\(\int \frac {x^6}{(1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1431]

Optimal result

Integrand size = 22, antiderivative size = 157 \[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {38 x}{15 \sqrt [4]{1+2 x^2}}-\frac {4}{15} x \left (1+2 x^2\right )^{3/4}+\frac {1}{9} x^3 \left (1+2 x^2\right )^{3/4}-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {19}{15} \sqrt {2} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2} x\right )\right |2\right ) \] Output:

38/15*x/(2*x^2+1)^(1/4)-4/15*x*(2*x^2+1)^(3/4)+1/9*x^3*(2*x^2+1)^(3/4)-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*arctanh 
(1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/x/(2*x^2+1)^(1/4))*2^(1/2)-19/15*2^(1/2)* 
EllipticE(sin(1/2*arctan(x*2^(1/2))),2^(1/2))
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 5.63 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {1}{45} x \left (19 x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-2 x^2,-x^2\right )+\frac {-12-19 x^2+10 x^4-\frac {36 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-2 x^2,-x^2\right )}{\left (1+x^2\right ) \left (-3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-2 x^2,-x^2\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-2 x^2,-x^2\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-2 x^2,-x^2\right )\right )\right )}}{\sqrt [4]{1+2 x^2}}\right ) \] Input:

Integrate[x^6/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

(x*(19*x^2*AppellF1[3/2, 1/4, 1, 5/2, -2*x^2, -x^2] + (-12 - 19*x^2 + 10*x 
^4 - (36*AppellF1[1/2, 1/4, 1, 3/2, -2*x^2, -x^2])/((1 + x^2)*(-3*AppellF1 
[1/2, 1/4, 1, 3/2, -2*x^2, -x^2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, -2*x^ 
2, -x^2] + AppellF1[3/2, 5/4, 1, 5/2, -2*x^2, -x^2]))))/(1 + 2*x^2)^(1/4)) 
)/45
 

Rubi [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.99, 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.

\(\displaystyle \int \frac {x^6}{\left (x^2+1\right ) \sqrt [4]{2 x^2+1}} \, dx\)

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {19}{15} \sqrt {2} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2} x\right )\right |2\right )+\frac {\text {arctanh}\left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {4}{15} \left (2 x^2+1\right )^{3/4} x+\frac {38 x}{15 \sqrt [4]{2 x^2+1}}+\frac {1}{9} \left (2 x^2+1\right )^{3/4} x^3\)

Input:

Int[x^6/((1 + x^2)*(1 + 2*x^2)^(1/4)),x]
 

Output:

(38*x)/(15*(1 + 2*x^2)^(1/4)) - (4*x*(1 + 2*x^2)^(3/4))/15 + (x^3*(1 + 2*x 
^2)^(3/4))/9 + ArcTan[(1 + Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))] 
/Sqrt[2] + ArcTanh[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))]/Sq 
rt[2] - (19*Sqrt[2]*EllipticE[ArcTan[Sqrt[2]*x]/2, 2])/15
 

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 [F]

\[\int \frac {x^{6}}{\left (x^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {1}{4}}}d x\]

Input:

int(x^6/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Output:

int(x^6/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Fricas [F]

\[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {x^{6}}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^6/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="fricas")
 

Output:

integral((2*x^2 + 1)^(3/4)*x^6/(2*x^4 + 3*x^2 + 1), x)
 

Sympy [F]

\[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {x^{6}}{\left (x^{2} + 1\right ) \sqrt [4]{2 x^{2} + 1}}\, dx \] Input:

integrate(x**6/(x**2+1)/(2*x**2+1)**(1/4),x)
 

Output:

Integral(x**6/((x**2 + 1)*(2*x**2 + 1)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {x^{6}}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^6/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="maxima")
 

Output:

integrate(x^6/((2*x^2 + 1)^(1/4)*(x^2 + 1)), x)
 

Giac [F]

\[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int { \frac {x^{6}}{{\left (2 \, x^{2} + 1\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^6/(x^2+1)/(2*x^2+1)^(1/4),x, algorithm="giac")
 

Output:

integrate(x^6/((2*x^2 + 1)^(1/4)*(x^2 + 1)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {x^6}{\left (x^2+1\right )\,{\left (2\,x^2+1\right )}^{1/4}} \,d x \] Input:

int(x^6/((x^2 + 1)*(2*x^2 + 1)^(1/4)),x)
 

Output:

int(x^6/((x^2 + 1)*(2*x^2 + 1)^(1/4)), x)
 

Reduce [F]

\[ \int \frac {x^6}{\left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {x^{6}}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{2}+\left (2 x^{2}+1\right )^{\frac {1}{4}}}d x \] Input:

int(x^6/(x^2+1)/(2*x^2+1)^(1/4),x)
 

Output:

int(x**6/((2*x**2 + 1)**(1/4)*x**2 + (2*x**2 + 1)**(1/4)),x)