3.81 \(\int \frac {x^5}{1+x^4} \, dx\)

Optimal. Leaf size=16 \[ \frac {x^2}{2}-\frac {1}{2} \tan ^{-1}\left (x^2\right ) \]

[Out]

1/2*x^2-1/2*arctan(x^2)

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Rubi [A]  time = 0.01, 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 = {275, 321, 203} \[ \frac {x^2}{2}-\frac {1}{2} \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5/(1 + x^4),x]

[Out]

x^2/2 - ArcTan[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 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 321

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \[ \frac {x^2}{2}-\frac {1}{2} \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(1 + x^4),x]

[Out]

x^2/2 - ArcTan[x^2]/2

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fricas [A]  time = 0.41, size = 12, normalized size = 0.75 \[ \frac {1}{2} \, x^{2} - \frac {1}{2} \, \arctan \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1),x, algorithm="fricas")

[Out]

1/2*x^2 - 1/2*arctan(x^2)

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giac [A]  time = 1.03, size = 12, normalized size = 0.75 \[ \frac {1}{2} \, x^{2} - \frac {1}{2} \, \arctan \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1),x, algorithm="giac")

[Out]

1/2*x^2 - 1/2*arctan(x^2)

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maple [A]  time = 0.00, size = 13, normalized size = 0.81 \[ \frac {x^{2}}{2}-\frac {\arctan \left (x^{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^4+1),x)

[Out]

1/2*x^2-1/2*arctan(x^2)

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maxima [A]  time = 1.47, size = 12, normalized size = 0.75 \[ \frac {1}{2} \, x^{2} - \frac {1}{2} \, \arctan \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1),x, algorithm="maxima")

[Out]

1/2*x^2 - 1/2*arctan(x^2)

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mupad [B]  time = 0.02, size = 12, normalized size = 0.75 \[ \frac {x^2}{2}-\frac {\mathrm {atan}\left (x^2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^4 + 1),x)

[Out]

x^2/2 - atan(x^2)/2

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sympy [A]  time = 0.09, size = 10, normalized size = 0.62 \[ \frac {x^{2}}{2} - \frac {\operatorname {atan}{\left (x^{2} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(x**4+1),x)

[Out]

x**2/2 - atan(x**2)/2

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