\(\int \frac {(-1+x^4) \sqrt [4]{x^2+x^6}}{1-x^4+x^8} \, dx\) [2362]

Optimal result

Integrand size = 29, antiderivative size = 187 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}}{1-x^4+x^8} \, dx=\frac {\arctan \left (\frac {\sqrt [8]{3} x}{\sqrt [4]{x^2+x^6}}\right )}{2\ 3^{3/8}}-\frac {\arctan \left (\frac {\sqrt {2} 3^{7/8} x \sqrt [4]{x^2+x^6}}{-3 x^2+3^{3/4} \sqrt {x^2+x^6}}\right )}{2 \sqrt {2} 3^{3/8}}-\frac {\text {arctanh}\left (\frac {\sqrt [8]{3} x}{\sqrt [4]{x^2+x^6}}\right )}{2\ 3^{3/8}}+\frac {\text {arctanh}\left (\frac {\frac {\sqrt [8]{3} x^2}{\sqrt {2}}+\frac {\sqrt {x^2+x^6}}{\sqrt {2} \sqrt [8]{3}}}{x \sqrt [4]{x^2+x^6}}\right )}{2 \sqrt {2} 3^{3/8}} \] Output:

1/6*arctan(3^(1/8)*x/(x^6+x^2)^(1/4))*3^(5/8)-1/12*arctan(2^(1/2)*3^(7/8)* 
x*(x^6+x^2)^(1/4)/(-3*x^2+3^(3/4)*(x^6+x^2)^(1/2)))*2^(1/2)*3^(5/8)-1/6*ar 
ctanh(3^(1/8)*x/(x^6+x^2)^(1/4))*3^(5/8)+1/12*arctanh((1/2*3^(1/8)*x^2*2^( 
1/2)+1/6*(x^6+x^2)^(1/2)*2^(1/2)*3^(7/8))/x/(x^6+x^2)^(1/4))*2^(1/2)*3^(5/ 
8)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.01 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}}{1-x^4+x^8} \, dx=\frac {\sqrt [4]{x^2+x^6} \left (2 \arctan \left (\frac {\sqrt [8]{3} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\sqrt {2} \arctan \left (\frac {\sqrt {2} 3^{7/8} \sqrt {x} \sqrt [4]{1+x^4}}{3 x-3^{3/4} \sqrt {1+x^4}}\right )-2 \text {arctanh}\left (\frac {\sqrt [8]{3} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} 3^{7/8} \sqrt {x} \sqrt [4]{1+x^4}}{3 x+3^{3/4} \sqrt {1+x^4}}\right )\right )}{4\ 3^{3/8} \sqrt {x} \sqrt [4]{1+x^4}} \] Input:

Integrate[((-1 + x^4)*(x^2 + x^6)^(1/4))/(1 - x^4 + x^8),x]
 

Output:

((x^2 + x^6)^(1/4)*(2*ArcTan[(3^(1/8)*Sqrt[x])/(1 + x^4)^(1/4)] + Sqrt[2]* 
ArcTan[(Sqrt[2]*3^(7/8)*Sqrt[x]*(1 + x^4)^(1/4))/(3*x - 3^(3/4)*Sqrt[1 + x 
^4])] - 2*ArcTanh[(3^(1/8)*Sqrt[x])/(1 + x^4)^(1/4)] + Sqrt[2]*ArcTanh[(Sq 
rt[2]*3^(7/8)*Sqrt[x]*(1 + x^4)^(1/4))/(3*x + 3^(3/4)*Sqrt[1 + x^4])]))/(4 
*3^(3/8)*Sqrt[x]*(1 + x^4)^(1/4))
 

Rubi [C] (warning: unable to verify)

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

Time = 1.51 (sec) , antiderivative size = 392, normalized size of antiderivative = 2.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2467, 25, 2035, 7279, 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.

