\(\int \frac {1+x^2+x^4}{(1-x^4) \sqrt [4]{x^3+x^5}} \, dx\) [2334]

Optimal result

Integrand size = 29, antiderivative size = 184 \[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {\left (x^3+x^5\right )^{3/4}}{x^2 \left (1+x^2\right )}+\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{4 \sqrt [4]{2}}-\frac {3 \arctan \left (\frac {2^{3/4} x \sqrt [4]{x^3+x^5}}{\sqrt {2} x^2-\sqrt {x^3+x^5}}\right )}{4\ 2^{3/4}}+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )}{4 \sqrt [4]{2}}+\frac {3 \text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^3+x^5}}{2^{3/4}}}{x \sqrt [4]{x^3+x^5}}\right )}{4\ 2^{3/4}} \] Output:

(x^5+x^3)^(3/4)/x^2/(x^2+1)+3/8*arctan(2^(1/4)*x/(x^5+x^3)^(1/4))*2^(3/4)- 
3/8*arctan(2^(3/4)*x*(x^5+x^3)^(1/4)/(2^(1/2)*x^2-(x^5+x^3)^(1/2)))*2^(1/4 
)+3/8*arctanh(2^(1/4)*x/(x^5+x^3)^(1/4))*2^(3/4)+3/8*arctanh((1/2*x^2*2^(3 
/4)+1/2*(x^5+x^3)^(1/2)*2^(1/4))/x/(x^5+x^3)^(1/4))*2^(1/4)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.23 \[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\frac {x^{3/4} \left (8 \sqrt [4]{x}+3\ 2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )-3 \sqrt [4]{2} \sqrt [4]{1+x^2} \arctan \left (\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{1+x^2}}{\sqrt {2} \sqrt {x}-\sqrt {1+x^2}}\right )+3\ 2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )+3 \sqrt [4]{2} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{x} \sqrt [4]{1+x^2}}{2 \sqrt {x}+\sqrt {2} \sqrt {1+x^2}}\right )\right )}{8 \sqrt [4]{x^3+x^5}} \] Input:

Integrate[(1 + x^2 + x^4)/((1 - x^4)*(x^3 + x^5)^(1/4)),x]
 

Output:

(x^(3/4)*(8*x^(1/4) + 3*2^(3/4)*(1 + x^2)^(1/4)*ArcTan[(2^(1/4)*x^(1/4))/( 
1 + x^2)^(1/4)] - 3*2^(1/4)*(1 + x^2)^(1/4)*ArcTan[(2^(3/4)*x^(1/4)*(1 + x 
^2)^(1/4))/(Sqrt[2]*Sqrt[x] - Sqrt[1 + x^2])] + 3*2^(3/4)*(1 + x^2)^(1/4)* 
ArcTanh[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] + 3*2^(1/4)*(1 + x^2)^(1/4)*Arc 
Tanh[(2*2^(1/4)*x^(1/4)*(1 + x^2)^(1/4))/(2*Sqrt[x] + Sqrt[2]*Sqrt[1 + x^2 
])]))/(8*(x^3 + x^5)^(1/4))
 

Rubi [C] (warning: unable to verify)

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

Time = 0.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.54, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2467, 1388, 2035, 7276, 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.

\(\displaystyle \int \frac {x^4+x^2+1}{\left (1-x^4\right ) \sqrt [4]{x^5+x^3}} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {x^{3/4} \sqrt [4]{x^2+1} \int \frac {x^4+x^2+1}{x^{3/4} \sqrt [4]{x^2+1} \left (1-x^4\right )}dx}{\sqrt [4]{x^5+x^3}}\)

