\(\int \frac {1}{x (1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1428]

Optimal result

Integrand size = 22, antiderivative size = 105 \[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\arctan \left (\sqrt [4]{1+2 x^2}\right )+\frac {\arctan \left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\text {arctanh}\left (\sqrt [4]{1+2 x^2}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Output:

arctan((2*x^2+1)^(1/4))+1/2*arctan(1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/(2*x^2+ 
1)^(1/4))*2^(1/2)-arctanh((2*x^2+1)^(1/4))+1/2*arctanh(2^(1/2)*(2*x^2+1)^( 
1/4)/(1+(2*x^2+1)^(1/2)))*2^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\arctan \left (\sqrt [4]{1+2 x^2}\right )-\frac {\arctan \left (\frac {-1+\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\text {arctanh}\left (\sqrt [4]{1+2 x^2}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Input:

Integrate[1/(x*(1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

ArcTan[(1 + 2*x^2)^(1/4)] - ArcTan[(-1 + Sqrt[1 + 2*x^2])/(Sqrt[2]*(1 + 2* 
x^2)^(1/4))]/Sqrt[2] - ArcTanh[(1 + 2*x^2)^(1/4)] + ArcTanh[(Sqrt[2]*(1 + 
2*x^2)^(1/4))/(1 + Sqrt[1 + 2*x^2])]/Sqrt[2]
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {349, 2009}

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.

\(\displaystyle \int \frac {1}{x \left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\)

\(\Big \downarrow \) 349

\(\displaystyle \int \left (\frac {1}{x \sqrt [4]{2 x^2+1}}-\frac {x}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \arctan \left (\sqrt [4]{2 x^2+1}\right )+\frac {\arctan \left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\text {arctanh}\left (\sqrt [4]{2 x^2+1}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}\)

Input:

Int[1/(x*(1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

ArcTan[(1 + 2*x^2)^(1/4)] + ArcTan[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*(1 + 2*x 
^2)^(1/4))]/Sqrt[2] - ArcTanh[(1 + 2*x^2)^(1/4)] + ArcTanh[(1 + Sqrt[1 + 2 
*x^2])/(Sqrt[2]*(1 + 2*x^2)^(1/4))]/Sqrt[2]
 

Defintions of rubi rules used

rule 349

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(1/4)*(c + d*x^2)), x], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In 
tegerQ[m/2])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 4.84 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {\sqrt {2 x^{2}+1}-\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}{\sqrt {2 x^{2}+1}+\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}\right ) \sqrt {2}}{4}-\frac {\arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1\right ) \sqrt {2}}{2}-\frac {\arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-1\right ) \sqrt {2}}{2}+\frac {\ln \left (-1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right )}{2}+\arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}}\right )-\frac {\ln \left (1+\left (2 x^{2}+1\right )^{\frac {1}{4}}\right )}{2}\)	\(140\)
trager	\(\frac {\ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}-\sqrt {2 x^{2}+1}-x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}-1}{x^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {2 x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (2 x^{2}+1\right )^{\frac {1}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{2}}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (-\frac {-\sqrt {2 x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\left (2 x^{2}+1\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\left (2 x^{2}+1\right )^{\frac {3}{4}}+\sqrt {2 x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}{x^{2}+1}\right )}{2}\)	\(307\)

Input:

int(1/x/(x^2+1)/(2*x^2+1)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

-1/4*ln(((2*x^2+1)^(1/2)-(2*x^2+1)^(1/4)*2^(1/2)+1)/((2*x^2+1)^(1/2)+(2*x^ 
2+1)^(1/4)*2^(1/2)+1))*2^(1/2)-1/2*arctan((2*x^2+1)^(1/4)*2^(1/2)+1)*2^(1/ 
2)-1/2*arctan((2*x^2+1)^(1/4)*2^(1/2)-1)*2^(1/2)+1/2*ln(-1+(2*x^2+1)^(1/4) 
)+arctan((2*x^2+1)^(1/4))-1/2*ln(1+(2*x^2+1)^(1/4))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.35 \[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \arctan \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \] Input:

integrate(1/x/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
 

Output:

-1/2*sqrt(2)*arctan(sqrt(2)*(2*x^2 + 1)^(1/4) + 1) - 1/2*sqrt(2)*arctan(sq 
rt(2)*(2*x^2 + 1)^(1/4) - 1) + 1/4*sqrt(2)*log(sqrt(2)*(2*x^2 + 1)^(1/4) + 
 sqrt(2*x^2 + 1) + 1) - 1/4*sqrt(2)*log(-sqrt(2)*(2*x^2 + 1)^(1/4) + sqrt( 
2*x^2 + 1) + 1) + arctan((2*x^2 + 1)^(1/4)) - 1/2*log((2*x^2 + 1)^(1/4) + 
1) + 1/2*log((2*x^2 + 1)^(1/4) - 1)
 

Sympy [F]

\[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{x \left (x^{2} + 1\right ) \sqrt [4]{2 x^{2} + 1}}\, dx \] Input:

integrate(1/x/(x**2+1)/(2*x**2+1)**(1/4),x)
 

Output:

Integral(1/(x*(x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )} x} \,d x } \] Input:

integrate(1/x/(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), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.45 \[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \arctan \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) \] Input:

integrate(1/x/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^2 + 1)^(1/4))) - 1/2*sqr 
t(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^(1/4))) + 1/4*sqrt(2)*lo 
g(sqrt(2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1) - 1/4*sqrt(2)*log(-sqrt 
(2)*(2*x^2 + 1)^(1/4) + sqrt(2*x^2 + 1) + 1) + arctan((2*x^2 + 1)^(1/4)) - 
 1/2*log((2*x^2 + 1)^(1/4) + 1) + 1/2*log((2*x^2 + 1)^(1/4) - 1)
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\mathrm {atan}\left ({\left (2\,x^2+1\right )}^{1/4}\right )+\mathrm {atan}\left ({\left (2\,x^2+1\right )}^{1/4}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(1/(x*(x^2 + 1)*(2*x^2 + 1)^(1/4)),x)
 

Output:

atan((2*x^2 + 1)^(1/4)) + atan((2*x^2 + 1)^(1/4)*1i)*1i - 2^(1/2)*atan(2^( 
1/2)*(2*x^2 + 1)^(1/4)*(1/2 - 1i/2))*(1/2 - 1i/2) - 2^(1/2)*atan(2^(1/2)*( 
2*x^2 + 1)^(1/4)*(1/2 + 1i/2))*(1/2 + 1i/2)
 

Reduce [F]

\[ \int \frac {1}{x \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{3}+\left (2 x^{2}+1\right )^{\frac {1}{4}} x}d x \] Input:

int(1/x/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Output:

int(1/((2*x**2 + 1)**(1/4)*x**3 + (2*x**2 + 1)**(1/4)*x),x)