3.280 \(\int \frac{1}{x^7 (1+2 x^4+x^8)} \, dx\)

Optimal. Leaf size=37 \[ \frac{1}{4 x^6 \left (x^4+1\right )}+\frac{5}{4 x^2}-\frac{5}{12 x^6}+\frac{5}{4} \tan ^{-1}\left (x^2\right ) \]

[Out]

-5/(12*x^6) + 5/(4*x^2) + 1/(4*x^6*(1 + x^4)) + (5*ArcTan[x^2])/4

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Rubi [A]  time = 0.0157296, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {28, 275, 290, 325, 203} \[ \frac{1}{4 x^6 \left (x^4+1\right )}+\frac{5}{4 x^2}-\frac{5}{12 x^6}+\frac{5}{4} \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + 2*x^4 + x^8)),x]

[Out]

-5/(12*x^6) + 5/(4*x^2) + 1/(4*x^6*(1 + x^4)) + (5*ArcTan[x^2])/4

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 290

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

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 203

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

Rubi steps

\begin{align*} \int \frac{1}{x^7 \left (1+2 x^4+x^8\right )} \, dx &=\int \frac{1}{x^7 \left (1+x^4\right )^2} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{1}{4 x^6 \left (1+x^4\right )}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{5}{12 x^6}+\frac{1}{4 x^6 \left (1+x^4\right )}-\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+x^2\right )} \, dx,x,x^2\right )\\ &=-\frac{5}{12 x^6}+\frac{5}{4 x^2}+\frac{1}{4 x^6 \left (1+x^4\right )}+\frac{5}{4} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,x^2\right )\\ &=-\frac{5}{12 x^6}+\frac{5}{4 x^2}+\frac{1}{4 x^6 \left (1+x^4\right )}+\frac{5}{4} \tan ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0107823, size = 33, normalized size = 0.89 \[ \frac{x^2}{4 \left (x^4+1\right )}+\frac{1}{x^2}-\frac{1}{6 x^6}-\frac{5}{4} \tan ^{-1}\left (\frac{1}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + 2*x^4 + x^8)),x]

[Out]

-1/(6*x^6) + x^(-2) + x^2/(4*(1 + x^4)) - (5*ArcTan[x^(-2)])/4

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Maple [A]  time = 0.011, size = 28, normalized size = 0.8 \begin{align*} -{\frac{1}{6\,{x}^{6}}}+{x}^{-2}+{\frac{{x}^{2}}{4\,{x}^{4}+4}}+{\frac{5\,\arctan \left ({x}^{2} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^8+2*x^4+1),x)

[Out]

-1/6/x^6+1/x^2+1/4*x^2/(x^4+1)+5/4*arctan(x^2)

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Maxima [A]  time = 1.4855, size = 41, normalized size = 1.11 \begin{align*} \frac{15 \, x^{8} + 10 \, x^{4} - 2}{12 \,{\left (x^{10} + x^{6}\right )}} + \frac{5}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

1/12*(15*x^8 + 10*x^4 - 2)/(x^10 + x^6) + 5/4*arctan(x^2)

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Fricas [A]  time = 1.42714, size = 96, normalized size = 2.59 \begin{align*} \frac{15 \, x^{8} + 10 \, x^{4} + 15 \,{\left (x^{10} + x^{6}\right )} \arctan \left (x^{2}\right ) - 2}{12 \,{\left (x^{10} + x^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

1/12*(15*x^8 + 10*x^4 + 15*(x^10 + x^6)*arctan(x^2) - 2)/(x^10 + x^6)

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Sympy [A]  time = 0.18952, size = 29, normalized size = 0.78 \begin{align*} \frac{5 \operatorname{atan}{\left (x^{2} \right )}}{4} + \frac{15 x^{8} + 10 x^{4} - 2}{12 x^{10} + 12 x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**8+2*x**4+1),x)

[Out]

5*atan(x**2)/4 + (15*x**8 + 10*x**4 - 2)/(12*x**10 + 12*x**6)

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Giac [A]  time = 1.14105, size = 42, normalized size = 1.14 \begin{align*} \frac{x^{2}}{4 \,{\left (x^{4} + 1\right )}} + \frac{6 \, x^{4} - 1}{6 \, x^{6}} + \frac{5}{4} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

1/4*x^2/(x^4 + 1) + 1/6*(6*x^4 - 1)/x^6 + 5/4*arctan(x^2)