\(\int \frac {x^8}{(1+x^2) (1+2 x^2)^{3/4}} \, dx\) [1471]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {125}{231} x \sqrt [4]{1+2 x^2}-\frac {16}{77} x^3 \sqrt [4]{1+2 x^2}+\frac {1}{11} x^5 \sqrt [4]{1+2 x^2}-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}-\frac {125}{231} \sqrt {2} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\sqrt {2} x\right ),2\right ) \] Output:

125/231*x*(2*x^2+1)^(1/4)-16/77*x^3*(2*x^2+1)^(1/4)+1/11*x^5*(2*x^2+1)^(1/ 
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*a 
rctanh(1/2*(1-(2*x^2+1)^(1/2))*2^(1/2)/x/(2*x^2+1)^(1/4))*2^(1/2)-125/231* 
2^(1/2)*InverseJacobiAM(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 = 6.05 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06 \[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\frac {1}{693} x \left (-356 x^2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},-2 x^2,-x^2\right )+\frac {3 \left (125+202 x^2-75 x^4+42 x^6+\frac {375 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-2 x^2,-x^2\right )}{\left (1+x^2\right ) \left (-3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},-2 x^2,-x^2\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},-2 x^2,-x^2\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},-2 x^2,-x^2\right )\right )\right )}\right )}{\left (1+2 x^2\right )^{3/4}}\right ) \] Input:

Integrate[x^8/((1 + x^2)*(1 + 2*x^2)^(3/4)),x]
 

Output:

(x*(-356*x^2*AppellF1[3/2, 3/4, 1, 5/2, -2*x^2, -x^2] + (3*(125 + 202*x^2 
- 75*x^4 + 42*x^6 + (375*AppellF1[1/2, 3/4, 1, 3/2, -2*x^2, -x^2])/((1 + x 
^2)*(-3*AppellF1[1/2, 3/4, 1, 3/2, -2*x^2, -x^2] + x^2*(2*AppellF1[3/2, 3/ 
4, 2, 5/2, -2*x^2, -x^2] + 3*AppellF1[3/2, 7/4, 1, 5/2, -2*x^2, -x^2]))))) 
/(1 + 2*x^2)^(3/4)))/693
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 159, 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 = {352, 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.

\(\displaystyle \int \frac {x^8}{\left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 352

\(\displaystyle \int \left (\frac {x^2}{\left (2 x^2+1\right )^{3/4}}+\frac {1}{\left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}}-\frac {1}{\left (2 x^2+1\right )^{3/4}}+\frac {x^6}{\left (2 x^2+1\right )^{3/4}}-\frac {x^4}{\left (2 x^2+1\right )^{3/4}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}-\frac {125}{231} \sqrt {2} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\sqrt {2} x\right ),2\right )-\frac {\text {arctanh}\left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {125}{231} \sqrt [4]{2 x^2+1} x+\frac {1}{11} \sqrt [4]{2 x^2+1} x^5-\frac {16}{77} \sqrt [4]{2 x^2+1} x^3\)

Input:

Int[x^8/((1 + x^2)*(1 + 2*x^2)^(3/4)),x]
 

Output:

(125*x*(1 + 2*x^2)^(1/4))/231 - (16*x^3*(1 + 2*x^2)^(1/4))/77 + (x^5*(1 + 
2*x^2)^(1/4))/11 + 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] - (125*Sqrt[2]*EllipticF[ArcTan[Sqrt[2]*x]/2, 2])/231
 

Defintions of rubi rules used

rule 352

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol 
] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(3/4)*(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]

\[\int \frac {x^{8}}{\left (x^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}d x\]

Input:

int(x^8/(x^2+1)/(2*x^2+1)^(3/4),x)
 

Output:

int(x^8/(x^2+1)/(2*x^2+1)^(3/4),x)
 

Fricas [F]

\[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^8/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {x^{8}}{\left (x^{2} + 1\right ) \left (2 x^{2} + 1\right )^{\frac {3}{4}}}\, dx \] Input:

integrate(x**8/(x**2+1)/(2*x**2+1)**(3/4),x)
 

Output:

Integral(x**8/((x**2 + 1)*(2*x**2 + 1)**(3/4)), x)
 

Maxima [F]

\[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^8/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="maxima")
 

Output:

integrate(x^8/((2*x^2 + 1)^(3/4)*(x^2 + 1)), x)
 

Giac [F]

\[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {x^{8}}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )}} \,d x } \] Input:

integrate(x^8/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="giac")
 

Output:

integrate(x^8/((2*x^2 + 1)^(3/4)*(x^2 + 1)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {x^8}{\left (x^2+1\right )\,{\left (2\,x^2+1\right )}^{3/4}} \,d x \] Input:

int(x^8/((x^2 + 1)*(2*x^2 + 1)^(3/4)),x)
 

Output:

int(x^8/((x^2 + 1)*(2*x^2 + 1)^(3/4)), x)
 

Reduce [F]

\[ \int \frac {x^8}{\left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {x^{8}}{\left (2 x^{2}+1\right )^{\frac {3}{4}} x^{2}+\left (2 x^{2}+1\right )^{\frac {3}{4}}}d x \] Input:

int(x^8/(x^2+1)/(2*x^2+1)^(3/4),x)
 

Output:

int(x**8/((2*x**2 + 1)**(3/4)*x**2 + (2*x**2 + 1)**(3/4)),x)