3.5.7 \(\int \frac {x^9}{2+x^5+x^{10}} \, dx\) [407]

Optimal. Leaf size=37 \[ -\frac {\tan ^{-1}\left (\frac {1+2 x^5}{\sqrt {7}}\right )}{5 \sqrt {7}}+\frac {1}{10} \log \left (2+x^5+x^{10}\right ) \]

[Out]

1/10*ln(x^10+x^5+2)-1/35*arctan(1/7*(2*x^5+1)*7^(1/2))*7^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1371, 648, 632, 210, 642} \begin {gather*} \frac {1}{10} \log \left (x^{10}+x^5+2\right )-\frac {\text {ArcTan}\left (\frac {2 x^5+1}{\sqrt {7}}\right )}{5 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(2 + x^5 + x^10),x]

[Out]

-1/5*ArcTan[(1 + 2*x^5)/Sqrt[7]]/Sqrt[7] + Log[2 + x^5 + x^10]/10

Rule 210

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

Rule 632

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 642

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 648

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

Rule 1371

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

Rubi steps

\begin {align*} \int \frac {x^9}{2+x^5+x^{10}} \, dx &=\frac {1}{5} \text {Subst}\left (\int \frac {x}{2+x+x^2} \, dx,x,x^5\right )\\ &=-\left (\frac {1}{10} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,x^5\right )\right )+\frac {1}{10} \text {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,x^5\right )\\ &=\frac {1}{10} \log \left (2+x^5+x^{10}\right )+\frac {1}{5} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 x^5\right )\\ &=-\frac {\tan ^{-1}\left (\frac {1+2 x^5}{\sqrt {7}}\right )}{5 \sqrt {7}}+\frac {1}{10} \log \left (2+x^5+x^{10}\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 37, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {1+2 x^5}{\sqrt {7}}\right )}{5 \sqrt {7}}+\frac {1}{10} \log \left (2+x^5+x^{10}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(2 + x^5 + x^10),x]

[Out]

-1/5*ArcTan[(1 + 2*x^5)/Sqrt[7]]/Sqrt[7] + Log[2 + x^5 + x^10]/10

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Maple [A]
time = 0.02, size = 31, normalized size = 0.84

method	result	size
default	\(\frac {\ln \left (x^{10}+x^{5}+2\right )}{10}-\frac {\arctan \left (\frac {\left (2 x^{5}+1\right ) \sqrt {7}}{7}\right ) \sqrt {7}}{35}\)	\(31\)
risch	\(\frac {\ln \left (4 x^{10}+4 x^{5}+8\right )}{10}-\frac {\arctan \left (\frac {\left (2 x^{5}+1\right ) \sqrt {7}}{7}\right ) \sqrt {7}}{35}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^10+x^5+2),x,method=_RETURNVERBOSE)

[Out]

1/10*ln(x^10+x^5+2)-1/35*arctan(1/7*(2*x^5+1)*7^(1/2))*7^(1/2)

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Maxima [A]
time = 0.51, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) + \frac {1}{10} \, \log \left (x^{10} + x^{5} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^10+x^5+2),x, algorithm="maxima")

[Out]

-1/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1)) + 1/10*log(x^10 + x^5 + 2)

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Fricas [A]
time = 0.34, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) + \frac {1}{10} \, \log \left (x^{10} + x^{5} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^10+x^5+2),x, algorithm="fricas")

[Out]

-1/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1)) + 1/10*log(x^10 + x^5 + 2)

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Sympy [A]
time = 0.05, size = 37, normalized size = 1.00 \begin {gather*} \frac {\log {\left (x^{10} + x^{5} + 2 \right )}}{10} - \frac {\sqrt {7} \operatorname {atan}{\left (\frac {2 \sqrt {7} x^{5}}{7} + \frac {\sqrt {7}}{7} \right )}}{35} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(x**10+x**5+2),x)

[Out]

log(x**10 + x**5 + 2)/10 - sqrt(7)*atan(2*sqrt(7)*x**5/7 + sqrt(7)/7)/35

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Giac [A]
time = 4.91, size = 30, normalized size = 0.81 \begin {gather*} -\frac {1}{35} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (2 \, x^{5} + 1\right )}\right ) + \frac {1}{10} \, \log \left (x^{10} + x^{5} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(x^10+x^5+2),x, algorithm="giac")

[Out]

-1/35*sqrt(7)*arctan(1/7*sqrt(7)*(2*x^5 + 1)) + 1/10*log(x^10 + x^5 + 2)

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Mupad [B]
time = 1.35, size = 32, normalized size = 0.86 \begin {gather*} \frac {\ln \left (x^{10}+x^5+2\right )}{10}-\frac {\sqrt {7}\,\mathrm {atan}\left (\frac {2\,\sqrt {7}\,x^5}{7}+\frac {\sqrt {7}}{7}\right )}{35} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(x^5 + x^10 + 2),x)

[Out]

log(x^5 + x^10 + 2)/10 - (7^(1/2)*atan(7^(1/2)/7 + (2*7^(1/2)*x^5)/7))/35

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