\(\Big \downarrow \) 1388

\(\displaystyle \frac {x^{3/4} \sqrt [4]{x^2+1} \int \frac {x^4+x^2+1}{x^{3/4} \left (1-x^2\right ) \left (x^2+1\right )^{5/4}}dx}{\sqrt [4]{x^5+x^3}}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \int \frac {x^4+x^2+1}{\left (1-x^2\right ) \left (x^2+1\right )^{5/4}}d\sqrt [4]{x}}{\sqrt [4]{x^5+x^3}}\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \int \left (-\frac {x^2}{\left (x^2+1\right )^{5/4}}+\frac {3}{\left (1-x^2\right ) \left (x^2+1\right )^{5/4}}-\frac {2}{\left (x^2+1\right )^{5/4}}\right )d\sqrt [4]{x}}{\sqrt [4]{x^5+x^3}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 x^{3/4} \sqrt [4]{x^2+1} \left (3 \sqrt [4]{x} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^2,-x^2\right )-2 \sqrt [4]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {5}{4},\frac {9}{8},-x^2\right )-\frac {1}{9} x^{9/4} \operatorname {Hypergeometric2F1}\left (\frac {9}{8},\frac {5}{4},\frac {17}{8},-x^2\right )\right )}{\sqrt [4]{x^5+x^3}}\)

Input:

Int[(1 + x^2 + x^4)/((1 - x^4)*(x^3 + x^5)^(1/4)),x]
 

Output:

(4*x^(3/4)*(1 + x^2)^(1/4)*(3*x^(1/4)*AppellF1[1/8, 1, 5/4, 9/8, x^2, -x^2 
] - 2*x^(1/4)*Hypergeometric2F1[1/8, 5/4, 9/8, -x^2] - (x^(9/4)*Hypergeome 
tric2F1[9/8, 5/4, 17/8, -x^2])/9))/(x^3 + x^5)^(1/4)
 

Defintions of rubi rules used

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

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 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [A] (verified)

Time = 19.83 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.53

method	result	size
pseudoelliptic	\(\frac {-6 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+3 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-3 \ln \left (\frac {-2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}{2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-6 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-6 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+16 x}{16 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\)	\(282\)
risch	\(\text {Expression too large to display}\)	\(733\)
trager	\(\text {Expression too large to display}\)	\(738\)

Input:

int((x^4+x^2+1)/(-x^4+1)/(x^5+x^3)^(1/4),x,method=_RETURNVERBOSE)
 

Output:

1/16*(-6*arctan(1/2*2^(3/4)/x*(x^3*(x^2+1))^(1/4))*2^(3/4)*(x^3*(x^2+1))^( 
1/4)+3*ln((-2^(1/4)*x-(x^3*(x^2+1))^(1/4))/(2^(1/4)*x-(x^3*(x^2+1))^(1/4)) 
)*2^(3/4)*(x^3*(x^2+1))^(1/4)-3*ln((-2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2) 
*x^2+(x^3*(x^2+1))^(1/2))/(2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3* 
(x^2+1))^(1/2)))*2^(1/4)*(x^3*(x^2+1))^(1/4)-6*arctan((2^(1/4)*(x^3*(x^2+1 
))^(1/4)+x)/x)*2^(1/4)*(x^3*(x^2+1))^(1/4)-6*arctan((2^(1/4)*(x^3*(x^2+1)) 
^(1/4)-x)/x)*2^(1/4)*(x^3*(x^2+1))^(1/4)+16*x)/(x^3*(x^2+1))^(1/4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 708 vs. \(2 (143) = 286\).

Time = 7.69 (sec) , antiderivative size = 708, normalized size of antiderivative = 3.85 \[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx =\text {Too large to display} \] Input:

integrate((x^4+x^2+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="fricas")
 

Output:

