Integrand size = 13, antiderivative size = 64 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {15}{16 a^{12} x^2}+\frac {1}{8 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5}{16 a^8 x^2 \left (a^4+x^4\right )}-\frac {15 \arctan \left (\frac {x^2}{a^2}\right )}{16 a^{14}} \] Output:
-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*arcta n(x^2/a^2)/a^14
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=\frac {-\frac {a^2 \left (8 a^8+25 a^4 x^4+15 x^8\right )}{x^2 \left (a^4+x^4\right )^2}+15 \arctan \left (1-\frac {\sqrt {2} x}{a}\right )+15 \arctan \left (1+\frac {\sqrt {2} x}{a}\right )}{16 a^{14}} \] Input:
Integrate[1/(x^3*(a^4 + x^4)^3),x]
Output:
(-((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 + (Sqrt[2]*x)/a])/(16*a^14)
Time = 0.18 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {807, 253, 253, 264, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (a^4+x^4\right )^3}dx^2\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \int \frac {1}{x^4 \left (a^4+x^4\right )^2}dx^2}{4 a^4}+\frac {1}{4 a^4 x^2 \left (a^4+x^4\right )^2}\right )\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (\frac {3 \int \frac {1}{x^4 \left (a^4+x^4\right )}dx^2}{2 a^4}+\frac {1}{2 a^4 x^2 \left (a^4+x^4\right )}\right )}{4 a^4}+\frac {1}{4 a^4 x^2 \left (a^4+x^4\right )^2}\right )\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {1}{2} \left (\frac {5 \left (\frac {3 \left (-\frac {\int \frac {1}{a^4+x^4}dx^2}{a^4}-\frac {1}{a^4 x^2}\right )}{2 a^4}+\frac {1}{2 a^4 x^2 \left (a^4+x^4\right )}\right )}{4 a^4}+\frac {1}{4 a^4 x^2 \left (a^4+x^4\right )^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4 a^4 x^2 \left (a^4+x^4\right )^2}+\frac {5 \left (\frac {1}{2 a^4 x^2 \left (a^4+x^4\right )}+\frac {3 \left (-\frac {1}{a^4 x^2}-\frac {\arctan \left (\frac {x^2}{a^2}\right )}{a^6}\right )}{2 a^4}\right )}{4 a^4}\right )\) |
Input:
Int[1/(x^3*(a^4 + x^4)^3),x]
Output:
(1/(4*a^4*x^2*(a^4 + x^4)^2) + (5*(1/(2*a^4*x^2*(a^4 + x^4)) + (3*(-(1/(a^ 4*x^2)) - ArcTan[x^2/a^2]/a^6))/(2*a^4)))/(4*a^4))/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83
method | result | size |
default | \(-\frac {1}{2 a^{12} x^{2}}-\frac {\frac {\frac {9}{8} a^{4} x^{2}+\frac {7}{8} x^{6}}{\left (a^{4}+x^{4}\right )^{2}}+\frac {15 \arctan \left (\frac {x^{2}}{a^{2}}\right )}{8 a^{2}}}{2 a^{12}}\) | \(53\) |
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 | \(\frac {15 i \ln \left (-i a^{2}+x^{2}\right ) x^{10}+30 i \ln \left (-i a^{2}+x^{2}\right ) x^{6} a^{4}+15 i \ln \left (-i a^{2}+x^{2}\right ) x^{2} a^{8}-15 i \ln \left (i a^{2}+x^{2}\right ) x^{10}-30 i \ln \left (i a^{2}+x^{2}\right ) x^{6} a^{4}-15 i \ln \left (i a^{2}+x^{2}\right ) x^{2} a^{8}-30 x^{8} a^{2}-50 x^{4} a^{6}-16 a^{10}}{32 a^{14} x^{2} \left (a^{4}+x^{4}\right )^{2}}\) | \(154\) |
Input:
int(1/x^3/(a^4+x^4)^3,x,method=_RETURNVERBOSE)
Output:
-1/2/a^12/x^2-1/2/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)
Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {8 \, a^{10} + 25 \, a^{6} x^{4} + 15 \, a^{2} x^{8} + 15 \, {\left (a^{8} x^{2} + 2 \, a^{4} x^{6} + x^{10}\right )} \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, {\left (a^{22} x^{2} + 2 \, a^{18} x^{6} + a^{14} x^{10}\right )}} \] Input:
integrate(1/x^3/(a^4+x^4)^3,x, algorithm="fricas")
Output:
-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)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=\frac {- 8 a^{8} - 25 a^{4} x^{4} - 15 x^{8}}{16 a^{20} x^{2} + 32 a^{16} x^{6} + 16 a^{12} x^{10}} + \frac {\frac {15 i \log {\left (- i a^{2} + x^{2} \right )}}{32} - \frac {15 i \log {\left (i a^{2} + x^{2} \right )}}{32}}{a^{14}} \] Input:
integrate(1/x**3/(a**4+x**4)**3,x)
Output:
(-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
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {8 \, a^{8} + 25 \, a^{4} x^{4} + 15 \, x^{8}}{16 \, {\left (a^{20} x^{2} + 2 \, a^{16} x^{6} + a^{12} x^{10}\right )}} - \frac {15 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, a^{14}} \] Input:
integrate(1/x^3/(a^4+x^4)^3,x, algorithm="maxima")
Output:
-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
Time = 0.12 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {9 \, a^{4} x^{2} + 7 \, x^{6}}{16 \, {\left (a^{4} + x^{4}\right )}^{2} a^{12}} - \frac {15 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{16 \, a^{14}} - \frac {1}{2 \, a^{12} x^{2}} \] Input:
integrate(1/x^3/(a^4+x^4)^3,x, algorithm="giac")
Output:
-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^1 4 - 1/2/(a^12*x^2)
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=-\frac {15\,\mathrm {atan}\left (\frac {x^2}{a^2}\right )}{16\,a^{14}}-\frac {\frac {a^{10}}{2}+\frac {25\,a^6\,x^4}{16}+\frac {15\,a^2\,x^8}{16}}{a^{14}\,x^2\,{\left (a^4+x^4\right )}^2} \] Input:
int(1/(x^3*(a^4 + x^4)^3),x)
Output:
- (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)
Time = 0.16 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.98 \[ \int \frac {1}{x^3 \left (a^4+x^4\right )^3} \, dx=\frac {15 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{8} x^{2}+30 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{4} x^{6}+15 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) x^{10}+15 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{8} x^{2}+30 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{4} x^{6}+15 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) x^{10}-8 a^{10}-25 a^{6} x^{4}-15 a^{2} x^{8}}{16 a^{14} x^{2} \left (a^{8}+2 a^{4} x^{4}+x^{8}\right )} \] Input:
int(1/x^3/(a^4+x^4)^3,x)
Output:
(15*atan((sqrt(2)*a - 2*x)/(sqrt(2)*a))*a**8*x**2 + 30*atan((sqrt(2)*a - 2 *x)/(sqrt(2)*a))*a**4*x**6 + 15*atan((sqrt(2)*a - 2*x)/(sqrt(2)*a))*x**10 + 15*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a**8*x**2 + 30*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a**4*x**6 + 15*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*x**10 - 8*a**10 - 25*a**6*x**4 - 15*a**2*x**8)/(16*a**14*x**2*(a**8 + 2*a**4*x* *4 + x**8))