2)*x^6*arctan(-sqrt(2)*x^2 + sqrt(2)*sqrt(x^4 + sqrt(2)*x^2 + 1) - 1) + 12*sqrt(2)*x^6*arctan(-s
qrt(2)*x^2 + sqrt(2)*sqrt(x^4 - sqrt(2)*x^2 + 1) + 1) - 3*sqrt(2)*x^6*log(x^4 + sqrt(2)*x^2 + 1) + 3*sqrt(2)*x
^6*log(x^4 - sqrt(2)*x^2 + 1) - 8)/x^6

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giac [A]  time = 0.17, size = 85, normalized size = 0.85 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) - \frac {1}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+1),x, algorithm="giac")

[Out]

-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) - 1/6/x^6

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maple [A]  time = 0.00, size = 71, normalized size = 0.71 \[ -\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x^{2}-1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x^{2}+1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {x^{4}+\sqrt {2}\, x^{2}+1}{x^{4}-\sqrt {2}\, x^{2}+1}\right )}{16}-\frac {1}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^8+1),x)

[Out]

-1/16*2^(1/2)*ln((x^4+2^(1/2)*x^2+1)/(x^4-2^(1/2)*x^2+1))-1/8*2^(1/2)*arctan(2^(1/2)*x^2+1)-1/8*2^(1/2)*arctan
(2^(1/2)*x^2-1)-1/6/x^6

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maxima [A]  time = 2.40, size = 85, normalized size = 0.85 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) - \frac {1}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+1),x, algorithm="maxima")

[Out]

-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) - 1/6/x^6

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mupad [B]  time = 1.05, size = 42, normalized size = 0.42 \[ -\frac {1}{6\,x^6}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(x^8 + 1)),x)

[Out]

- 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) -
1/(6*x^6)

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sympy [A]  time = 0.22, size = 87, normalized size = 0.87 \[ \frac {\sqrt {2} \log {\left (x^{4} - \sqrt {2} x^{2} + 1 \right )}}{16} - \frac {\sqrt {2} \log {\left (x^{4} + \sqrt {2} x^{2} + 1 \right )}}{16} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} - 1 \right )}}{8} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} + 1 \right )}}{8} - \frac {1}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**8+1),x)

[Out]

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 - 1/(6*x**6)

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