3.377 \(\int \frac {1}{x^7 (1+3 x^4+x^8)} \, dx\)

Optimal. Leaf size=97 \[ -\frac {1}{6 x^6}+\frac {3}{2 x^2}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (123+55 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

[Out]

-1/6/x^6+3/2/x^2-1/2*arctan(x^2*2^(1/2)/(3+5^(1/2))^(1/2))*(5/2-11/10*5^(1/2))+1/2*arctan(x^2*(1/2+1/2*5^(1/2)
))*(5/2+11/10*5^(1/2))

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Rubi [A]  time = 0.13, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 = {1359, 1123, 1281, 1166, 203} \[ \frac {3}{2 x^2}-\frac {1}{6 x^6}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{10} \left (123+55 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(1 + 3*x^4 + x^8)),x]

[Out]

-1/(6*x^6) + 3/(2*x^2) - (Sqrt[(123 - 55*Sqrt[5])/10]*ArcTan[Sqrt[2/(3 + Sqrt[5])]*x^2])/2 + (Sqrt[(123 + 55*S
qrt[5])/10]*ArcTan[Sqrt[(3 + Sqrt[5])/2]*x^2])/2

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])

Rule 1123

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*x^2 +
c*x^4)^(p + 1))/(a*d*(m + 1)), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1166

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

Rule 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1359

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (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) + c*x^((2*n)/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b,
c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {1}{x^7 \left (1+3 x^4+x^8\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^4 \left (1+3 x^2+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {-9-3 x^2}{x^2 \left (1+3 x^2+x^4\right )} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}+\frac {3}{2 x^2}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {-24-9 x^2}{1+3 x^2+x^4} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}+\frac {3}{2 x^2}-\frac {1}{20} \left (-15+7 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}+\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )+\frac {1}{20} \left (15+7 \sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {3}{2}-\frac {\sqrt {5}}{2}+x^2} \, dx,x,x^2\right )\\ &=-\frac {1}{6 x^6}+\frac {3}{2 x^2}-\frac {1}{2} \sqrt {\frac {1}{10} \left (123-55 \sqrt {5}\right )} \tan ^{-1}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{20} \sqrt {1230+550 \sqrt {5}} \tan ^{-1}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 73, normalized size = 0.75 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8+3 \text {$\#$1}^4+1\& ,\frac {3 \text {$\#$1}^4 \log (x-\text {$\#$1})+8 \log (x-\text {$\#$1})}{2 \text {$\#$1}^6+3 \text {$\#$1}^2}\& \right ]-\frac {1}{6 x^6}+\frac {3}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(1 + 3*x^4 + x^8)),x]

[Out]

-1/6*1/x^6 + 3/(2*x^2) + RootSum[1 + 3*#1^4 + #1^8 & , (8*Log[x - #1] + 3*Log[x - #1]*#1^4)/(3*#1^2 + 2*#1^6)
& ]/4

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fricas [B]  time = 0.88, size = 180, normalized size = 1.86 \[ \frac {3 \, \sqrt {10} x^{6} \sqrt {-55 \, \sqrt {5} + 123} \arctan \left (\frac {1}{40} \, \sqrt {10} \sqrt {2 \, x^{4} + \sqrt {5} + 3} {\left (7 \, \sqrt {5} \sqrt {2} + 15 \, \sqrt {2}\right )} \sqrt {-55 \, \sqrt {5} + 123} - \frac {1}{20} \, \sqrt {10} {\left (7 \, \sqrt {5} x^{2} + 15 \, x^{2}\right )} \sqrt {-55 \, \sqrt {5} + 123}\right ) - 3 \, \sqrt {10} x^{6} \sqrt {55 \, \sqrt {5} + 123} \arctan \left (\frac {1}{40} \, {\left (\sqrt {10} \sqrt {2 \, x^{4} - \sqrt {5} + 3} {\left (7 \, \sqrt {5} \sqrt {2} - 15 \, \sqrt {2}\right )} - 2 \, \sqrt {10} {\left (7 \, \sqrt {5} x^{2} - 15 \, x^{2}\right )}\right )} \sqrt {55 \, \sqrt {5} + 123}\right ) + 45 \, x^{4} - 5}{30 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="fricas")

[Out]

1/30*(3*sqrt(10)*x^6*sqrt(-55*sqrt(5) + 123)*arctan(1/40*sqrt(10)*sqrt(2*x^4 + sqrt(5) + 3)*(7*sqrt(5)*sqrt(2)
+ 15*sqrt(2))*sqrt(-55*sqrt(5) + 123) - 1/20*sqrt(10)*(7*sqrt(5)*x^2 + 15*x^2)*sqrt(-55*sqrt(5) + 123)) - 3*s
qrt(10)*x^6*sqrt(55*sqrt(5) + 123)*arctan(1/40*(sqrt(10)*sqrt(2*x^4 - sqrt(5) + 3)*(7*sqrt(5)*sqrt(2) - 15*sqr
t(2)) - 2*sqrt(10)*(7*sqrt(5)*x^2 - 15*x^2))*sqrt(55*sqrt(5) + 123)) + 45*x^4 - 5)/x^6

