Integrand size = 19, antiderivative size = 84 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}} \] Output:
1/2*arctan(2^(1/2)*x*(2*x^2+1)^(1/4)/(1+(2*x^2+1)^(1/2)))*2^(1/2)-1/2*arct anh(1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/x/(2*x^2+1)^(1/4))*2^(1/2)
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {-\arctan \left (\frac {-x^2+\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{x^2+\sqrt {1+2 x^2}}\right )}{2 \sqrt {2}} \] Input:
Integrate[1/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
Output:
(-ArcTan[(-x^2 + Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))] + ArcTanh [(Sqrt[2]*x*(1 + 2*x^2)^(1/4))/(x^2 + Sqrt[1 + 2*x^2])])/(2*Sqrt[2])
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {308}
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.
\(\displaystyle \int \frac {1}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\) |
\(\Big \downarrow \) 308 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}\) |
Input:
Int[1/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
Output:
-(ArcTan[(1 + Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))]/Sqrt[2]) - A rcTanh[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))]/Sqrt[2]
Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit h[{q = Rt[b^2/a, 4]}, Simp[(-b/(2*a*d*q))*ArcTan[(b + q^2*Sqrt[a + b*x^2])/ (q^3*x*(a + b*x^2)^(1/4))], x] - Simp[(b/(2*a*d*q))*ArcTanh[(b - q^2*Sqrt[a + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))], x]] /; FreeQ[{a, b, c, d}, x] && EqQ [b*c - 2*a*d, 0] && PosQ[b^2/a]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.16 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.85
method | result | size |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}-\sqrt {2 x^{2}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x}{x^{2}+1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}-\sqrt {2 x^{2}+1}\, x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}\) | \(155\) |
Input:
int(1/(x^2+1)/(2*x^2+1)^(1/4),x,method=_RETURNVERBOSE)
Output:
-1/2*RootOf(_Z^4+1)*ln(-(RootOf(_Z^4+1)*(2*x^2+1)^(3/4)+RootOf(_Z^4+1)^3*( 2*x^2+1)^(1/4)-(2*x^2+1)^(1/2)*x-RootOf(_Z^4+1)^2*x)/(x^2+1))-1/2*RootOf(_ Z^4+1)^3*ln(-(RootOf(_Z^4+1)^3*(2*x^2+1)^(3/4)-(2*x^2+1)^(1/2)*x+RootOf(_Z ^4+1)^2*x+RootOf(_Z^4+1)*(2*x^2+1)^(1/4))/(x^2+1))
Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (68) = 136\).
Time = 1.55 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.73 \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {x^{4} + \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} x^{3} + \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} x + x^{2} + \sqrt {2 \, x^{2} + 1} {\left (x^{2} + 1\right )}}{x^{4} - 2 \, x^{2} - 1}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {x^{4} - \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} x^{3} - \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} x + x^{2} + \sqrt {2 \, x^{2} + 1} {\left (x^{2} + 1\right )}}{x^{4} - 2 \, x^{2} - 1}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\frac {x^{2} + \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} x + \sqrt {2 \, x^{2} + 1}}{x^{2} + 1}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\frac {x^{2} - \sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} x + \sqrt {2 \, x^{2} + 1}}{x^{2} + 1}\right ) \] Input:
integrate(1/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
Output:
-1/4*sqrt(2)*arctan((x^4 + sqrt(2)*(2*x^2 + 1)^(1/4)*x^3 + sqrt(2)*(2*x^2 + 1)^(3/4)*x + x^2 + sqrt(2*x^2 + 1)*(x^2 + 1))/(x^4 - 2*x^2 - 1)) - 1/4*s qrt(2)*arctan(-(x^4 - sqrt(2)*(2*x^2 + 1)^(1/4)*x^3 - sqrt(2)*(2*x^2 + 1)^ (3/4)*x + x^2 + sqrt(2*x^2 + 1)*(x^2 + 1))/(x^4 - 2*x^2 - 1)) + 1/8*sqrt(2 )*log((x^2 + sqrt(2)*(2*x^2 + 1)^(1/4)*x + sqrt(2*x^2 + 1))/(x^2 + 1)) - 1 /8*sqrt(2)*log((x^2 - sqrt(2)*(2*x^2 + 1)^(1/4)*x + sqrt(2*x^2 + 1))/(x^2 + 1))
\[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (x^{2} + 1\right ) \sqrt [4]{2 x^{2} + 1}}\, dx \] Input:
integrate(1/(x**2+1)/(2*x**2+1)**(1/4),x)
Output:
Integral(1/((x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
\[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:
integrate(1/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((2*x^2 + 1)^(1/4)*(x^2 + 1)), x)
\[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:
integrate(1/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
Output:
integrate(1/((2*x^2 + 1)^(1/4)*(x^2 + 1)), x)
Timed out. \[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (x^2+1\right )\,{\left (2\,x^2+1\right )}^{1/4}} \,d x \] Input:
int(1/((x^2 + 1)*(2*x^2 + 1)^(1/4)),x)
Output:
int(1/((x^2 + 1)*(2*x^2 + 1)^(1/4)), x)
\[ \int \frac {1}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}}d x \] Input:
int(1/(x^2+1)/(2*x^2+1)^(1/4),x)
Output:
int(1/((2*x**2 + 1)**(1/4)*x**2 + (2*x**2 + 1)**(1/4)),x)