Integrand size = 20, antiderivative size = 193 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{1+x^4}}\right ) \]
-1/20*(-10+10*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1/ 4))-1/20*(10+10*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4+1)^(1 /4))-1/20*(-10+10*5^(1/2))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4+1 )^(1/4))-1/20*(10+10*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4 +1)^(1/4))
Time = 0.45 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=-\frac {\sqrt {-1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {1+\sqrt {5}} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {-1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )+\sqrt {1+\sqrt {5}} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {10}} \]
-1/2*(Sqrt[-1 + Sqrt[5]]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4) ] + Sqrt[1 + Sqrt[5]]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + Sqrt[-1 + Sqrt[5]]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)] + S qrt[1 + Sqrt[5]]*ArcTanh[(Sqrt[(1 + Sqrt[5])/2]*x)/(1 + x^4)^(1/4)])/Sqrt[ 10]
Time = 0.30 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1758, 902, 756, 216, 219}
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.
| \(\displaystyle \int \frac {1}{\sqrt [4]{x^4+1} \left (x^8+x^4-1\right )} \, dx\) |
| \(\Big \downarrow \) 1758 |
| \(\displaystyle \frac {2 \int \frac {1}{\sqrt [4]{x^4+1} \left (2 x^4-\sqrt {5}+1\right )}dx}{\sqrt {5}}-\frac {2 \int \frac {1}{\sqrt [4]{x^4+1} \left (2 x^4+\sqrt {5}+1\right )}dx}{\sqrt {5}}\) |
| \(\Big \downarrow \) 902 |
| \(\displaystyle \frac {2 \int \frac {1}{-\frac {\left (-1-\sqrt {5}\right ) x^4}{x^4+1}-\sqrt {5}+1}d\frac {x}{\sqrt [4]{x^4+1}}}{\sqrt {5}}-\frac {2 \int \frac {1}{-\frac {\left (-1+\sqrt {5}\right ) x^4}{x^4+1}+\sqrt {5}+1}d\frac {x}{\sqrt [4]{x^4+1}}}{\sqrt {5}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {2 \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}-\frac {\int \frac {1}{\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}+\sqrt {3-\sqrt {5}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}+\frac {\int \frac {1}{\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}+\sqrt {3+\sqrt {5}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}\right )}{\sqrt {5}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 \left (-\frac {\int \frac {1}{\sqrt {3-\sqrt {5}}-\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}-\frac {\arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {\int \frac {1}{\sqrt {3+\sqrt {5}}-\frac {\sqrt {2} x^2}{\sqrt {x^4+1}}}d\frac {x}{\sqrt [4]{x^4+1}}}{2 \sqrt {2}}+\frac {\arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}\right )}{\sqrt {5}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 \left (-\frac {\arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3-\sqrt {5}}}\right )}{\sqrt {5}}-\frac {2 \left (\frac {\arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4} \sqrt [4]{3+\sqrt {5}}}\right )}{\sqrt {5}}\) |
(-2*(ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)]/(2*2^(3/4)*(3 + S qrt[5])^(1/4)) + ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x)/(1 + x^4)^(1/4)]/(2*2 ^(3/4)*(3 + Sqrt[5])^(1/4))))/Sqrt[5] + (2*(-1/2*ArcTan[(((3 + Sqrt[5])/2) ^(1/4)*x)/(1 + x^4)^(1/4)]/(2^(3/4)*(3 - Sqrt[5])^(1/4)) - ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x)/(1 + x^4)^(1/4)]/(2*2^(3/4)*(3 - Sqrt[5])^(1/4))))/Sq rt[5]
3.25.3.3.1 Defintions of rubi rules used
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[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[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b , c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_ )), x_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/r) Int[(d + e*x ^n)^q/(b - r + 2*c*x^n), x], x] - Simp[2*(c/r) Int[(d + e*x^n)^q/(b + r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && Ne Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[q]
Time = 11.24 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(-\frac {\sqrt {5}\, \left (\left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )\right ) \left (\sqrt {5}+1\right ) \sqrt {-2+2 \sqrt {5}}+\sqrt {2+2 \sqrt {5}}\, \left (\sqrt {5}-1\right ) \left (\operatorname {arctanh}\left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )-\arctan \left (\frac {2 \left (x^{4}+1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )\right )\right )}{40}\) | \(131\) |
| trager | \(\text {Expression too large to display}\) | \(1671\) |
-1/40*5^(1/2)*((arctanh(2/(2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))-arctan(2/(2 +2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4)))*(5^(1/2)+1)*(-2+2*5^(1/2))^(1/2)+(2+2* 5^(1/2))^(1/2)*(5^(1/2)-1)*(arctanh(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4) )-arctan(2/(-2+2*5^(1/2))^(1/2)/x*(x^4+1)^(1/4))))
Leaf count of result is larger than twice the leaf count of optimal. 1189 vs. \(2 (125) = 250\).
