Integrand size = 16, antiderivative size = 79 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=-\frac {1}{4} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \arctan \left (\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x^4\right )-\frac {1}{4} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x^4\right ) \] Output:
-1/40*(10+10*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x^4)-1/40*(-10+ 10*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x^4)
Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=-\frac {\arctan \left (\sqrt {\frac {2}{-1+\sqrt {5}}} x^4\right )}{2 \sqrt {10 \left (-1+\sqrt {5}\right )}}-\frac {\text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x^4\right )}{2 \sqrt {10 \left (1+\sqrt {5}\right )}} \] Input:
Integrate[x^3/(-1 - x^8 + x^16),x]
Output:
-1/2*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x^4]/Sqrt[10*(-1 + Sqrt[5])] - ArcTanh[ Sqrt[2/(1 + Sqrt[5])]*x^4]/(2*Sqrt[10*(1 + Sqrt[5])])
Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1695, 1406, 216, 220}
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.
\(\displaystyle \int \frac {x^3}{x^{16}-x^8-1} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{4} \int \frac {1}{x^{16}-x^8-1}dx^4\) |
\(\Big \downarrow \) 1406 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {1}{x^8+\frac {1}{2} \left (-1-\sqrt {5}\right )}dx^4}{\sqrt {5}}-\frac {\int \frac {1}{x^8+\frac {1}{2} \left (-1+\sqrt {5}\right )}dx^4}{\sqrt {5}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {1}{x^8+\frac {1}{2} \left (-1-\sqrt {5}\right )}dx^4}{\sqrt {5}}-\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \arctan \left (\sqrt {\frac {2}{\sqrt {5}-1}} x^4\right )\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{4} \left (-\sqrt {\frac {2}{5 \left (\sqrt {5}-1\right )}} \arctan \left (\sqrt {\frac {2}{\sqrt {5}-1}} x^4\right )-\sqrt {\frac {2}{5 \left (1+\sqrt {5}\right )}} \text {arctanh}\left (\sqrt {\frac {2}{1+\sqrt {5}}} x^4\right )\right )\) |
Input:
Int[x^3/(-1 - x^8 + x^16),x]
Output:
(-(Sqrt[2/(5*(-1 + Sqrt[5]))]*ArcTan[Sqrt[2/(-1 + Sqrt[5])]*x^4]) - Sqrt[2 /(5*(1 + Sqrt[5]))]*ArcTanh[Sqrt[2/(1 + Sqrt[5])]*x^4])/4
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 2 - 4*a*c, 2]}, Simp[c/q Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q I nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c , 0] && PosQ[b^2 - 4*a*c]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (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) + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && IntegerQ[m]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.43
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (25 \textit {\_Z}^{4}+5 \textit {\_Z}^{2}-1\right )}{\sum }\textit {\_R} \ln \left (x^{4}-5 \textit {\_R}^{3}-3 \textit {\_R} \right )\right )}{8}\) | \(34\) |
default | \(-\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {2 x^{4}}{\sqrt {2+2 \sqrt {5}}}\right )}{10 \sqrt {2+2 \sqrt {5}}}-\frac {\sqrt {5}\, \arctan \left (\frac {2 x^{4}}{\sqrt {-2+2 \sqrt {5}}}\right )}{10 \sqrt {-2+2 \sqrt {5}}}\) | \(60\) |
Input:
int(x^3/(x^16-x^8-1),x,method=_RETURNVERBOSE)
Output:
1/8*sum(_R*ln(x^4-5*_R^3-3*_R),_R=RootOf(25*_Z^4+5*_Z^2-1))
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.22 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=-\frac {1}{4} \, \sqrt {\frac {1}{10} \, \sqrt {5} + \frac {1}{10}} \arctan \left (\sqrt {5} x^{4} \sqrt {\frac {1}{10} \, \sqrt {5} + \frac {1}{10}}\right ) - \frac {1}{8} \, \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} \log \left (2 \, x^{4} + {\left (\sqrt {5} + 5\right )} \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}}\right ) + \frac {1}{8} \, \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}} \log \left (2 \, x^{4} - {\left (\sqrt {5} + 5\right )} \sqrt {\frac {1}{10} \, \sqrt {5} - \frac {1}{10}}\right ) \] Input:
integrate(x^3/(x^16-x^8-1),x, algorithm="fricas")
Output:
-1/4*sqrt(1/10*sqrt(5) + 1/10)*arctan(sqrt(5)*x^4*sqrt(1/10*sqrt(5) + 1/10 )) - 1/8*sqrt(1/10*sqrt(5) - 1/10)*log(2*x^4 + (sqrt(5) + 5)*sqrt(1/10*sqr t(5) - 1/10)) + 1/8*sqrt(1/10*sqrt(5) - 1/10)*log(2*x^4 - (sqrt(5) + 5)*sq rt(1/10*sqrt(5) - 1/10))
