Integrand size = 13, antiderivative size = 83 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=-\frac {x^{10}}{16 \left (a^4+x^4\right )^4}-\frac {5 x^6}{96 \left (a^4+x^4\right )^3}-\frac {5 x^2}{128 \left (a^4+x^4\right )^2}+\frac {5 x^2}{256 a^4 \left (a^4+x^4\right )}+\frac {5 \arctan \left (\frac {x^2}{a^2}\right )}{256 a^6} \]
-1/16*x^10/(a^4+x^4)^4-5/96*x^6/(a^4+x^4)^3-5/128*x^2/(a^4+x^4)^2+5/256*x^ 2/a^4/(a^4+x^4)+5/256*arctan(x^2/a^2)/a^6
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=\frac {-\frac {a^2 x^2 \left (15 a^{12}+55 a^8 x^4+73 a^4 x^8-15 x^{12}\right )}{\left (a^4+x^4\right )^4}+15 \arctan \left (\frac {x^2}{a^2}\right )}{768 a^6} \]
(-((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)
Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {807, 252, 252, 252, 215, 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.
\(\displaystyle \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {x^{12}}{\left (a^4+x^4\right )^5}dx^2\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{8} \int \frac {x^8}{\left (a^4+x^4\right )^4}dx^2-\frac {x^{10}}{8 \left (a^4+x^4\right )^4}\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{8} \left (\frac {1}{2} \int \frac {x^4}{\left (a^4+x^4\right )^3}dx^2-\frac {x^6}{6 \left (a^4+x^4\right )^3}\right )-\frac {x^{10}}{8 \left (a^4+x^4\right )^4}\right )\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{\left (a^4+x^4\right )^2}dx^2-\frac {x^2}{4 \left (a^4+x^4\right )^2}\right )-\frac {x^6}{6 \left (a^4+x^4\right )^3}\right )-\frac {x^{10}}{8 \left (a^4+x^4\right )^4}\right )\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {\int \frac {1}{a^4+x^4}dx^2}{2 a^4}+\frac {x^2}{2 a^4 \left (a^4+x^4\right )}\right )-\frac {x^2}{4 \left (a^4+x^4\right )^2}\right )-\frac {x^6}{6 \left (a^4+x^4\right )^3}\right )-\frac {x^{10}}{8 \left (a^4+x^4\right )^4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {5}{8} \left (\frac {1}{2} \left (\frac {1}{4} \left (\frac {x^2}{2 a^4 \left (a^4+x^4\right )}+\frac {\arctan \left (\frac {x^2}{a^2}\right )}{2 a^6}\right )-\frac {x^2}{4 \left (a^4+x^4\right )^2}\right )-\frac {x^6}{6 \left (a^4+x^4\right )^3}\right )-\frac {x^{10}}{8 \left (a^4+x^4\right )^4}\right )\) |
(-1/8*x^10/(a^4 + x^4)^4 + (5*(-1/6*x^6/(a^4 + x^4)^3 + (-1/4*x^2/(a^4 + x ^4)^2 + (x^2/(2*a^4*(a^4 + x^4)) + ArcTan[x^2/a^2]/(2*a^6))/4)/2))/8)/2
3.3.7.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[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 = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.66
method | result | size |
risch | \(\frac {-\frac {5 a^{8} x^{2}}{256}-\frac {55 a^{4} x^{6}}{768}-\frac {73 x^{10}}{768}+\frac {5 x^{14}}{256 a^{4}}}{\left (a^{4}+x^{4}\right )^{4}}+\frac {5 \arctan \left (\frac {x^{2}}{a^{2}}\right )}{256 a^{6}}\) | \(55\) |
default | \(\frac {\frac {5 x^{14}}{128 a^{4}}-\frac {73 x^{10}}{384}-\frac {55 a^{4} x^{6}}{384}-\frac {5 a^{8} x^{2}}{128}}{2 \left (a^{4}+x^{4}\right )^{4}}+\frac {5 \arctan \left (\frac {x^{2}}{a^{2}}\right )}{256 a^{6}}\) | \(56\) |
parallelrisch | \(-\frac {-60 i \ln \left (i a^{2}+x^{2}\right ) x^{4} a^{12}+60 i \ln \left (-i a^{2}+x^{2}\right ) x^{4} a^{12}-60 i \ln \left (i a^{2}+x^{2}\right ) x^{12} a^{4}+15 i \ln \left (-i a^{2}+x^{2}\right ) x^{16}-90 i \ln \left (i a^{2}+x^{2}\right ) x^{8} a^{8}+90 i \ln \left (-i a^{2}+x^{2}\right ) x^{8} a^{8}+15 i \ln \left (-i a^{2}+x^{2}\right ) a^{16}+60 i \ln \left (-i a^{2}+x^{2}\right ) x^{12} a^{4}-15 i \ln \left (i a^{2}+x^{2}\right ) a^{16}-15 i \ln \left (i a^{2}+x^{2}\right ) x^{16}-30 x^{14} a^{2}+146 x^{10} a^{6}+110 x^{6} a^{10}+30 x^{2} a^{14}}{1536 a^{6} \left (a^{4}+x^{4}\right )^{4}}\) | \(236\) |
(-5/256*a^8*x^2-55/768*a^4*x^6-73/768*x^10+5/256/a^4*x^14)/(a^4+x^4)^4+5/2 56*arctan(x^2/a^2)/a^6
Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.36 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=-\frac {15 \, a^{14} x^{2} + 55 \, a^{10} x^{6} + 73 \, a^{6} x^{10} - 15 \, a^{2} x^{14} - 15 \, {\left (a^{16} + 4 \, a^{12} x^{4} + 6 \, a^{8} x^{8} + 4 \, a^{4} x^{12} + x^{16}\right )} \arctan \left (\frac {x^{2}}{a^{2}}\right )}{768 \, {\left (a^{22} + 4 \, a^{18} x^{4} + 6 \, a^{14} x^{8} + 4 \, a^{10} x^{12} + a^{6} x^{16}\right )}} \]
-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)
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.23 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=\frac {- 15 a^{12} x^{2} - 55 a^{8} x^{6} - 73 a^{4} x^{10} + 15 x^{14}}{768 a^{20} + 3072 a^{16} x^{4} + 4608 a^{12} x^{8} + 3072 a^{8} x^{12} + 768 a^{4} x^{16}} + \frac {- \frac {5 i \log {\left (- i a^{2} + x^{2} \right )}}{512} + \frac {5 i \log {\left (i a^{2} + x^{2} \right )}}{512}}{a^{6}} \]
(-15*a**12*x**2 - 55*a**8*x**6 - 73*a**4*x**10 + 15*x**14)/(768*a**20 + 30 72*a**16*x**4 + 4608*a**12*x**8 + 3072*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
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=-\frac {15 \, a^{12} x^{2} + 55 \, a^{8} x^{6} + 73 \, a^{4} x^{10} - 15 \, x^{14}}{768 \, {\left (a^{20} + 4 \, a^{16} x^{4} + 6 \, a^{12} x^{8} + 4 \, a^{8} x^{12} + a^{4} x^{16}\right )}} + \frac {5 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{256 \, a^{6}} \]
-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
Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=\frac {5 \, \arctan \left (\frac {x^{2}}{a^{2}}\right )}{256 \, a^{6}} - \frac {15 \, a^{12} x^{2} + 55 \, a^{8} x^{6} + 73 \, a^{4} x^{10} - 15 \, x^{14}}{768 \, {\left (a^{4} + x^{4}\right )}^{4} a^{4}} \]
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)
Time = 0.15 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=\frac {5\,\mathrm {atan}\left (\frac {x^2}{a^2}\right )}{256\,a^6}-\frac {\frac {73\,x^{10}}{768}+\frac {55\,a^4\,x^6}{768}+\frac {5\,a^8\,x^2}{256}-\frac {5\,x^{14}}{256\,a^4}}{a^{16}+4\,a^{12}\,x^4+6\,a^8\,x^8+4\,a^4\,x^{12}+x^{16}} \]
(5*atan(x^2/a^2))/(256*a^6) - ((73*x^10)/768 + (55*a^4*x^6)/768 + (5*a^8*x ^2)/256 - (5*x^14)/(256*a^4))/(a^16 + x^16 + 4*a^4*x^12 + 6*a^8*x^8 + 4*a^ 12*x^4)
Time = 0.00 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.72 \[ \int \frac {x^{13}}{\left (a^4+x^4\right )^5} \, dx=\frac {-15 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{16}-60 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{12} x^{4}-90 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{8} x^{8}-60 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) a^{4} x^{12}-15 \mathit {atan} \left (\frac {\sqrt {2}\, a -2 x}{\sqrt {2}\, a}\right ) x^{16}-15 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{16}-60 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{12} x^{4}-90 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{8} x^{8}-60 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) a^{4} x^{12}-15 \mathit {atan} \left (\frac {\sqrt {2}\, a +2 x}{\sqrt {2}\, a}\right ) x^{16}-15 a^{14} x^{2}-55 a^{10} x^{6}-73 a^{6} x^{10}+15 a^{2} x^{14}}{768 a^{6} \left (a^{16}+4 a^{12} x^{4}+6 a^{8} x^{8}+4 a^{4} x^{12}+x^{16}\right )} \]
( - 15*atan((sqrt(2)*a - 2*x)/(sqrt(2)*a))*a**16 - 60*atan((sqrt(2)*a - 2* x)/(sqrt(2)*a))*a**12*x**4 - 90*atan((sqrt(2)*a - 2*x)/(sqrt(2)*a))*a**8*x **8 - 60*atan((sqrt(2)*a - 2*x)/(sqrt(2)*a))*a**4*x**12 - 15*atan((sqrt(2) *a - 2*x)/(sqrt(2)*a))*x**16 - 15*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a**1 6 - 60*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a**12*x**4 - 90*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a**8*x**8 - 60*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*a* *4*x**12 - 15*atan((sqrt(2)*a + 2*x)/(sqrt(2)*a))*x**16 - 15*a**14*x**2 - 55*a**10*x**6 - 73*a**6*x**10 + 15*a**2*x**14)/(768*a**6*(a**16 + 4*a**12* x**4 + 6*a**8*x**8 + 4*a**4*x**12 + x**16))