3.1496 \(\int \frac{1}{x^3 (1+x^8)} \, dx\)

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

[Out]

-1/(2*x^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 = 0.0656631, 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, 325, 297, 1162, 617, 204, 1165, 628} \[ -\frac{1}{2 x^2}-\frac{\log \left (x^4-\sqrt{2} x^2+1\right )}{8 \sqrt{2}}+\frac{\log \left (x^4+\sqrt{2} x^2+1\right )}{8 \sqrt{2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (\sqrt{2} x^2+1\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(1 + x^8)),x]

[Out]

-1/(2*x^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 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 297

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

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])

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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (1+x^8\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^2}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,x^2\right )\\ &=-\frac{1}{2 x^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{1}{2 x^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{1}{2 x^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{1}{2 x^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 [B]  time = 0.0341491, size = 208, normalized size = 2.08 \[ -\frac{1}{2 x^2}+\frac{\log \left (x^2-2 x \sin \left (\frac{\pi }{8}\right )+1\right )}{8 \sqrt{2}}+\frac{\log \left (x^2+2 x \sin \left (\frac{\pi }{8}\right )+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^2-2 x \cos \left (\frac{\pi }{8}\right )+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^2+2 x \cos \left (\frac{\pi }{8}\right )+1\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (\csc \left (\frac{\pi }{8}\right ) \left (x-\cos \left (\frac{\pi }{8}\right )\right )\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\csc \left (\frac{\pi }{8}\right ) \left (x+\cos \left (\frac{\pi }{8}\right )\right )\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (\sec \left (\frac{\pi }{8}\right ) \left (x-\sin \left (\frac{\pi }{8}\right )\right )\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\sec \left (\frac{\pi }{8}\right ) \left (x+\sin \left (\frac{\pi }{8}\right )\right )\right )}{4 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(1 + x^8)),x]

[Out]

-1/(2*x^2) - ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]/(4*Sqrt[2]) + ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]]/(4*Sqrt[2]) - A
rcTan[Sec[Pi/8]*(x - Sin[Pi/8])]/(4*Sqrt[2]) + ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]/(4*Sqrt[2]) - Log[1 + x^2 - 2
*x*Cos[Pi/8]]/(8*Sqrt[2]) - Log[1 + x^2 + 2*x*Cos[Pi/8]]/(8*Sqrt[2]) + Log[1 + x^2 - 2*x*Sin[Pi/8]]/(8*Sqrt[2]
) + Log[1 + x^2 + 2*x*Sin[Pi/8]]/(8*Sqrt[2])

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Maple [A]  time = 0.004, size = 71, normalized size = 0.7 \begin{align*} -{\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 ) }-{\frac{1}{2\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(x^8+1),x)

[Out]

-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)))-1/2/x^2

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Maxima [A]  time = 1.48423, size = 115, normalized size = 1.15 \begin{align*} -\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}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(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) - 1/2/x^2

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Fricas [A]  time = 1.35122, size = 343, normalized size = 3.43 \begin{align*} \frac{4 \, \sqrt{2} x^{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} + \sqrt{2} x^{2} + 1} - 1\right ) + 4 \, \sqrt{2} x^{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} - \sqrt{2} x^{2} + 1} + 1\right ) + \sqrt{2} x^{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) - \sqrt{2} x^{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) - 8}{16 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.171768, size = 87, normalized size = 0.87 \begin{align*} - \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}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(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/(2*x**2)

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Giac [A]  time = 1.19632, size = 131, normalized size = 1.31 \begin{align*} -\frac{1}{8} \, \sqrt{2} x^{4} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} + \sqrt{2}\right )}\right ) - \frac{1}{8} \, \sqrt{2} x^{4} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} - \sqrt{2}\right )}\right ) - \frac{1}{16} \, \sqrt{2} x^{4} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} x^{4} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) - \frac{1}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(x^8+1),x, algorithm="giac")

[Out]

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