Integrand size = 16, antiderivative size = 83 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {1}{3 x^3}-\frac {x}{4 \left (1+x^4\right )}+\frac {7 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{8 \sqrt {2}} \] Output:
-1/3/x^3-x/(4*x^4+4)-7/16*arctan(-1+x*2^(1/2))*2^(1/2)-7/16*arctan(1+x*2^( 1/2))*2^(1/2)-7/16*arctanh(2^(1/2)*x/(x^2+1))*2^(1/2)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{96} \left (-\frac {32}{x^3}-\frac {24 x}{1+x^4}+42 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-42 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+21 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-21 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \] Input:
Integrate[1/(x^4*(1 + 2*x^4 + x^8)),x]
Output:
(-32/x^3 - (24*x)/(1 + x^4) + 42*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 42*Sqrt[2 ]*ArcTan[1 + Sqrt[2]*x] + 21*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 21*Sqrt[2] *Log[1 + Sqrt[2]*x + x^2])/96
Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.40, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {1380, 819, 847, 755, 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.
\(\displaystyle \int \frac {1}{x^4 \left (x^8+2 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {1}{x^4 \left (x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {7}{4} \int \frac {1}{x^4 \left (x^4+1\right )}dx+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {7}{4} \left (-\int \frac {1}{x^4+1}dx-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {7}{4} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {7}{4} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {7}{4} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {7}{4} \left (-\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {7}{4} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {7}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {7}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {7}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{3 x^3}+\frac {1}{2} \left (\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )\right )+\frac {1}{4 x^3 \left (x^4+1\right )}\) |
Input:
Int[1/(x^4*(1 + 2*x^4 + x^8)),x]
Output:
1/(4*x^3*(1 + x^4)) + (7*(-1/3*1/x^3 + (ArcTan[1 - Sqrt[2]*x]/Sqrt[2] - Ar cTan[1 + Sqrt[2]*x]/Sqrt[2])/2 + (Log[1 - Sqrt[2]*x + x^2]/(2*Sqrt[2]) - L og[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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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[((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] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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 = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.47
method | result | size |
risch | \(\frac {-\frac {7 x^{4}}{12}-\frac {1}{3}}{x^{3} \left (x^{4}+1\right )}+\frac {7 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{16}\) | \(39\) |
default | \(-\frac {x}{4 \left (x^{4}+1\right )}-\frac {7 \sqrt {2}\, \left (\ln \left (\frac {x^{2}+x \sqrt {2}+1}{x^{2}-x \sqrt {2}+1}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{32}-\frac {1}{3 x^{3}}\) | \(68\) |
Input:
int(1/x^4/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)
Output:
(-7/12*x^4-1/3)/x^3/(x^4+1)+7/16*sum(_R*ln(x-_R),_R=RootOf(_Z^4+1))
Time = 0.07 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {56 \, x^{4} + 42 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \arctan \left (\sqrt {2} x + 1\right ) + 42 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \arctan \left (\sqrt {2} x - 1\right ) + 21 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - 21 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + 32}{96 \, {\left (x^{7} + x^{3}\right )}} \] Input:
integrate(1/x^4/(x^8+2*x^4+1),x, algorithm="fricas")
Output:
-1/96*(56*x^4 + 42*sqrt(2)*(x^7 + x^3)*arctan(sqrt(2)*x + 1) + 42*sqrt(2)* (x^7 + x^3)*arctan(sqrt(2)*x - 1) + 21*sqrt(2)*(x^7 + x^3)*log(x^2 + sqrt( 2)*x + 1) - 21*sqrt(2)*(x^7 + x^3)*log(x^2 - sqrt(2)*x + 1) + 32)/(x^7 + x ^3)
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 7 x^{4} - 4}{12 x^{7} + 12 x^{3}} + \frac {7 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \] Input:
integrate(1/x**4/(x**8+2*x**4+1),x)
Output:
(-7*x**4 - 4)/(12*x**7 + 12*x**3) + 7*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - 7*sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 - 7*sqrt(2)*atan(sqrt(2)*x - 1)/ 16 - 7*sqrt(2)*atan(sqrt(2)*x + 1)/16
Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {7 \, x^{4} + 4}{12 \, {\left (x^{7} + x^{3}\right )}} \] Input:
integrate(1/x^4/(x^8+2*x^4+1),x, algorithm="maxima")
Output:
-7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 7/16*sqrt(2)*arctan(1/ 2*sqrt(2)*(2*x - sqrt(2))) - 7/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 7/32* sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/12*(7*x^4 + 4)/(x^7 + x^3)
Time = 0.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{3 \, x^{3}} \] Input:
integrate(1/x^4/(x^8+2*x^4+1),x, algorithm="giac")
Output:
-7/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 7/16*sqrt(2)*arctan(1/ 2*sqrt(2)*(2*x - sqrt(2))) - 7/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 7/32* sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*x/(x^4 + 1) - 1/3/x^3
Time = 19.67 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {\frac {7\,x^4}{12}+\frac {1}{3}}{x^7+x^3}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}-\frac {7}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}+\frac {7}{16}{}\mathrm {i}\right ) \] Input:
int(1/(x^4*(2*x^4 + x^8 + 1)),x)
Output:
- 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(7/16 + 7i/16) - 2^(1/2)*atan(2^(1/ 2)*x*(1/2 + 1i/2))*(7/16 - 7i/16) - ((7*x^4)/12 + 1/3)/(x^3 + x^7)
Time = 0.21 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.99 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {42 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{7}+42 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}-2 x}{\sqrt {2}}\right ) x^{3}-42 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{7}-42 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {2}+2 x}{\sqrt {2}}\right ) x^{3}+21 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{7}+21 \sqrt {2}\, \mathrm {log}\left (-\sqrt {2}\, x +x^{2}+1\right ) x^{3}-21 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{7}-21 \sqrt {2}\, \mathrm {log}\left (\sqrt {2}\, x +x^{2}+1\right ) x^{3}-56 x^{4}-32}{96 x^{3} \left (x^{4}+1\right )} \] Input:
int(1/x^4/(x^8+2*x^4+1),x)
Output:
(42*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**7 + 42*sqrt(2)*atan((sqrt(2) - 2*x)/sqrt(2))*x**3 - 42*sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**7 - 42* sqrt(2)*atan((sqrt(2) + 2*x)/sqrt(2))*x**3 + 21*sqrt(2)*log( - sqrt(2)*x + x**2 + 1)*x**7 + 21*sqrt(2)*log( - sqrt(2)*x + x**2 + 1)*x**3 - 21*sqrt(2 )*log(sqrt(2)*x + x**2 + 1)*x**7 - 21*sqrt(2)*log(sqrt(2)*x + x**2 + 1)*x* *3 - 56*x**4 - 32)/(96*x**3*(x**4 + 1))