Integrand size = 18, antiderivative size = 411 \[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=-\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2-b}}-2 x}{\sqrt {2+\sqrt {2-b}}}\right )}{4 \sqrt {2+\sqrt {2-b}}}-\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2-b}}-2 x}{\sqrt {2-\sqrt {2-b}}}\right )}{4 \sqrt {2-\sqrt {2-b}}}+\frac {\arctan \left (\frac {\sqrt {2-\sqrt {2-b}}+2 x}{\sqrt {2+\sqrt {2-b}}}\right )}{4 \sqrt {2+\sqrt {2-b}}}+\frac {\arctan \left (\frac {\sqrt {2+\sqrt {2-b}}+2 x}{\sqrt {2-\sqrt {2-b}}}\right )}{4 \sqrt {2-\sqrt {2-b}}}-\frac {\log \left (1-\sqrt {2-\sqrt {2-b}} x+x^2\right )}{8 \sqrt {2-\sqrt {2-b}}}+\frac {\log \left (1+\sqrt {2-\sqrt {2-b}} x+x^2\right )}{8 \sqrt {2-\sqrt {2-b}}}-\frac {\log \left (1-\sqrt {2+\sqrt {2-b}} x+x^2\right )}{8 \sqrt {2+\sqrt {2-b}}}+\frac {\log \left (1+\sqrt {2+\sqrt {2-b}} x+x^2\right )}{8 \sqrt {2+\sqrt {2-b}}} \]
-1/4*arctan((-2*x+(2+(2-b)^(1/2))^(1/2))/(2-(2-b)^(1/2))^(1/2))/(2-(2-b)^( 1/2))^(1/2)+1/4*arctan((2*x+(2+(2-b)^(1/2))^(1/2))/(2-(2-b)^(1/2))^(1/2))/ (2-(2-b)^(1/2))^(1/2)-1/8*ln(1+x^2-x*(2-(2-b)^(1/2))^(1/2))/(2-(2-b)^(1/2) )^(1/2)+1/8*ln(1+x^2+x*(2-(2-b)^(1/2))^(1/2))/(2-(2-b)^(1/2))^(1/2)-1/4*ar ctan((-2*x+(2-(2-b)^(1/2))^(1/2))/(2+(2-b)^(1/2))^(1/2))/(2+(2-b)^(1/2))^( 1/2)+1/4*arctan((2*x+(2-(2-b)^(1/2))^(1/2))/(2+(2-b)^(1/2))^(1/2))/(2+(2-b )^(1/2))^(1/2)-1/8*ln(1+x^2-x*(2+(2-b)^(1/2))^(1/2))/(2+(2-b)^(1/2))^(1/2) +1/8*ln(1+x^2+x*(2+(2-b)^(1/2))^(1/2))/(2+(2-b)^(1/2))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.13 \[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\frac {1}{4} \text {RootSum}\left [1+b \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ] \]
Time = 0.67 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.21, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1749, 1407, 1142, 25, 1083, 217, 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.
