3.408 \(\int \frac {x^4}{2+x^5+x^{10}} \, dx\)

Optimal. Leaf size=23 \[ \frac {2 \tan ^{-1}\left (\frac {2 x^5+1}{\sqrt {7}}\right )}{5 \sqrt {7}} \]

[Out]

2/35*arctan(1/7*(2*x^5+1)*7^(1/2))*7^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1352, 618, 204} \[ \frac {2 \tan ^{-1}\left (\frac {2 x^5+1}{\sqrt {7}}\right )}{5 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(2 + x^5 + x^10),x]

[Out]

(2*ArcTan[(1 + 2*x^5)/Sqrt[7]])/(5*Sqrt[7])

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1352

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

Rubi steps

\begin {align*} \int \frac {x^4}{2+x^5+x^{10}} \, dx &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,x^5\right )\\ &=-\left (\frac {2}{5} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 x^5\right )\right )\\ &=\frac {2 \tan ^{-1}\left (\frac {1+2 x^5}{\sqrt {7}}\right )}{5 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.00 \[ \frac {2 \tan ^{-1}\left (\frac {2 x^5+1}{\sqrt {7}}\right )}{5 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(2 + x^5 + x^10),x]

[Out]

(2*ArcTan[(1 + 2*x^5)/Sqrt[7]])/(5*Sqrt[7])

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fricas [A]  time = 0.96, size = 18, normalized size = 0.78 \[ \frac {2}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10+x^5+2),x, algorithm="fricas")

[Out]

2/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1))

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giac [A]  time = 2.96, size = 18, normalized size = 0.78 \[ \frac {2}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10+x^5+2),x, algorithm="giac")

[Out]

2/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1))

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maple [A]  time = 0.00, size = 19, normalized size = 0.83 \[ \frac {2 \sqrt {7}\, \arctan \left (\frac {\left (2 x^{5}+1\right ) \sqrt {7}}{7}\right )}{35} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^10+x^5+2),x)

[Out]

2/35*7^(1/2)*arctan(1/7*(2*x^5+1)*7^(1/2))

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maxima [A]  time = 2.09, size = 18, normalized size = 0.78 \[ \frac {2}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(x^10+x^5+2),x, algorithm="maxima")

[Out]

2/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1))

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mupad [B]  time = 1.33, size = 20, normalized size = 0.87 \[ \frac {2\,\sqrt {7}\,\mathrm {atan}\left (\frac {2\,\sqrt {7}\,x^5}{7}+\frac {\sqrt {7}}{7}\right )}{35} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(x^5 + x^10 + 2),x)

[Out]

(2*7^(1/2)*atan(7^(1/2)/7 + (2*7^(1/2)*x^5)/7))/35

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sympy [A]  time = 0.13, size = 27, normalized size = 1.17 \[ \frac {2 \sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x^{5}}{7} + \frac {\sqrt {7}}{7} \right )}}{35} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(x**10+x**5+2),x)

[Out]

2*sqrt(7)*atan(2*sqrt(7)*x**5/7 + sqrt(7)/7)/35

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