Time = 15.09 (sec) , antiderivative size = 1189, normalized size of antiderivative = 6.16 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\text {Too large to display} \]
-1/80*sqrt(10)*sqrt(-sqrt(5) + 1)*log((sqrt(10)*sqrt(x^4 + 1)*(5*x^2 - sqr t(5)*(2*x^6 + x^2))*sqrt(-sqrt(5) + 1) + sqrt(10)*(5*x^8 + 5*x^4 - sqrt(5) *(2*x^4 + 1))*sqrt(-sqrt(5) + 1) + 10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3 /4) + 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/80*sqrt(10)*sqrt(-sqrt(5) + 1)*log(-(sqrt(10)*sqrt(x^4 + 1)*(5*x^ 2 - sqrt(5)*(2*x^6 + x^2))*sqrt(-sqrt(5) + 1) + sqrt(10)*(5*x^8 + 5*x^4 - sqrt(5)*(2*x^4 + 1))*sqrt(-sqrt(5) + 1) - 10*(2*x^5 - sqrt(5)*x + x)*(x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 - sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x^8 + x^4 - 1)) + 1/80*sqrt(10)*sqrt(-sqrt(5) - 1)*log((sqrt(10)*sqrt(x^4 + 1) *(5*x^2 + sqrt(5)*(2*x^6 + x^2))*sqrt(-sqrt(5) - 1) - sqrt(10)*(5*x^8 + 5* x^4 + sqrt(5)*(2*x^4 + 1))*sqrt(-sqrt(5) - 1) + 10*(2*x^5 + sqrt(5)*x + x) *(x^4 + 1)^(3/4) - 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4)) /(x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(-sqrt(5) - 1)*log(-(sqrt(10)*sqrt(x ^4 + 1)*(5*x^2 + sqrt(5)*(2*x^6 + x^2))*sqrt(-sqrt(5) - 1) - sqrt(10)*(5*x ^8 + 5*x^4 + sqrt(5)*(2*x^4 + 1))*sqrt(-sqrt(5) - 1) - 10*(2*x^5 + sqrt(5) *x + x)*(x^4 + 1)^(3/4) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1) ^(1/4))/(x^8 + x^4 - 1)) - 1/80*sqrt(10)*sqrt(sqrt(5) + 1)*log((10*(2*x^5 + sqrt(5)*x + x)*(x^4 + 1)^(3/4) + (sqrt(10)*sqrt(x^4 + 1)*(5*x^2 + sqrt(5 )*(2*x^6 + x^2)) + sqrt(10)*(5*x^8 + 5*x^4 + sqrt(5)*(2*x^4 + 1)))*sqrt(sq rt(5) + 1) + 10*(x^7 + 3*x^3 + sqrt(5)*(x^7 + x^3))*(x^4 + 1)^(1/4))/(x...
\[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {1}{\sqrt [4]{x^{4} + 1} \left (x^{8} + x^{4} - 1\right )}\, dx \]
\[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} + x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (-1+x^4+x^8\right )} \, dx=\int \frac {1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+x^4-1\right )} \,d x \]