\(\int \frac {x^5}{(1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1425]

Optimal result

Integrand size = 22, antiderivative size = 111 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=-\frac {1}{2} \left (1+2 x^2\right )^{3/4}+\frac {1}{14} \left (1+2 x^2\right )^{7/4}-\frac {\arctan \left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Output:

-1/2*(2*x^2+1)^(3/4)+1/14*(2*x^2+1)^(7/4)-1/2*arctan(1/2*(1-(2*x^2+1)^(1/2 
))*2^(1/2)/(2*x^2+1)^(1/4))*2^(1/2)-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 = 98, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{7} \left (-3+x^2\right ) \left (1+2 x^2\right )^{3/4}+\frac {\arctan \left (\frac {-1+\sqrt {1+2 x^2}}{\sqrt {2} \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}} \] Input:

Integrate[x^5/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

((-3 + x^2)*(1 + 2*x^2)^(3/4))/7 + ArcTan[(-1 + Sqrt[1 + 2*x^2])/(Sqrt[2]* 
(1 + 2*x^2)^(1/4))]/Sqrt[2] - ArcTanh[(Sqrt[2]*(1 + 2*x^2)^(1/4))/(1 + Sqr 
t[1 + 2*x^2])]/Sqrt[2]
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 109, 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.

\(\displaystyle \int \frac {x^5}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\)

\(\Big \downarrow \) 349

\(\displaystyle \int \left (\frac {x}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}}-\frac {x}{\sqrt [4]{2 x^2+1}}+\frac {x^3}{\sqrt [4]{2 x^2+1}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\arctan \left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {1}{14} \left (2 x^2+1\right )^{7/4}-\frac {1}{2} \left (2 x^2+1\right )^{3/4}\)

Input:

Int[x^5/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

-1/2*(1 + 2*x^2)^(3/4) + (1 + 2*x^2)^(7/4)/14 - ArcTan[(1 - Sqrt[1 + 2*x^2 
])/(Sqrt[2]*(1 + 2*x^2)^(1/4))]/Sqrt[2] - ArcTanh[(1 + Sqrt[1 + 2*x^2])/(S 
qrt[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 = 1.89 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(\frac {x^{2} \left (2 x^{2}+1\right )^{\frac {3}{4}}}{7}-\frac {3 \left (2 x^{2}+1\right )^{\frac {3}{4}}}{7}+\frac {\ln \left (\frac {\sqrt {2 x^{2}+1}-\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}{\sqrt {2 x^{2}+1}+\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1}\right ) \sqrt {2}}{4}+\frac {\arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}+1\right ) \sqrt {2}}{2}+\frac {\arctan \left (\left (2 x^{2}+1\right )^{\frac {1}{4}} \sqrt {2}-1\right ) \sqrt {2}}{2}\)	\(127\)
risch	\(\frac {\left (x^{2}-3\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}{7}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}+x^{2}}{x^{2}+1}\right )}{2}\)	\(169\)
trager	\(\left (\frac {x^{2}}{7}-\frac {3}{7}\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{2} \sqrt {2 x^{2}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )^{3} \left (2 x^{2}+1\right )^{\frac {1}{4}}-x^{2}}{x^{2}+1}\right )}{2}\)	\(174\)

Input:

int(x^5/(x^2+1)/(2*x^2+1)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/7*x^2*(2*x^2+1)^(3/4)-3/7*(2*x^2+1)^(3/4)+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)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.08 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{7} \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} - 3\right )} + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} - 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) \] Input:

integrate(x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
 

Output:

1/7*(2*x^2 + 1)^(3/4)*(x^2 - 3) + 1/2*sqrt(2)*arctan(sqrt(2)*(2*x^2 + 1)^( 
1/4) + 1) + 1/2*sqrt(2)*arctan(sqrt(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)
 

Sympy [F]

\[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {x^{5}}{\left (x^{2} + 1\right ) \sqrt [4]{2 x^{2} + 1}}\, dx \] Input:

integrate(x**5/(x**2+1)/(2*x**2+1)**(1/4),x)
 

Output:

Integral(x**5/((x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.23 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{14} \, {\left (2 \, x^{2} + 1\right )}^{\frac {7}{4}} + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{2} \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} \] Input:

integrate(x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="maxima")
 

Output:

1/14*(2*x^2 + 1)^(7/4) + 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^ 
2 + 1)^(1/4))) + 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^ 
(1/4))) - 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) - 1/2 
*(2*x^2 + 1)^(3/4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.23 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{14} \, {\left (2 \, x^{2} + 1\right )}^{\frac {7}{4}} + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} {\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} + \sqrt {2 \, x^{2} + 1} + 1\right ) - \frac {1}{2} \, {\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} \] Input:

integrate(x^5/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
 

Output:

1/14*(2*x^2 + 1)^(7/4) + 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*(2*x^ 
2 + 1)^(1/4))) + 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(2*x^2 + 1)^ 
(1/4))) - 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) - 1/2 
*(2*x^2 + 1)^(3/4)
 

Mupad [B] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.64 \[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {{\left (2\,x^2+1\right )}^{7/4}}{14}-\frac {{\left (2\,x^2+1\right )}^{3/4}}{2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,{\left (2\,x^2+1\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right ) \] Input:

int(x^5/((x^2 + 1)*(2*x^2 + 1)^(1/4)),x)
 

Output:

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) - (2*x^2 + 1)^ 
(3/4)/2 + (2*x^2 + 1)^(7/4)/14
 

Reduce [F]

\[ \int \frac {x^5}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {x^{5}}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}}d x \] Input:

int(x^5/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Output:

int(x**5/((2*x**2 + 1)**(1/4)*x**2 + (2*x**2 + 1)**(1/4)),x)