| \(\displaystyle \int \frac {x^4+1}{b x^4+x^8+1} \, dx\) |
| \(\Big \downarrow \) 1749 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4-\sqrt {2-b} x^2+1}dx+\frac {1}{2} \int \frac {1}{x^4+\sqrt {2-b} x^2+1}dx\) |
| \(\Big \downarrow \) 1407 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2-\sqrt {2-b}}-x}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\int \frac {x+\sqrt {2-\sqrt {2-b}}}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\int \frac {\sqrt {\sqrt {2-b}+2}-x}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\int \frac {x+\sqrt {\sqrt {2-b}+2}}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {2-\sqrt {2-b}} \int \frac {1}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx-\frac {1}{2} \int -\frac {\sqrt {2-\sqrt {2-b}}-2 x}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\frac {1}{2} \sqrt {2-\sqrt {2-b}} \int \frac {1}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {2-\sqrt {2-b}}}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {\sqrt {2-b}+2} \int \frac {1}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx-\frac {1}{2} \int -\frac {\sqrt {\sqrt {2-b}+2}-2 x}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\frac {1}{2} \sqrt {\sqrt {2-b}+2} \int \frac {1}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {\sqrt {2-b}+2}}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {2-\sqrt {2-b}} \int \frac {1}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx+\frac {1}{2} \int \frac {\sqrt {2-\sqrt {2-b}}-2 x}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\frac {1}{2} \sqrt {2-\sqrt {2-b}} \int \frac {1}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {2-\sqrt {2-b}}}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\frac {1}{2} \sqrt {\sqrt {2-b}+2} \int \frac {1}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx+\frac {1}{2} \int \frac {\sqrt {\sqrt {2-b}+2}-2 x}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\frac {1}{2} \sqrt {\sqrt {2-b}+2} \int \frac {1}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx+\frac {1}{2} \int \frac {2 x+\sqrt {\sqrt {2-b}+2}}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {2-\sqrt {2-b}}-2 x}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx-\sqrt {2-\sqrt {2-b}} \int \frac {1}{-\left (2 x-\sqrt {2-\sqrt {2-b}}\right )^2-\sqrt {2-b}-2}d\left (2 x-\sqrt {2-\sqrt {2-b}}\right )}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\frac {1}{2} \int \frac {2 x+\sqrt {2-\sqrt {2-b}}}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx-\sqrt {2-\sqrt {2-b}} \int \frac {1}{-\left (2 x+\sqrt {2-\sqrt {2-b}}\right )^2-\sqrt {2-b}-2}d\left (2 x+\sqrt {2-\sqrt {2-b}}\right )}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\sqrt {2-b}+2}-2 x}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx-\sqrt {\sqrt {2-b}+2} \int \frac {1}{-\left (2 x-\sqrt {\sqrt {2-b}+2}\right )^2+\sqrt {2-b}-2}d\left (2 x-\sqrt {\sqrt {2-b}+2}\right )}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\frac {1}{2} \int \frac {2 x+\sqrt {\sqrt {2-b}+2}}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx-\sqrt {\sqrt {2-b}+2} \int \frac {1}{-\left (2 x+\sqrt {\sqrt {2-b}+2}\right )^2+\sqrt {2-b}-2}d\left (2 x+\sqrt {\sqrt {2-b}+2}\right )}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {2-\sqrt {2-b}}-2 x}{x^2-\sqrt {2-\sqrt {2-b}} x+1}dx+\frac {\sqrt {2-\sqrt {2-b}} \arctan \left (\frac {2 x-\sqrt {2-\sqrt {2-b}}}{\sqrt {\sqrt {2-b}+2}}\right )}{\sqrt {\sqrt {2-b}+2}}}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\frac {1}{2} \int \frac {2 x+\sqrt {2-\sqrt {2-b}}}{x^2+\sqrt {2-\sqrt {2-b}} x+1}dx+\frac {\sqrt {2-\sqrt {2-b}} \arctan \left (\frac {\sqrt {2-\sqrt {2-b}}+2 x}{\sqrt {\sqrt {2-b}+2}}\right )}{\sqrt {\sqrt {2-b}+2}}}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\frac {1}{2} \int \frac {\sqrt {\sqrt {2-b}+2}-2 x}{x^2-\sqrt {\sqrt {2-b}+2} x+1}dx+\frac {\sqrt {\sqrt {2-b}+2} \arctan \left (\frac {2 x-\sqrt {\sqrt {2-b}+2}}{\sqrt {2-\sqrt {2-b}}}\right )}{\sqrt {2-\sqrt {2-b}}}}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\frac {1}{2} \int \frac {2 x+\sqrt {\sqrt {2-b}+2}}{x^2+\sqrt {\sqrt {2-b}+2} x+1}dx+\frac {\sqrt {\sqrt {2-b}+2} \arctan \left (\frac {\sqrt {\sqrt {2-b}+2}+2 x}{\sqrt {2-\sqrt {2-b}}}\right )}{\sqrt {2-\sqrt {2-b}}}}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {\sqrt {2-\sqrt {2-b}} \arctan \left (\frac {2 x-\sqrt {2-\sqrt {2-b}}}{\sqrt {\sqrt {2-b}+2}}\right )}{\sqrt {\sqrt {2-b}+2}}-\frac {1}{2} \log \left (-\sqrt {2-\sqrt {2-b}} x+x^2+1\right )}{2 \sqrt {2-\sqrt {2-b}}}+\frac {\frac {\sqrt {2-\sqrt {2-b}} \arctan \left (\frac {\sqrt {2-\sqrt {2-b}}+2 x}{\sqrt {\sqrt {2-b}+2}}\right )}{\sqrt {\sqrt {2-b}+2}}+\frac {1}{2} \log \left (\sqrt {2-\sqrt {2-b}} x+x^2+1\right )}{2 \sqrt {2-\sqrt {2-b}}}\right )+\frac {1}{2} \left (\frac {\frac {\sqrt {\sqrt {2-b}+2} \arctan \left (\frac {2 x-\sqrt {\sqrt {2-b}+2}}{\sqrt {2-\sqrt {2-b}}}\right )}{\sqrt {2-\sqrt {2-b}}}-\frac {1}{2} \log \left (-\sqrt {\sqrt {2-b}+2} x+x^2+1\right )}{2 \sqrt {\sqrt {2-b}+2}}+\frac {\frac {\sqrt {\sqrt {2-b}+2} \arctan \left (\frac {\sqrt {\sqrt {2-b}+2}+2 x}{\sqrt {2-\sqrt {2-b}}}\right )}{\sqrt {2-\sqrt {2-b}}}+\frac {1}{2} \log \left (\sqrt {\sqrt {2-b}+2} x+x^2+1\right )}{2 \sqrt {\sqrt {2-b}+2}}\right )\) |
(((Sqrt[2 - Sqrt[2 - b]]*ArcTan[(-Sqrt[2 - Sqrt[2 - b]] + 2*x)/Sqrt[2 + Sq rt[2 - b]]])/Sqrt[2 + Sqrt[2 - b]] - Log[1 - Sqrt[2 - Sqrt[2 - b]]*x + x^2 ]/2)/(2*Sqrt[2 - Sqrt[2 - b]]) + ((Sqrt[2 - Sqrt[2 - b]]*ArcTan[(Sqrt[2 - Sqrt[2 - b]] + 2*x)/Sqrt[2 + Sqrt[2 - b]]])/Sqrt[2 + Sqrt[2 - b]] + Log[1 + Sqrt[2 - Sqrt[2 - b]]*x + x^2]/2)/(2*Sqrt[2 - Sqrt[2 - b]]))/2 + (((Sqrt [2 + Sqrt[2 - b]]*ArcTan[(-Sqrt[2 + Sqrt[2 - b]] + 2*x)/Sqrt[2 - Sqrt[2 - b]]])/Sqrt[2 - Sqrt[2 - b]] - Log[1 - Sqrt[2 + Sqrt[2 - b]]*x + x^2]/2)/(2 *Sqrt[2 + Sqrt[2 - b]]) + ((Sqrt[2 + Sqrt[2 - b]]*ArcTan[(Sqrt[2 + Sqrt[2 - b]] + 2*x)/Sqrt[2 - Sqrt[2 - b]]])/Sqrt[2 - Sqrt[2 - b]] + Log[1 + Sqrt[ 2 + Sqrt[2 - b]]*x + x^2]/2)/(2*Sqrt[2 + Sqrt[2 - b]]))/2
3.1.9.3.1 Defintions of rubi rules used
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_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/ c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) Int[(r - x)/(q - r* x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(r + x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x^(n/2) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && IGtQ[n/2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.10
| method | result | size |
| default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+\textit {\_Z}^{4} b +1\right )}{\sum }\frac {\left (\textit {\_R}^{4}+1\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7}+\textit {\_R}^{3} b}\right )}{4}\) | \(42\) |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+\textit {\_Z}^{4} b +1\right )}{\sum }\frac {\left (\textit {\_R}^{4}+1\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7}+\textit {\_R}^{3} b}\right )}{4}\) | \(42\) |
Leaf count of result is larger than twice the leaf count of optimal. 1177 vs. \(2 (321) = 642\).
