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

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

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Rubi [A]  time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.68 \[ \frac {1}{16} \left (\tan ^{-1}\left (x^2\right )+\frac {\left (x^4-1\right ) x^2}{\left (x^4+1\right )^2}\right ) \]

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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IntegrateAlgebraic [A]  time = 0.02, size = 28, normalized size = 0.76 \[ \frac {1}{16} \tan ^{-1}\left (x^2\right )+\frac {\left (x^4-1\right ) x^2}{16 \left (x^4+1\right )^2} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/(1 + x^4)^3,x]

[Out]

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

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fricas [A]  time = 1.37, size = 38, normalized size = 1.03 \[ \frac {x^{6} - x^{2} + {\left (x^{8} + 2 \, x^{4} + 1\right )} \arctan \left (x^{2}\right )}{16 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}} \]

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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giac [A]  time = 1.01, size = 40, normalized size = 1.08 \[ \frac {x^{2} - \frac {1}{x^{2}}}{16 \, {\left ({\left (x^{2} - \frac {1}{x^{2}}\right )}^{2} + 4\right )}} + \frac {1}{32} \, \arctan \left (\frac {x^{4} - 1}{2 \, x^{2}}\right ) \]

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)

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maple [A]  time = 0.27, size = 27, normalized size = 0.73

method	result	size
meijerg	\(-\frac {x^{2} \left (-3 x^{4}+3\right )}{48 \left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\)	\(27\)
risch	\(\frac {\frac {1}{16} x^{6}-\frac {1}{16} x^{2}}{\left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\)	\(27\)
default	\(\frac {\frac {1}{8} x^{6}-\frac {1}{8} x^{2}}{2 \left (x^{4}+1\right )^{2}}+\frac {\arctan \left (x^{2}\right )}{16}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*x^2*(-3*x^4+3)/(x^4+1)^2+1/16*arctan(x^2)

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maxima [A]  time = 0.95, size = 30, normalized size = 0.81 \[ \frac {x^{6} - x^{2}}{16 \, {\left (x^{8} + 2 \, x^{4} + 1\right )}} + \frac {1}{16} \, \arctan \left (x^{2}\right ) \]

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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mupad [B]  time = 0.04, size = 32, normalized size = 0.86 \[ \frac {\mathrm {atan}\left (x^2\right )}{16}-\frac {\frac {x^2}{16}-\frac {x^6}{16}}{x^8+2\,x^4+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^4 + 1)^3,x)

[Out]

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

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sympy [A]  time = 0.17, size = 24, normalized size = 0.65 \[ \frac {x^{6} - x^{2}}{16 x^{8} + 32 x^{4} + 16} + \frac {\operatorname {atan}{\left (x^{2} \right )}}{16} \]

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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