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giac [A]  time = 0.51, size = 77, normalized size = 0.79 \[ \frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} - 5\right )} + 8 \, \sqrt {5} - 40\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} + 1}\right ) + \frac {1}{20} \, {\left (3 \, x^{4} {\left (\sqrt {5} + 5\right )} + 8 \, \sqrt {5} + 40\right )} \arctan \left (\frac {2 \, x^{2}}{\sqrt {5} - 1}\right ) + \frac {9 \, x^{4} - 1}{6 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="giac")

[Out]

1/20*(3*x^4*(sqrt(5) - 5) + 8*sqrt(5) - 40)*arctan(2*x^2/(sqrt(5) + 1)) + 1/20*(3*x^4*(sqrt(5) + 5) + 8*sqrt(5
) + 40)*arctan(2*x^2/(sqrt(5) - 1)) + 1/6*(9*x^4 - 1)/x^6

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maple [B]  time = 0.03, size = 122, normalized size = 1.26 \[ \frac {7 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{5 \left (-2+2 \sqrt {5}\right )}+\frac {3 \arctan \left (\frac {4 x^{2}}{-2+2 \sqrt {5}}\right )}{-2+2 \sqrt {5}}-\frac {7 \sqrt {5}\, \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{5 \left (2+2 \sqrt {5}\right )}+\frac {3 \arctan \left (\frac {4 x^{2}}{2+2 \sqrt {5}}\right )}{2+2 \sqrt {5}}+\frac {3}{2 x^{2}}-\frac {1}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(x^8+3*x^4+1),x)

[Out]

-1/6/x^6+3/2/x^2-7/5*5^(1/2)/(2+2*5^(1/2))*arctan(4/(2+2*5^(1/2))*x^2)+3/(2+2*5^(1/2))*arctan(4/(2+2*5^(1/2))*
x^2)+7/5*5^(1/2)/(-2+2*5^(1/2))*arctan(4/(-2+2*5^(1/2))*x^2)+3/(-2+2*5^(1/2))*arctan(4/(-2+2*5^(1/2))*x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {9 \, x^{4} - 1}{6 \, x^{6}} + \int \frac {{\left (3 \, x^{4} + 8\right )} x}{x^{8} + 3 \, x^{4} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(x^8+3*x^4+1),x, algorithm="maxima")

[Out]

1/6*(9*x^4 - 1)/x^6 + integrate((3*x^4 + 8)*x/(x^8 + 3*x^4 + 1), x)

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mupad [B]  time = 0.12, size = 136, normalized size = 1.40 \[ 2\,\mathrm {atanh}\left (\frac {3327500\,x^2\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}-2550075}-\frac {1488300\,\sqrt {5}\,x^2\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}-2550075}\right )\,\sqrt {\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}-2\,\mathrm {atanh}\left (\frac {3327500\,x^2\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}+2550075}+\frac {1488300\,\sqrt {5}\,x^2\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}}{1140425\,\sqrt {5}+2550075}\right )\,\sqrt {-\frac {11\,\sqrt {5}}{32}-\frac {123}{160}}+\frac {\frac {3\,x^4}{2}-\frac {1}{6}}{x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(3*x^4 + x^8 + 1)),x)

[Out]

2*atanh((3327500*x^2*((11*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) - 2550075) - (1488300*5^(1/2)*x^2*((1
1*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) - 2550075))*((11*5^(1/2))/32 - 123/160)^(1/2) - 2*atanh((3327
500*x^2*(- (11*5^(1/2))/32 - 123/160)^(1/2))/(1140425*5^(1/2) + 2550075) + (1488300*5^(1/2)*x^2*(- (11*5^(1/2)
)/32 - 123/160)^(1/2))/(1140425*5^(1/2) + 2550075))*(- (11*5^(1/2))/32 - 123/160)^(1/2) + ((3*x^4)/2 - 1/6)/x^
6

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sympy [A]  time = 0.27, size = 65, normalized size = 0.67 \[ 2 \left (\frac {11 \sqrt {5}}{40} + \frac {5}{8}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{-1 + \sqrt {5}} \right )} - 2 \left (\frac {5}{8} - \frac {11 \sqrt {5}}{40}\right ) \operatorname {atan}{\left (\frac {2 x^{2}}{1 + \sqrt {5}} \right )} + \frac {9 x^{4} - 1}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(x**8+3*x**4+1),x)

[Out]

2*(11*sqrt(5)/40 + 5/8)*atan(2*x**2/(-1 + sqrt(5))) - 2*(5/8 - 11*sqrt(5)/40)*atan(2*x**2/(1 + sqrt(5))) + (9*
x**4 - 1)/(6*x**6)

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