Time = 0.28 (sec) , antiderivative size = 1177, normalized size of antiderivative = 2.86 \[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\text {Too large to display} \]
-1/4*sqrt(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12* b + 8)) + b)/(b^2 + 4*b + 4)))*log(1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b - 2)*sqrt(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt(( b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4))) + x) + 1/4*sqrt(sq rt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b) /(b^2 + 4*b + 4)))*log(-1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 1 2*b + 8)) - b - 2)*sqrt(sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4))) + x) - 1/4*sqrt(-sqrt(1/2)*sq rt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4* b + 4)))*log(1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b - 2)*sqrt(-sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4))) + x) + 1/4*sqrt(-sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b)/(b^2 + 4*b + 4)))*l og(-1/2*((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b - 2)*s qrt(-sqrt(1/2)*sqrt(-((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8 )) + b)/(b^2 + 4*b + 4))) + x) + 1/4*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)* sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4)))*log(1/2*((b^ 2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) + b + 2)*sqrt(sqrt(1/2 )*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2)/(b^3 + 6*b^2 + 12*b + 8)) - b)/(b^2 + 4*b + 4))) + x) - 1/4*sqrt(sqrt(1/2)*sqrt(((b^2 + 4*b + 4)*sqrt((b - 2...
Time = 1.81 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.18 \[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\operatorname {RootSum} {\left (t^{8} \cdot \left (65536 b^{4} + 524288 b^{3} + 1572864 b^{2} + 2097152 b + 1048576\right ) + t^{4} \cdot \left (256 b^{3} + 1024 b^{2} + 1024 b\right ) + 1, \left ( t \mapsto t \log {\left (1024 t^{5} b^{2} + 4096 t^{5} b + 4096 t^{5} + 4 t b + 4 t + x \right )} \right )\right )} \]
RootSum(_t**8*(65536*b**4 + 524288*b**3 + 1572864*b**2 + 2097152*b + 10485 76) + _t**4*(256*b**3 + 1024*b**2 + 1024*b) + 1, Lambda(_t, _t*log(1024*_t **5*b**2 + 4096*_t**5*b + 4096*_t**5 + 4*_t*b + 4*_t + x)))
\[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\int { \frac {x^{4} + 1}{x^{8} + b x^{4} + 1} \,d x } \]
\[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\int { \frac {x^{4} + 1}{x^{8} + b x^{4} + 1} \,d x } \]
Time = 9.56 (sec) , antiderivative size = 5341, normalized size of antiderivative = 13.00 \[ \int \frac {1+x^4}{1+b x^4+x^8} \, dx=\text {Too large to display} \]
- atan((((-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(512*(32*b + 24 *b^2 + 8*b^3 + b^4 + 16)))^(1/4)*(((-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4* b^2 + b^3)/(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(1/4)*(262144*b + 196 608*b^2 - 196608*b^3 - 49152*b^4 + 49152*b^5 + 4096*b^6 - 4096*b^7 - 26214 4) + x*(32768*b + 65536*b^2 - 32768*b^3 - 20480*b^4 + 10240*b^5 + 2048*b^6 - 1024*b^7 - 65536))*(-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(5 12*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(3/4) - 256*b + 64*b^3 - 16*b^4 + 256) + x*(32*b - 48*b^2 + 24*b^3 - 4*b^4))*(-(4*b + ((b - 2)*(b + 2)^5)^(1 /2) + 4*b^2 + b^3)/(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(1/4)*1i - (( -(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(512*(32*b + 24*b^2 + 8*b ^3 + b^4 + 16)))^(1/4)*(((-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3) /(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(1/4)*(262144*b + 196608*b^2 - 196608*b^3 - 49152*b^4 + 49152*b^5 + 4096*b^6 - 4096*b^7 - 262144) - x*(32 768*b + 65536*b^2 - 32768*b^3 - 20480*b^4 + 10240*b^5 + 2048*b^6 - 1024*b^ 7 - 65536))*(-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(3/4) - 256*b + 64*b^3 - 16*b^4 + 256) - x*( 32*b - 48*b^2 + 24*b^3 - 4*b^4))*(-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^ 2 + b^3)/(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(1/4)*1i)/(((-(4*b + (( b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(512*(32*b + 24*b^2 + 8*b^3 + b^4 + 16)))^(1/4)*(((-(4*b + ((b - 2)*(b + 2)^5)^(1/2) + 4*b^2 + b^3)/(512*(...