3.14.61 \(\int \frac {x^8}{1+x^6} \, dx\) [1361]

Optimal. Leaf size=16 \[ \frac {x^3}{3}-\frac {1}{3} \tan ^{-1}\left (x^3\right ) \]

[Out]

1/3*x^3-1/3*arctan(x^3)

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Rubi [A]
time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {281, 327, 209} \begin {gather*} \frac {x^3}{3}-\frac {\text {ArcTan}\left (x^3\right )}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(1 + x^6),x]

[Out]

x^3/3 - ArcTan[x^3]/3

Rule 209

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

Rule 281

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 327

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

Rubi steps

\begin {align*} \int \frac {x^8}{1+x^6} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,x^3\right )\\ &=\frac {x^3}{3}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^3\right )\\ &=\frac {x^3}{3}-\frac {1}{3} \tan ^{-1}\left (x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 16, normalized size = 1.00 \begin {gather*} \frac {x^3}{3}-\frac {1}{3} \tan ^{-1}\left (x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(1 + x^6),x]

[Out]

x^3/3 - ArcTan[x^3]/3

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Maple [A]
time = 0.16, size = 13, normalized size = 0.81

method	result	size
default	\(\frac {x^{3}}{3}-\frac {\arctan \left (x^{3}\right )}{3}\)	\(13\)
meijerg	\(\frac {x^{3}}{3}-\frac {\arctan \left (x^{3}\right )}{3}\)	\(13\)
risch	\(\frac {x^{3}}{3}-\frac {\arctan \left (x^{3}\right )}{3}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3-1/3*arctan(x^3)

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Maxima [A]
time = 0.50, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{3} \, x^{3} - \frac {1}{3} \, \arctan \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6+1),x, algorithm="maxima")

[Out]

1/3*x^3 - 1/3*arctan(x^3)

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Fricas [A]
time = 0.36, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{3} \, x^{3} - \frac {1}{3} \, \arctan \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6+1),x, algorithm="fricas")

[Out]

1/3*x^3 - 1/3*arctan(x^3)

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Sympy [A]
time = 0.03, size = 10, normalized size = 0.62 \begin {gather*} \frac {x^{3}}{3} - \frac {\operatorname {atan}{\left (x^{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(x**6+1),x)

[Out]

x**3/3 - atan(x**3)/3

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Giac [A]
time = 1.35, size = 12, normalized size = 0.75 \begin {gather*} \frac {1}{3} \, x^{3} - \frac {1}{3} \, \arctan \left (x^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(x^6+1),x, algorithm="giac")

[Out]

1/3*x^3 - 1/3*arctan(x^3)

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Mupad [B]
time = 1.03, size = 12, normalized size = 0.75 \begin {gather*} \frac {x^3}{3}-\frac {\mathrm {atan}\left (x^3\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(x^6 + 1),x)

[Out]

x^3/3 - atan(x^3)/3

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