Integrand size = 16, antiderivative size = 78 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=-\frac {x^3}{4 \left (1+x^4\right )}-\frac {3 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {3 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}} \] Output:
-1/4*x^3/(x^4+1)+3/16*arctan(-1+x*2^(1/2))*2^(1/2)+3/16*arctan(1+x*2^(1/2) )*2^(1/2)-3/16*arctanh(2^(1/2)*x/(x^2+1))*2^(1/2)
Time = 0.06 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=\frac {1}{32} \left (-\frac {8 x^3}{1+x^4}-6 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )+6 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+3 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-3 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:
Integrate[x^6/(1 + 2*x^4 + x^8),x]
Output:
((-8*x^3)/(1 + x^4) - 6*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] + 3*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 3*Sqrt[2]*Log[1 + Sqr t[2]*x + x^2])/32
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {1380, 817, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {x^6}{x^8+2 x^4+1} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {x^6}{\left (x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {3}{4} \int \frac {x^2}{x^4+1}dx-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx+\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )\right )-\frac {x^3}{4 \left (x^4+1\right )}\) |
Input:
Int[x^6/(1 + 2*x^4 + x^8),x]
Output:
-1/4*x^3/(1 + x^4) + (3*((-(ArcTan[1 - Sqrt[2]*x]/Sqrt[2]) + ArcTan[1 + Sq rt[2]*x]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*x + x^2]/(2*Sqrt[2]) - Log[1 + Sqrt [2]*x + x^2]/(2*Sqrt[2]))/2))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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] - Simp[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]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.45
method | result | size |
risch | \(-\frac {x^{3}}{4 \left (x^{4}+1\right )}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{16}\) | \(35\) |
default | \(-\frac {x^{3}}{4 \left (x^{4}+1\right )}+\frac {3 \sqrt {2}\, \left (\ln \left (\frac {x^{2}-x \sqrt {2}+1}{x^{2}+x \sqrt {2}+1}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{32}\) | \(65\) |
Input:
int(x^6/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)
Output:
-1/4*x^3/(x^4+1)+3/16*sum(1/_R*ln(x-_R),_R=RootOf(_Z^4+1))
Time = 0.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=-\frac {8 \, x^{3} - 6 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x + 1\right ) - 6 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (\sqrt {2} x - 1\right ) + 3 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - 3 \, \sqrt {2} {\left (x^{4} + 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right )}{32 \, {\left (x^{4} + 1\right )}} \] Input:
integrate(x^6/(x^8+2*x^4+1),x, algorithm="fricas")
Output:
-1/32*(8*x^3 - 6*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x + 1) - 6*sqrt(2)*(x^4 + 1)*arctan(sqrt(2)*x - 1) + 3*sqrt(2)*(x^4 + 1)*log(x^2 + sqrt(2)*x + 1) - 3*sqrt(2)*(x^4 + 1)*log(x^2 - sqrt(2)*x + 1))/(x^4 + 1)
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.15 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=- \frac {x^{3}}{4 x^{4} + 4} + \frac {3 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {3 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {3 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {3 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \] Input:
integrate(x**6/(x**8+2*x**4+1),x)
Output:
-x**3/(4*x**4 + 4) + 3*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 3*sqrt(2)*lo g(x**2 + sqrt(2)*x + 1)/32 + 3*sqrt(2)*atan(sqrt(2)*x - 1)/16 + 3*sqrt(2)* atan(sqrt(2)*x + 1)/16
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=-\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {3}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {3}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {3}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \] Input:
integrate(x^6/(x^8+2*x^4+1),x, algorithm="maxima")
Output:
-1/4*x^3/(x^4 + 1) + 3/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 3/ 16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 3/32*sqrt(2)*log(x^2 + sq rt(2)*x + 1) + 3/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1)
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=-\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {3}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {3}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {3}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {3}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \] Input:
integrate(x^6/(x^8+2*x^4+1),x, algorithm="giac")
Output:
-1/4*x^3/(x^4 + 1) + 3/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 3/ 16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 3/32*sqrt(2)*log(x^2 + sq rt(2)*x + 1) + 3/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1)
Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.60 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=-\frac {x^3}{4\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{16}-\frac {3}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {3}{16}+\frac {3}{16}{}\mathrm {i}\right ) \] Input:
int(x^6/(2*x^4 + x^8 + 1),x)
Output:
2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(3/16 - 3i/16) + 2^(1/2)*atan(2^(1/2) *x*(1/2 + 1i/2))*(3/16 + 3i/16) - x^3/(4*(x^4 + 1))
Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.92 \[ \int \frac {x^6}{1+2 x^4+x^8} \, dx=\frac {-6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{4}-6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right )+6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{4}+6 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right )+3 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{4}+3 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right )-3 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{4}-3 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right )-8 x^{3}}{32 x^{4}+32} \] Input:
int(x^6/(x^8+2*x^4+1),x)
Output:
( - 6*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**4 - 6*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2)) + 6*sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**4 + 6*sqrt(2 )*atan((sqrt(2) + 2*x)/sqrt(2)) + 3*sqrt(2)*log( - sqrt(2)*x + x**2 + 1)*x **4 + 3*sqrt(2)*log( - sqrt(2)*x + x**2 + 1) - 3*sqrt(2)*log(sqrt(2)*x + x **2 + 1)*x**4 - 3*sqrt(2)*log(sqrt(2)*x + x**2 + 1) - 8*x**3)/(32*(x**4 + 1))