3.207 \(\int \frac{x^{13}}{(a^4+x^4)^5} \, dx\)

Optimal. Leaf size=83 \[ -\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}+\frac{5 x^2}{256 a^4 \left (a^4+x^4\right )}-\frac{5 x^2}{128 \left (a^4+x^4\right )^2}+\frac{5 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{256 a^6} \]

[Out]

-x^10/(16*(a^4 + x^4)^4) - (5*x^6)/(96*(a^4 + x^4)^3) - (5*x^2)/(128*(a^4 + x^4)^2) + (5*x^2)/(256*a^4*(a^4 +
x^4)) + (5*ArcTan[x^2/a^2])/(256*a^6)

________________________________________________________________________________________

Rubi [A]  time = 0.0391206, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {275, 288, 199, 203} \[ -\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}+\frac{5 x^2}{256 a^4 \left (a^4+x^4\right )}-\frac{5 x^2}{128 \left (a^4+x^4\right )^2}+\frac{5 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{256 a^6} \]

Antiderivative was successfully verified.

[In]

Int[x^13/(a^4 + x^4)^5,x]

[Out]

-x^10/(16*(a^4 + x^4)^4) - (5*x^6)/(96*(a^4 + x^4)^3) - (5*x^2)/(128*(a^4 + x^4)^2) + (5*x^2)/(256*a^4*(a^4 +
x^4)) + (5*ArcTan[x^2/a^2])/(256*a^6)

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^{13}}{\left (a^4+x^4\right )^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^6}{\left (a^4+x^2\right )^5} \, dx,x,x^2\right )\\ &=-\frac{x^{10}}{16 \left (a^4+x^4\right )^4}+\frac{5}{16} \operatorname{Subst}\left (\int \frac{x^4}{\left (a^4+x^2\right )^4} \, dx,x,x^2\right )\\ &=-\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}+\frac{5}{32} \operatorname{Subst}\left (\int \frac{x^2}{\left (a^4+x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}-\frac{5 x^2}{128 \left (a^4+x^4\right )^2}+\frac{5}{128} \operatorname{Subst}\left (\int \frac{1}{\left (a^4+x^2\right )^2} \, dx,x,x^2\right )\\ &=-\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}-\frac{5 x^2}{128 \left (a^4+x^4\right )^2}+\frac{5 x^2}{256 a^4 \left (a^4+x^4\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a^4+x^2} \, dx,x,x^2\right )}{256 a^4}\\ &=-\frac{x^{10}}{16 \left (a^4+x^4\right )^4}-\frac{5 x^6}{96 \left (a^4+x^4\right )^3}-\frac{5 x^2}{128 \left (a^4+x^4\right )^2}+\frac{5 x^2}{256 a^4 \left (a^4+x^4\right )}+\frac{5 \tan ^{-1}\left (\frac{x^2}{a^2}\right )}{256 a^6}\\ \end{align*}

Mathematica [A]  time = 0.0218465, size = 62, normalized size = 0.75 \[ \frac{15 \tan ^{-1}\left (\frac{x^2}{a^2}\right )-\frac{a^2 x^2 \left (55 a^8 x^4+73 a^4 x^8+15 a^{12}-15 x^{12}\right )}{\left (a^4+x^4\right )^4}}{768 a^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^13/(a^4 + x^4)^5,x]

[Out]

(-((a^2*x^2*(15*a^12 + 55*a^8*x^4 + 73*a^4*x^8 - 15*x^12))/(a^4 + x^4)^4) + 15*ArcTan[x^2/a^2])/(768*a^6)

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 56, normalized size = 0.7 \begin{align*}{\frac{1}{2\, \left ({a}^{4}+{x}^{4} \right ) ^{4}} \left ({\frac{5\,{x}^{14}}{128\,{a}^{4}}}-{\frac{73\,{x}^{10}}{384}}-{\frac{55\,{x}^{6}{a}^{4}}{384}}-{\frac{5\,{a}^{8}{x}^{2}}{128}} \right ) }+{\frac{5}{256\,{a}^{6}}\arctan \left ({\frac{{x}^{2}}{{a}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(a^4+x^4)^5,x)

[Out]