\(\displaystyle \int \frac {\left (x^4-1\right ) \sqrt [4]{x^6+x^2}}{x^8-x^4+1} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{x^6+x^2} \int -\frac {\sqrt {x} \left (1-x^4\right ) \sqrt [4]{x^4+1}}{x^8-x^4+1}dx}{\sqrt {x} \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [4]{x^6+x^2} \int \frac {\sqrt {x} \left (1-x^4\right ) \sqrt [4]{x^4+1}}{x^8-x^4+1}dx}{\sqrt {x} \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt [4]{x^6+x^2} \int \frac {x \left (1-x^4\right ) \sqrt [4]{x^4+1}}{x^8-x^4+1}d\sqrt {x}}{\sqrt {x} \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 7279

\(\displaystyle -\frac {2 \sqrt [4]{x^6+x^2} \int \left (\frac {x \sqrt [4]{x^4+1}}{x^8-x^4+1}-\frac {x^5 \sqrt [4]{x^4+1}}{x^8-x^4+1}\right )d\sqrt {x}}{\sqrt {x} \sqrt [4]{x^4+1}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt [4]{x^6+x^2} \left (\frac {2 x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},-\frac {1}{4},1,\frac {11}{8},-x^4,\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \sqrt {3} \left (\sqrt {3}+i\right )}+\frac {2 x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},\frac {3}{4},1,\frac {11}{8},-x^4,\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \sqrt {3} \left (\sqrt {3}+i\right )}+\frac {2 i x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},\frac {3}{4},1,\frac {11}{8},-x^4,\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2 x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},1,-\frac {1}{4},\frac {11}{8},\frac {2 x^4}{1+i \sqrt {3}},-x^4\right )}{3 \sqrt {3} \left (-\sqrt {3}+i\right )}-\frac {2 x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {3}{4},\frac {11}{8},\frac {2 x^4}{1+i \sqrt {3}},-x^4\right )}{3 \sqrt {3} \left (-\sqrt {3}+i\right )}-\frac {2 i x^{3/2} \operatorname {AppellF1}\left (\frac {3}{8},1,\frac {3}{4},\frac {11}{8},\frac {2 x^4}{1+i \sqrt {3}},-x^4\right )}{3 \sqrt {3}}-\frac {1}{3} x^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},\frac {3}{4},\frac {11}{8},-x^4\right )\right )}{\sqrt {x} \sqrt [4]{x^4+1}}\)

Input:

Int[((-1 + x^4)*(x^2 + x^6)^(1/4))/(1 - x^4 + x^8),x]
 

Output:

(-2*(x^2 + x^6)^(1/4)*((2*x^(3/2)*AppellF1[3/8, -1/4, 1, 11/8, -x^4, (2*x^ 
4)/(1 - I*Sqrt[3])])/(3*Sqrt[3]*(I + Sqrt[3])) + (((2*I)/3)*x^(3/2)*Appell 
F1[3/8, 3/4, 1, 11/8, -x^4, (2*x^4)/(1 - I*Sqrt[3])])/Sqrt[3] + (2*x^(3/2) 
*AppellF1[3/8, 3/4, 1, 11/8, -x^4, (2*x^4)/(1 - I*Sqrt[3])])/(3*Sqrt[3]*(I 
 + Sqrt[3])) - (2*x^(3/2)*AppellF1[3/8, 1, -1/4, 11/8, (2*x^4)/(1 + I*Sqrt 
[3]), -x^4])/(3*Sqrt[3]*(I - Sqrt[3])) - (((2*I)/3)*x^(3/2)*AppellF1[3/8, 
1, 3/4, 11/8, (2*x^4)/(1 + I*Sqrt[3]), -x^4])/Sqrt[3] - (2*x^(3/2)*AppellF 
1[3/8, 1, 3/4, 11/8, (2*x^4)/(1 + I*Sqrt[3]), -x^4])/(3*Sqrt[3]*(I - Sqrt[ 
3])) - (x^(3/2)*Hypergeometric2F1[3/8, 3/4, 11/8, -x^4])/3))/(Sqrt[x]*(1 + 
 x^4)^(1/4))
 

Defintions of rubi rules used

rule 25

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 57.84 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.20

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-3\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3}}\right )}{4}\)	\(37\)
trager	\(\text {Expression too large to display}\)	\(1490\)

Input:

int((x^4-1)*(x^6+x^2)^(1/4)/(x^8-x^4+1),x,method=_RETURNVERBOSE)
 

Output:

1/4*sum(ln((-_R*x+(x^2*(x^4+1))^(1/4))/x)/_R^3,_R=RootOf(_Z^8-3))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.27 (sec) , antiderivative size = 1185, normalized size of antiderivative = 6.34 \[ \int \frac {\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}}{1-x^4+x^8} \, dx=\text {Too large to display} \] Input:

integrate((x^4-1)*(x^6+x^2)^(1/4)/(x^8-x^4+1),x, algorithm="fricas")
 

