\(\int \frac {x^2}{(-1+x^4) \sqrt [4]{x^2+x^6}} \, dx\) [2184]

Optimal result

Integrand size = 22, antiderivative size = 162 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}-\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{4\ 2^{3/4}} \] Output:

-1/8*arctan(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3/4)-1/8*arctan(2^(3/4)*x*(x^6+x 
^2)^(1/4)/(2^(1/2)*x^2-(x^6+x^2)^(1/2)))*2^(1/4)-1/8*arctanh(2^(1/4)*x/(x^ 
6+x^2)^(1/4))*2^(3/4)+1/8*arctanh((1/2*x^2*2^(3/4)+1/2*(x^6+x^2)^(1/2)*2^( 
1/4))/x/(x^6+x^2)^(1/4))*2^(1/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.09 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\sqrt {x} \sqrt [4]{1+x^4} \left (\sqrt {2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )-\text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{4\ 2^{3/4} \sqrt [4]{x^2+x^6}} \] Input:

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

Output:

-1/4*(Sqrt[x]*(1 + x^4)^(1/4)*(Sqrt[2]*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^4)^ 
(1/4)] + ArcTan[(2^(3/4)*Sqrt[x]*(1 + x^4)^(1/4))/(Sqrt[2]*x - Sqrt[1 + x^ 
4])] + Sqrt[2]*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^4)^(1/4)] - ArcTanh[(2*2^( 
1/4)*Sqrt[x]*(1 + x^4)^(1/4))/(2*x + Sqrt[2]*Sqrt[1 + x^4])]))/(2^(3/4)*(x 
^2 + x^6)^(1/4))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.28, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1948, 25, 966, 1012}

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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 966

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

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {2 x^3 \sqrt [4]{x^4+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},x^4,-x^4\right )}{5 \sqrt [4]{x^6+x^2}}\)

Input:

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

Output:

(-2*x^3*(1 + x^4)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, x^4, -x^4])/(5*(x^2 + 
x^6)^(1/4))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 966

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*( 
m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*x)^( 
1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && FractionQ[m] && IntegerQ[p]
 

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [A] (verified)

Time = 16.93 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.26

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) \sqrt {2}+\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right )\right )}{16}\)	\(204\)
trager	\(\text {Expression too large to display}\)	\(636\)

Input:

int(x^2/(x^4-1)/(x^6+x^2)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

-1/16*2^(1/4)*(ln((-2^(1/4)*x-(x^2*(x^4+1))^(1/4))/(2^(1/4)*x-(x^2*(x^4+1) 
)^(1/4)))*2^(1/2)-2*arctan(1/2*2^(3/4)/x*(x^2*(x^4+1))^(1/4))*2^(1/2)+ln(( 
-2^(3/4)*(x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^(1/2))/(2^(3/4)*( 
x^2*(x^4+1))^(1/4)*x+2^(1/2)*x^2+(x^2*(x^4+1))^(1/2)))+2*arctan((2^(1/4)*( 
x^2*(x^4+1))^(1/4)-x)/x)+2*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (123) = 246\).

Time = 8.96 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.48 \[ \int \frac {x^2}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\frac {1}{32} \cdot 8^{\frac {3}{4}} \arctan \left (-\frac {8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )} - 4 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{8 \, {\left (x^{5} + x\right )}}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} + 1\right )} - 2 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}}}{4 \, {\left (x^{5} + x\right )}}\right ) + \frac {1}{128} \cdot 8^{\frac {3}{4}} \log \left (\frac {x^{5} + 8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, x^{3} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x + 2 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}\right ) - \frac {1}{128} \cdot 8^{\frac {3}{4}} \log \left (\frac {x^{5} - 8^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, x^{3} + 4 \, \sqrt {2} \sqrt {x^{6} + x^{2}} x - 2 \cdot 8^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + x}{x^{5} + 2 \, x^{3} + x}\right ) - \frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} + \sqrt {2} {\left (x^{5} + 2 \, x^{3} + x\right )} + 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} - 2 \, x^{3} + x}\right ) + \frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (-\frac {4 \cdot 2^{\frac {1}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \cdot 2^{\frac {3}{4}} {\left (x^{6} + x^{2}\right )}^{\frac {3}{4}} - \sqrt {2} {\left (x^{5} + 2 \, x^{3} + x\right )} - 4 \, \sqrt {x^{6} + x^{2}} x}{x^{5} - 2 \, x^{3} + x}\right ) \] Input:

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

Output:

1/32*8^(3/4)*arctan(-1/8*(8^(3/4)*(x^6 + x^2)^(1/4)*(x^4 + 1) - 4*8^(1/4)* 
(x^6 + x^2)^(3/4))/(x^5 + x)) - 1/16*2^(3/4)*arctan(-1/4*(2^(3/4)*(x^6 + x 
^2)^(1/4)*(x^4 + 1) - 2*2^(1/4)*(x^6 + x^2)^(3/4))/(x^5 + x)) + 1/128*8^(3 
/4)*log((x^5 + 8^(3/4)*(x^6 + x^2)^(1/4)*x^2 + 2*x^3 + 4*sqrt(2)*sqrt(x^6 
+ x^2)*x + 2*8^(1/4)*(x^6 + x^2)^(3/4) + x)/(x^5 + 2*x^3 + x)) - 1/128*8^( 
3/4)*log((x^5 - 8^(3/4)*(x^6 + x^2)^(1/4)*x^2 + 2*x^3 + 4*sqrt(2)*sqrt(x^6 
 + x^2)*x - 2*8^(1/4)*(x^6 + x^2)^(3/4) + x)/(x^5 + 2*x^3 + x)) - 1/32*2^( 
3/4)*log((4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 2*2^(3/4)*(x^6 + x^2)^(3/4) + 
sqrt(2)*(x^5 + 2*x^3 + x) + 4*sqrt(x^6 + x^2)*x)/(x^5 - 2*x^3 + x)) + 1/32 
*2^(3/4)*log(-(4*2^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 2*2^(3/4)*(x^6 + x^2)^(3/ 
4) - sqrt(2)*(x^5 + 2*x^3 + x) - 4*sqrt(x^6 + x^2)*x)/(x^5 - 2*x^3 + x))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(x)*(x**4 + 1)**(1/4)*x)/(sqrt(x**4 + 1)*x**4 - sqrt(x**4 + 1)),x 
)