3.170 \(\int \frac{x^5}{(1+x^4)^3} \, dx\)

Optimal. Leaf size=37 \[ \frac{x^2}{16 \left (x^4+1\right )}-\frac{x^2}{8 \left (x^4+1\right )^2}+\frac{1}{16} \tan ^{-1}\left (x^2\right ) \]

[Out]

-x^2/(8*(1 + x^4)^2) + x^2/(16*(1 + x^4)) + ArcTan[x^2]/16

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Rubi [A]  time = 0.0129613, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {275, 288, 199, 203} \[ \frac{x^2}{16 \left (x^4+1\right )}-\frac{x^2}{8 \left (x^4+1\right )^2}+\frac{1}{16} \tan ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5/(1 + x^4)^3,x]

[Out]

-x^2/(8*(1 + x^4)^2) + x^2/(16*(1 + x^4)) + ArcTan[x^2]/16

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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])

Rubi steps

\begin{align*} \int \frac{x^5}{\left (1+x^4\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac{x^2}{8 \left (1+x^4\right )^2}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{x^2}{8 \left (1+x^4\right )^2}+\frac{x^2}{16 \left (1+x^4\right )}+\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,x^2\right )\\ &=-\frac{x^2}{8 \left (1+x^4\right )^2}+\frac{x^2}{16 \left (1+x^4\right )}+\frac{1}{16} \tan ^{-1}\left (x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0104496, size = 25, normalized size = 0.68 \[ \frac{1}{16} \left (\frac{\left (x^4-1\right ) x^2}{\left (x^4+1\right )^2}+\tan ^{-1}\left (x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(1 + x^4)^3,x]

[Out]

((x^2*(-1 + x^4))/(1 + x^4)^2 + ArcTan[x^2])/16

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Maple [A]  time = 0.007, size = 28, normalized size = 0.8 \begin{align*}{\frac{1}{2\, \left ({x}^{4}+1 \right ) ^{2}} \left ({\frac{{x}^{6}}{8}}-{\frac{{x}^{2}}{8}} \right ) }+{\frac{\arctan \left ({x}^{2} \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^4+1)^3,x)

[Out]

1/2*(1/8*x^6-1/8*x^2)/(x^4+1)^2+1/16*arctan(x^2)

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Maxima [A]  time = 1.44623, size = 41, normalized size = 1.11 \begin{align*} \frac{x^{6} - x^{2}}{16 \,{\left (x^{8} + 2 \, x^{4} + 1\right )}} + \frac{1}{16} \, \arctan \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1)^3,x, algorithm="maxima")

[Out]

1/16*(x^6 - x^2)/(x^8 + 2*x^4 + 1) + 1/16*arctan(x^2)

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Fricas [A]  time = 1.94351, size = 92, normalized size = 2.49 \begin{align*} \frac{x^{6} - x^{2} +{\left (x^{8} + 2 \, x^{4} + 1\right )} \arctan \left (x^{2}\right )}{16 \,{\left (x^{8} + 2 \, x^{4} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1)^3,x, algorithm="fricas")

[Out]

1/16*(x^6 - x^2 + (x^8 + 2*x^4 + 1)*arctan(x^2))/(x^8 + 2*x^4 + 1)

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Sympy [A]  time = 0.148692, size = 24, normalized size = 0.65 \begin{align*} \frac{x^{6} - x^{2}}{16 x^{8} + 32 x^{4} + 16} + \frac{\operatorname{atan}{\left (x^{2} \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(x**4+1)**3,x)

[Out]

(x**6 - x**2)/(16*x**8 + 32*x**4 + 16) + atan(x**2)/16

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Giac [A]  time = 1.06708, size = 54, normalized size = 1.46 \begin{align*} \frac{x^{2} - \frac{1}{x^{2}}}{16 \,{\left ({\left (x^{2} - \frac{1}{x^{2}}\right )}^{2} + 4\right )}} + \frac{1}{32} \, \arctan \left (\frac{x^{4} - 1}{2 \, x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(x^4+1)^3,x, algorithm="giac")

[Out]

1/16*(x^2 - 1/x^2)/((x^2 - 1/x^2)^2 + 4) + 1/32*arctan(1/2*(x^4 - 1)/x^2)