Integrand size = 19, antiderivative size = 53 \[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \] Output:
1/4*arctan(x*2^(1/2)/(x^4+1)^(1/2))*2^(1/2)+1/4*arctanh(x*2^(1/2)/(x^4+1)^ (1/2))*2^(1/2)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \] Input:
Integrate[Sqrt[1 + x^4]/(1 - x^4),x]
Output:
(ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]] + ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]])/( 2*Sqrt[2])
Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {920, 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.
\(\displaystyle \int \frac {\sqrt {x^4+1}}{1-x^4} \, dx\) |
\(\Big \downarrow \) 920 |
\(\displaystyle \int \frac {1}{1-\frac {4 x^4}{\left (x^4+1\right )^2}}d\frac {x}{\sqrt {x^4+1}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{2} \int \frac {1}{1-\frac {2 x^2}{x^4+1}}d\frac {x}{\sqrt {x^4+1}}+\frac {1}{2} \int \frac {1}{\frac {2 x^2}{x^4+1}+1}d\frac {x}{\sqrt {x^4+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \int \frac {1}{1-\frac {2 x^2}{x^4+1}}d\frac {x}{\sqrt {x^4+1}}+\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}\) |
Input:
Int[Sqrt[1 + x^4]/(1 - x^4),x]
Output:
ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(2*Sqrt[2]) + ArcTanh[(Sqrt[2]*x)/Sqrt[1 + x^4]]/(2*Sqrt[2])
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[Sqrt[(a_) + (b_.)*(x_)^4]/((c_) + (d_.)*(x_)^4), x_Symbol] :> Simp[a/c Subst[Int[1/(1 - 4*a*b*x^4), x], x, x/Sqrt[a + b*x^4]], x] /; FreeQ[{a, b , c, d}, x] && EqQ[b*c + a*d, 0] && PosQ[a*b]
Time = 1.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {\left (2 \arctan \left (\frac {x \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right ) \sqrt {2}}{8}\) | \(62\) |
pseudoelliptic | \(\frac {\left (2 \arctan \left (\frac {x \sqrt {2}}{\sqrt {x^{4}+1}}\right )+\operatorname {arctanh}\left (\frac {\left (x^{2}-x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )-\operatorname {arctanh}\left (\frac {\left (x^{2}+x +1\right ) \sqrt {2}}{\sqrt {x^{4}+1}}\right )\right ) \sqrt {2}}{8}\) | \(62\) |
elliptic | \(\frac {\left (-\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}+\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{2}\right ) \sqrt {2}}{2}\) | \(65\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{4}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (-1+x \right ) \left (1+x \right )}\right )}{4}\) | \(73\) |
Input:
int((x^4+1)^(1/2)/(-x^4+1),x,method=_RETURNVERBOSE)
Output:
1/8*(2*arctan(x*2^(1/2)/(x^4+1)^(1/2))+arctanh((x^2-x+1)*2^(1/2)/(x^4+1)^( 1/2))-arctanh((x^2+x+1)*2^(1/2)/(x^4+1)^(1/2)))*2^(1/2)
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\frac {x^{4} + 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) \] Input:
integrate((x^4+1)^(1/2)/(-x^4+1),x, algorithm="fricas")
Output:
1/4*sqrt(2)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) + 1/8*sqrt(2)*log((x^4 + 2*sqr t(2)*sqrt(x^4 + 1)*x + 2*x^2 + 1)/(x^4 - 2*x^2 + 1))
\[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=- \int \frac {\sqrt {x^{4} + 1}}{x^{4} - 1}\, dx \] Input:
integrate((x**4+1)**(1/2)/(-x**4+1),x)
Output:
-Integral(sqrt(x**4 + 1)/(x**4 - 1), x)
\[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=\int { -\frac {\sqrt {x^{4} + 1}}{x^{4} - 1} \,d x } \] Input:
integrate((x^4+1)^(1/2)/(-x^4+1),x, algorithm="maxima")
Output:
-integrate(sqrt(x^4 + 1)/(x^4 - 1), x)
\[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=\int { -\frac {\sqrt {x^{4} + 1}}{x^{4} - 1} \,d x } \] Input:
integrate((x^4+1)^(1/2)/(-x^4+1),x, algorithm="giac")
Output:
integrate(-sqrt(x^4 + 1)/(x^4 - 1), x)
Timed out. \[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=-\int \frac {\sqrt {x^4+1}}{x^4-1} \,d x \] Input:
int(-(x^4 + 1)^(1/2)/(x^4 - 1),x)
Output:
-int((x^4 + 1)^(1/2)/(x^4 - 1), x)
\[ \int \frac {\sqrt {1+x^4}}{1-x^4} \, dx=-\left (\int \frac {\sqrt {x^{4}+1}}{x^{4}-1}d x \right ) \] Input:
int((x^4+1)^(1/2)/(-x^4+1),x)
Output:
- int(sqrt(x**4 + 1)/(x**4 - 1),x)