\(\int \frac {1}{x^3 (a^4+x^4)^3} \, dx\) [175]

Optimal result

Integrand size = 13, antiderivative size = 64 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {15}{16 a^{12} x^2}+\frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac {15 \arctan \left (\frac {x^2}{a^2}\right )}{16 a^{14}} \]

[Out]

-15/16/a^12/x^2+1/8/a^4/x^2/(a^4+x^4)^2+5/16/a^8/x^2/(a^4+x^4)-15/16*arctan(x^2/a^2)/a^14

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {281, 296, 331, 209} \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {15}{16 a^{12} x^2}+\frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}-\frac {15 \arctan \left (\frac {x^2}{a^2}\right )}{16 a^{14}}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )} \]

[In]

Int[1/(x^3*(a^4 + x^4)^3),x]

[Out]

-15/(16*a^12*x^2) + 1/(8*a^4*x^2*(a^4 + x^4)^2) + 5/(16*a^8*x^2*(a^4 + x^4)) - (15*ArcTan[x^2/a^2])/(16*a^14)

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 296

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

Rule 331

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 \left (a^4+x^2\right )^3} \, dx,x,x^2\right ) \\ & = \frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5 \text {Subst}\left (\int \frac {1}{x^2 \left (a^4+x^2\right )^2} \, dx,x,x^2\right )}{8 a^4} \\ & = \frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )}+\frac {15 \text {Subst}\left (\int \frac {1}{x^2 \left (a^4+x^2\right )} \, dx,x,x^2\right )}{16 a^8} \\ & = -\frac {15}{16 a^{12} x^2}+\frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac {15 \text {Subst}\left (\int \frac {1}{a^4+x^2} \, dx,x,x^2\right )}{16 a^{12}} \\ & = -\frac {15}{16 a^{12} x^2}+\frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac {15 \arctan \left (\frac {x^2}{a^2}\right )}{16 a^{14}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=\frac {-\frac {a^2 \left (8 a^8+25 a^4 x^4+15 x^8\right )}{x^2 \left (a^4+x^4\right )^2}+15 \arctan \left (1-\frac {\sqrt {2} x}{a}\right )+15 \arctan \left (1+\frac {\sqrt {2} x}{a}\right )}{16 a^{14}} \]

[In]

Integrate[1/(x^3*(a^4 + x^4)^3),x]

[Out]

(-((a^2*(8*a^8 + 25*a^4*x^4 + 15*x^8))/(x^2*(a^4 + x^4)^2)) + 15*ArcTan[1 - (Sqrt[2]*x)/a] + 15*ArcTan[1 + (Sq
rt[2]*x)/a])/(16*a^14)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {\frac {\frac {9}{8} a^{4} x^{2}+\frac {7}{8} x^{6}}{\left (a^{4}+x^{4}\right )^{2}}+\frac {15 \arctan \left (\frac {x^{2}}{a^{2}}\right )}{8 a^{2}}}{2 a^{12}}-\frac {1}{2 a^{12} x^{2}}\)	\(53\)
risch	\(\frac {-\frac {15 x^{8}}{16 a^{12}}-\frac {25 x^{4}}{16 a^{8}}-\frac {1}{2 a^{4}}}{x^{2} \left (a^{4}+x^{4}\right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{28} \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\left (-5 \textit {\_R}^{2} a^{28}-4\right ) x^{2}-a^{16} \textit {\_R} \right )\right )}{32}\)	\(76\)
parallelrisch	\(\frac {15 i \ln \left (-i a^{2}+x^{2}\right ) x^{10}+30 i \ln \left (-i a^{2}+x^{2}\right ) x^{6} a^{4}+15 i \ln \left (-i a^{2}+x^{2}\right ) x^{2} a^{8}-15 i \ln \left (i a^{2}+x^{2}\right ) x^{10}-30 i \ln \left (i a^{2}+x^{2}\right ) x^{6} a^{4}-15 i \ln \left (i a^{2}+x^{2}\right ) x^{2} a^{8}-30 x^{8} a^{2}-50 x^{4} a^{6}-16 a^{10}}{32 a^{14} x^{2} \left (a^{4}+x^{4}\right )^{2}}\)	\(154\)

[In]

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

[Out]

-1/2/a^12*((9/8*a^4*x^2+7/8*x^6)/(a^4+x^4)^2+15/8*arctan(x^2/a^2)/a^2)-1/2/a^12/x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {8 \, a^{10} + 25 \, a^{6} x^{4} + 15 \, a^{2} x^{8} + 15 \, {\left (a^{8} x^{2} + 2 \, a^{4} x^{6} + x^{10}\right )} \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, {\left (a^{22} x^{2} + 2 \, a^{18} x^{6} + a^{14} x^{10}\right )}} \]

[In]

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

[Out]

-1/16*(8*a^10 + 25*a^6*x^4 + 15*a^2*x^8 + 15*(a^8*x^2 + 2*a^4*x^6 + x^10)*arctan(x^2/a^2))/(a^22*x^2 + 2*a^18*
x^6 + a^14*x^10)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=\frac {- 8 a^{8} - 25 a^{4} x^{4} - 15 x^{8}}{16 a^{20} x^{2} + 32 a^{16} x^{6} + 16 a^{12} x^{10}} + \frac {\frac {15 i \log {\left (- i a^{2} + x^{2} \right )}}{32} - \frac {15 i \log {\left (i a^{2} + x^{2} \right )}}{32}}{a^{14}} \]

[In]

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

[Out]

(-8*a**8 - 25*a**4*x**4 - 15*x**8)/(16*a**20*x**2 + 32*a**16*x**6 + 16*a**12*x**10) + (15*I*log(-I*a**2 + x**2
)/32 - 15*I*log(I*a**2 + x**2)/32)/a**14

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {8 \, a^{8} + 25 \, a^{4} x^{4} + 15 \, x^{8}}{16 \, {\left (a^{20} x^{2} + 2 \, a^{16} x^{6} + a^{12} x^{10}\right )}} - \frac {15 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, a^{14}} \]

[In]

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

[Out]

-1/16*(8*a^8 + 25*a^4*x^4 + 15*x^8)/(a^20*x^2 + 2*a^16*x^6 + a^12*x^10) - 15/16*arctan(x^2/a^2)/a^14

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {9 \, a^{4} x^{2} + 7 \, x^{6}}{16 \, {\left (a^{4} + x^{4}\right )}^{2} a^{12}} - \frac {15 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, a^{14}} - \frac {1}{2 \, a^{12} x^{2}} \]

[In]

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

[Out]

-1/16*(9*a^4*x^2 + 7*x^6)/((a^4 + x^4)^2*a^12) - 15/16*arctan(x^2/a^2)/a^14 - 1/2/(a^12*x^2)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {15\,\mathrm {atan}\left (\frac {x^2}{a^2}\right )}{16\,a^{14}}-\frac {\frac {a^{10}}{2}+\frac {25\,a^6\,x^4}{16}+\frac {15\,a^2\,x^8}{16}}{a^{14}\,x^2\,{\left (a^4+x^4\right )}^2} \]

[In]

int(1/(x^3*(a^4 + x^4)^3),x)

[Out]

- (15*atan(x^2/a^2))/(16*a^14) - (a^10/2 + (15*a^2*x^8)/16 + (25*a^6*x^4)/16)/(a^14*x^2*(a^4 + x^4)^2)