Output:

(1/432*I - 1/432)*27^(7/8)*sqrt(2)*log(-(2*27^(3/4)*(I*x^7 + I*x^3) - (x^6 
 + x^2)^(3/4)*((9*I + 9)*27^(1/8)*sqrt(2)*x^2 + 27^(5/8)*sqrt(2)*(-(I + 1) 
*x^4 - I - 1)) - 18*sqrt(x^6 + x^2)*(x^5 - sqrt(3)*x^3 + x) + 3*27^(1/4)*( 
-I*x^9 - 5*I*x^5 - I*x) - (x^6 + x^2)^(1/4)*(-(I - 1)*27^(7/8)*sqrt(2)*x^4 
 - 3*27^(3/8)*sqrt(2)*(-(I - 1)*x^6 - (I - 1)*x^2)))/(x^9 - x^5 + x)) - (1 
/432*I - 1/432)*27^(7/8)*sqrt(2)*log(-(2*27^(3/4)*(I*x^7 + I*x^3) - (x^6 + 
 x^2)^(3/4)*(-(9*I + 9)*27^(1/8)*sqrt(2)*x^2 + 27^(5/8)*sqrt(2)*((I + 1)*x 
^4 + I + 1)) - 18*sqrt(x^6 + x^2)*(x^5 - sqrt(3)*x^3 + x) + 3*27^(1/4)*(-I 
*x^9 - 5*I*x^5 - I*x) - (x^6 + x^2)^(1/4)*((I - 1)*27^(7/8)*sqrt(2)*x^4 - 
3*27^(3/8)*sqrt(2)*((I - 1)*x^6 + (I - 1)*x^2)))/(x^9 - x^5 + x)) - (1/432 
*I + 1/432)*27^(7/8)*sqrt(2)*log(-(2*27^(3/4)*(-I*x^7 - I*x^3) - (x^6 + x^ 
2)^(3/4)*(-(9*I - 9)*27^(1/8)*sqrt(2)*x^2 + 27^(5/8)*sqrt(2)*((I - 1)*x^4 
+ I - 1)) - 18*sqrt(x^6 + x^2)*(x^5 - sqrt(3)*x^3 + x) + 3*27^(1/4)*(I*x^9 
 + 5*I*x^5 + I*x) - (x^6 + x^2)^(1/4)*((I + 1)*27^(7/8)*sqrt(2)*x^4 - 3*27 
^(3/8)*sqrt(2)*((I + 1)*x^6 + (I + 1)*x^2)))/(x^9 - x^5 + x)) + (1/432*I + 
 1/432)*27^(7/8)*sqrt(2)*log(-(2*27^(3/4)*(-I*x^7 - I*x^3) - (x^6 + x^2)^( 
3/4)*((9*I - 9)*27^(1/8)*sqrt(2)*x^2 + 27^(5/8)*sqrt(2)*(-(I - 1)*x^4 - I 
+ 1)) - 18*sqrt(x^6 + x^2)*(x^5 - sqrt(3)*x^3 + x) + 3*27^(1/4)*(I*x^9 + 5 
*I*x^5 + I*x) - (x^6 + x^2)^(1/4)*(-(I + 1)*27^(7/8)*sqrt(2)*x^4 - 3*27^(3 
/8)*sqrt(2)*(-(I + 1)*x^6 - (I + 1)*x^2)))/(x^9 - x^5 + x)) - 1/216*27^...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((x^4-1)*(x^6+x^2)^(1/4)/(x^8-x^4+1),x, algorithm="maxima")
 

Output:

integrate((x^6 + x^2)^(1/4)*(x^4 - 1)/(x^8 - x^4 + 1), x)
 

Giac [F]

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

integrate((x^4-1)*(x^6+x^2)^(1/4)/(x^8-x^4+1),x, algorithm="giac")
 

Output:

integrate((x^6 + x^2)^(1/4)*(x^4 - 1)/(x^8 - x^4 + 1), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(x)*(x**4 + 1)**(1/4)*x**4)/(x**8 - x**4 + 1),x) - int((sqrt(x)*( 
x**4 + 1)**(1/4))/(x**8 - x**4 + 1),x)