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.33 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=\operatorname {RootSum} {\left (102400 t^{4} + 320 t^{2} - 1, \left ( t \mapsto t \log {\left (- 2560 t^{3} - 24 t + x^{4} \right )} \right )\right )} \] Input:
integrate(x**3/(x**16-x**8-1),x)
Output:
RootSum(102400*_t**4 + 320*_t**2 - 1, Lambda(_t, _t*log(-2560*_t**3 - 24*_ t + x**4)))
\[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=\int { \frac {x^{3}}{x^{16} - x^{8} - 1} \,d x } \] Input:
integrate(x^3/(x^16-x^8-1),x, algorithm="maxima")
Output:
integrate(x^3/(x^16 - x^8 - 1), x)
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=-\frac {1}{40} \, \sqrt {10 \, \sqrt {5} + 10} \arctan \left (\frac {x^{4}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) - \frac {1}{80} \, \sqrt {10 \, \sqrt {5} - 10} \log \left (x^{4} + \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}\right ) + \frac {1}{80} \, \sqrt {10 \, \sqrt {5} - 10} \log \left ({\left | x^{4} - \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} \right |}\right ) \] Input:
integrate(x^3/(x^16-x^8-1),x, algorithm="giac")
Output:
-1/40*sqrt(10*sqrt(5) + 10)*arctan(x^4/sqrt(1/2*sqrt(5) - 1/2)) - 1/80*sqr t(10*sqrt(5) - 10)*log(x^4 + sqrt(1/2*sqrt(5) + 1/2)) + 1/80*sqrt(10*sqrt( 5) - 10)*log(abs(x^4 - sqrt(1/2*sqrt(5) + 1/2)))
Time = 9.71 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.73 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx=\frac {\sqrt {10}\,\mathrm {atanh}\left (\frac {13\,\sqrt {10}\,x^4\,\sqrt {-\sqrt {5}-1}}{2\,\left (13\,\sqrt {5}+29\right )}+\frac {29\,\sqrt {5}\,\sqrt {10}\,x^4\,\sqrt {-\sqrt {5}-1}}{10\,\left (13\,\sqrt {5}+29\right )}\right )\,\sqrt {-\sqrt {5}-1}}{40}-\frac {\sqrt {10}\,\mathrm {atanh}\left (\frac {13\,\sqrt {10}\,x^4\,\sqrt {\sqrt {5}-1}}{2\,\left (13\,\sqrt {5}-29\right )}-\frac {29\,\sqrt {5}\,\sqrt {10}\,x^4\,\sqrt {\sqrt {5}-1}}{10\,\left (13\,\sqrt {5}-29\right )}\right )\,\sqrt {\sqrt {5}-1}}{40} \] Input:
int(-x^3/(x^8 - x^16 + 1),x)
Output:
(10^(1/2)*atanh((13*10^(1/2)*x^4*(- 5^(1/2) - 1)^(1/2))/(2*(13*5^(1/2) + 2 9)) + (29*5^(1/2)*10^(1/2)*x^4*(- 5^(1/2) - 1)^(1/2))/(10*(13*5^(1/2) + 29 )))*(- 5^(1/2) - 1)^(1/2))/40 - (10^(1/2)*atanh((13*10^(1/2)*x^4*(5^(1/2) - 1)^(1/2))/(2*(13*5^(1/2) - 29)) - (29*5^(1/2)*10^(1/2)*x^4*(5^(1/2) - 1) ^(1/2))/(10*(13*5^(1/2) - 29)))*(5^(1/2) - 1)^(1/2))/40
Time = 0.27 (sec) , antiderivative size = 795, normalized size of antiderivative = 10.06 \[ \int \frac {x^3}{-1-x^8+x^{16}} \, dx =\text {Too large to display} \] Input:
int(x^3/(x^16-x^8-1),x)
Output:
( - 2*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*sqrt(5)*ata n(((sqrt(5) - 1)**(1/8)*sqrt( - sqrt(2) + 2)*2**(1/8) - 2*2**(1/4)*x)/(sqr t(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1/8))) - 10*sqrt(sqrt(2) + 2)*sqrt (sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*atan(((sqrt(5) - 1)**(1/8)*sqrt( - sqrt (2) + 2)*2**(1/8) - 2*2**(1/4)*x)/(sqrt(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)* 2**(1/8))) - 2*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*sq rt(5)*atan(((sqrt(5) - 1)**(1/8)*sqrt( - sqrt(2) + 2)*2**(1/8) + 2*2**(1/4 )*x)/(sqrt(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1/8))) - 10*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*atan(((sqrt(5) - 1)**(1/8)*sqr t( - sqrt(2) + 2)*2**(1/8) + 2*2**(1/4)*x)/(sqrt(sqrt(2) + 2)*(sqrt(5) - 1 )**(1/8)*2**(1/8))) + 2*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2 ) + 2)*sqrt(5)*atan((sqrt(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1/8) - 2*2 **(1/4)*x)/((sqrt(5) - 1)**(1/8)*sqrt( - sqrt(2) + 2)*2**(1/8))) + 10*sqrt (sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1/8) - 2*2**(1/4)*x)/((sqrt(5) - 1)**(1/8)*sqr t( - sqrt(2) + 2)*2**(1/8))) + 2*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*sqrt(5)*atan((sqrt(sqrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1 /8) + 2*2**(1/4)*x)/((sqrt(5) - 1)**(1/8)*sqrt( - sqrt(2) + 2)*2**(1/8))) + 10*sqrt(sqrt(2) + 2)*sqrt(sqrt(5) - 1)*sqrt( - sqrt(2) + 2)*atan((sqrt(s qrt(2) + 2)*(sqrt(5) - 1)**(1/8)*2**(1/8) + 2*2**(1/4)*x)/((sqrt(5) - 1...