1/2*(5/128/a^4*x^14-73/384*x^10-55/384*x^6*a^4-5/128*a^8*x^2)/(a^4+x^4)^4+5/256*arctan(x^2/a^2)/a^6

________________________________________________________________________________________

Maxima [A]  time = 1.54016, size = 112, normalized size = 1.35 \begin{align*} -\frac{15 \, a^{12} x^{2} + 55 \, a^{8} x^{6} + 73 \, a^{4} x^{10} - 15 \, x^{14}}{768 \,{\left (a^{20} + 4 \, a^{16} x^{4} + 6 \, a^{12} x^{8} + 4 \, a^{8} x^{12} + a^{4} x^{16}\right )}} + \frac{5 \, \arctan \left (\frac{x^{2}}{a^{2}}\right )}{256 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(a^4+x^4)^5,x, algorithm="maxima")

[Out]

-1/768*(15*a^12*x^2 + 55*a^8*x^6 + 73*a^4*x^10 - 15*x^14)/(a^20 + 4*a^16*x^4 + 6*a^12*x^8 + 4*a^8*x^12 + a^4*x
^16) + 5/256*arctan(x^2/a^2)/a^6

________________________________________________________________________________________

Fricas [A]  time = 1.76912, size = 263, normalized size = 3.17 \begin{align*} -\frac{15 \, a^{14} x^{2} + 55 \, a^{10} x^{6} + 73 \, a^{6} x^{10} - 15 \, a^{2} x^{14} - 15 \,{\left (a^{16} + 4 \, a^{12} x^{4} + 6 \, a^{8} x^{8} + 4 \, a^{4} x^{12} + x^{16}\right )} \arctan \left (\frac{x^{2}}{a^{2}}\right )}{768 \,{\left (a^{22} + 4 \, a^{18} x^{4} + 6 \, a^{14} x^{8} + 4 \, a^{10} x^{12} + a^{6} x^{16}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(a^4+x^4)^5,x, algorithm="fricas")

[Out]

-1/768*(15*a^14*x^2 + 55*a^10*x^6 + 73*a^6*x^10 - 15*a^2*x^14 - 15*(a^16 + 4*a^12*x^4 + 6*a^8*x^8 + 4*a^4*x^12
+ x^16)*arctan(x^2/a^2))/(a^22 + 4*a^18*x^4 + 6*a^14*x^8 + 4*a^10*x^12 + a^6*x^16)

________________________________________________________________________________________

Sympy [C]  time = 35.7445, size = 102, normalized size = 1.23 \begin{align*} \frac{- 15 a^{12} x^{2} - 55 a^{8} x^{6} - 73 a^{4} x^{10} + 15 x^{14}}{768 a^{20} + 3072 a^{16} x^{4} + 4608 a^{12} x^{8} + 3072 a^{8} x^{12} + 768 a^{4} x^{16}} + \frac{- \frac{5 i \log{\left (- i a^{2} + x^{2} \right )}}{512} + \frac{5 i \log{\left (i a^{2} + x^{2} \right )}}{512}}{a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(a**4+x**4)**5,x)

[Out]

(-15*a**12*x**2 - 55*a**8*x**6 - 73*a**4*x**10 + 15*x**14)/(768*a**20 + 3072*a**16*x**4 + 4608*a**12*x**8 + 30
72*a**8*x**12 + 768*a**4*x**16) + (-5*I*log(-I*a**2 + x**2)/512 + 5*I*log(I*a**2 + x**2)/512)/a**6

________________________________________________________________________________________

Giac [A]  time = 1.06157, size = 78, normalized size = 0.94 \begin{align*} \frac{5 \, \arctan \left (\frac{x^{2}}{a^{2}}\right )}{256 \, a^{6}} - \frac{15 \, a^{12} x^{2} + 55 \, a^{8} x^{6} + 73 \, a^{4} x^{10} - 15 \, x^{14}}{768 \,{\left (a^{4} + x^{4}\right )}^{4} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(a^4+x^4)^5,x, algorithm="giac")

[Out]

5/256*arctan(x^2/a^2)/a^6 - 1/768*(15*a^12*x^2 + 55*a^8*x^6 + 73*a^4*x^10 - 15*x^14)/((a^4 + x^4)^4*a^4)