\(\int \frac {1}{x^6 (1+x^2) \sqrt [4]{1+2 x^2}} \, dx\) [1437]

Optimal result

Integrand size = 22, antiderivative size = 177 \[ \int \frac {1}{x^6 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {34 x}{5 \sqrt [4]{1+2 x^2}}-\frac {\left (1+2 x^2\right )^{3/4}}{5 x^5}+\frac {4 \left (1+2 x^2\right )^{3/4}}{5 x^3}-\frac {17 \left (1+2 x^2\right )^{3/4}}{5 x}-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {17}{5} \sqrt {2} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2} x\right )\right |2\right ) \] Output:

34/5*x/(2*x^2+1)^(1/4)-1/5*(2*x^2+1)^(3/4)/x^5+4/5*(2*x^2+1)^(3/4)/x^3-17/ 
5*(2*x^2+1)^(3/4)/x-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)-17/5*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 = 11.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^6 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\frac {17}{15} x^3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},1,\frac {5}{2},-2 x^2,-x^2\right )+\frac {-1+2 x^2-9 x^4-34 x^6-\frac {36 x^6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-2 x^2,-x^2\right )}{\left (1+x^2\right ) \left (-3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},-2 x^2,-x^2\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},-2 x^2,-x^2\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},-2 x^2,-x^2\right )\right )\right )}}{5 x^5 \sqrt [4]{1+2 x^2}} \] Input:

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

Output:

(17*x^3*AppellF1[3/2, 1/4, 1, 5/2, -2*x^2, -x^2])/15 + (-1 + 2*x^2 - 9*x^4 
 - 34*x^6 - (36*x^6*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]))))/(5*x^5* 
(1 + 2*x^2)^(1/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 176, 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.

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

\(\Big \downarrow \) 349

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {17}{5} \sqrt {2} E\left (\left .\frac {1}{2} \arctan \left (\sqrt {2} x\right )\right |2\right )+\frac {\text {arctanh}\left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {34 x}{5 \sqrt [4]{2 x^2+1}}-\frac {17 \left (2 x^2+1\right )^{3/4}}{5 x}-\frac {\left (2 x^2+1\right )^{3/4}}{5 x^5}+\frac {4 \left (2 x^2+1\right )^{3/4}}{5 x^3}\)

Input:

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

Output:

(34*x)/(5*(1 + 2*x^2)^(1/4)) - (1 + 2*x^2)^(3/4)/(5*x^5) + (4*(1 + 2*x^2)^ 
(3/4))/(5*x^3) - (17*(1 + 2*x^2)^(3/4))/(5*x) + 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))]/Sqrt[2] - (17*Sqrt[2]*EllipticE[ArcTan[Sqrt 
[2]*x]/2, 2])/5
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((2*x^2 + 1)^(3/4)/(2*x^10 + 3*x^8 + x^6), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(1/x^6/(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^6), x)
 

Giac [F]

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

integrate(1/x^6/(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^6), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^6 \left (1+x^2\right ) \sqrt [4]{1+2 x^2}} \, dx=\int \frac {1}{\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{8}+\left (2 x^{2}+1\right )^{\frac {1}{4}} x^{6}}d x \] Input:

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

Output:

int(1/((2*x**2 + 1)**(1/4)*x**8 + (2*x**2 + 1)**(1/4)*x**6),x)