1/128*(6*8^(3/4)*(x^4 + x^2)*arctan((x^6 + 4*x^5 + 6*x^4 + 4*x^3 + 8^(3/4) 
*(x^5 + x^3)^(1/4)*(3*x^4 - 2*x^3 + 3*x^2) + 4*sqrt(2)*sqrt(x^5 + x^3)*(x^ 
3 + 2*x^2 + x) + 2*8^(1/4)*(x^5 + x^3)^(3/4)*(x^2 - 6*x + 1) + x^2)/(x^6 - 
 28*x^5 + 6*x^4 - 28*x^3 + x^2)) + 6*8^(3/4)*(x^4 + x^2)*arctan(-(x^6 + 4* 
x^5 + 6*x^4 + 4*x^3 - 8^(3/4)*(x^5 + x^3)^(1/4)*(3*x^4 - 2*x^3 + 3*x^2) + 
4*sqrt(2)*sqrt(x^5 + x^3)*(x^3 + 2*x^2 + x) - 2*8^(1/4)*(x^5 + x^3)^(3/4)* 
(x^2 - 6*x + 1) + x^2)/(x^6 - 28*x^5 + 6*x^4 - 28*x^3 + x^2)) + 24*2^(3/4) 
*(x^4 + x^2)*arctan(2*(2^(3/4)*(x^5 + x^3)^(1/4)*x^2 + 2^(1/4)*(x^5 + x^3) 
^(3/4))/(x^4 - 2*x^3 + x^2)) + 12*2^(3/4)*(x^4 + x^2)*log((4*sqrt(2)*(x^5 
+ x^3)^(1/4)*x^2 + 2^(3/4)*(x^4 + 2*x^3 + x^2) + 4*2^(1/4)*sqrt(x^5 + x^3) 
*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - 12*2^(3/4)*(x^4 + x^2)*lo 
g((4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - 2^(3/4)*(x^4 + 2*x^3 + x^2) - 4*2^(1/ 
4)*sqrt(x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) + 3*8^(3/ 
4)*(x^4 + x^2)*log(2*(4*8^(1/4)*(x^5 + x^3)^(1/4)*x^2 + 8^(3/4)*(x^5 + x^3 
)^(3/4) + sqrt(2)*(x^4 + 2*x^3 + x^2) + 8*sqrt(x^5 + x^3)*x)/(x^4 + 2*x^3 
+ x^2)) - 3*8^(3/4)*(x^4 + x^2)*log(-2*(4*8^(1/4)*(x^5 + x^3)^(1/4)*x^2 + 
8^(3/4)*(x^5 + x^3)^(3/4) - sqrt(2)*(x^4 + 2*x^3 + x^2) - 8*sqrt(x^5 + x^3 
)*x)/(x^4 + 2*x^3 + x^2)) + 128*(x^5 + x^3)^(3/4))/(x^4 + x^2)
 

Sympy [F]

\[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=- \int \frac {x^{2}}{x^{4} \sqrt [4]{x^{5} + x^{3}} - \sqrt [4]{x^{5} + x^{3}}}\, dx - \int \frac {x^{4}}{x^{4} \sqrt [4]{x^{5} + x^{3}} - \sqrt [4]{x^{5} + x^{3}}}\, dx - \int \frac {1}{x^{4} \sqrt [4]{x^{5} + x^{3}} - \sqrt [4]{x^{5} + x^{3}}}\, dx \] Input:

integrate((x**4+x**2+1)/(-x**4+1)/(x**5+x**3)**(1/4),x)
 

Output:

-Integral(x**2/(x**4*(x**5 + x**3)**(1/4) - (x**5 + x**3)**(1/4)), x) - In 
tegral(x**4/(x**4*(x**5 + x**3)**(1/4) - (x**5 + x**3)**(1/4)), x) - Integ 
ral(1/(x**4*(x**5 + x**3)**(1/4) - (x**5 + x**3)**(1/4)), x)
 

Maxima [F]

\[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} + x^{2} + 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \] Input:

integrate((x^4+x^2+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="maxima")
 

Output:

-integrate((x^4 + x^2 + 1)/((x^5 + x^3)^(1/4)*(x^4 - 1)), x)
 

Giac [F]

\[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int { -\frac {x^{4} + x^{2} + 1}{{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \] Input:

integrate((x^4+x^2+1)/(-x^4+1)/(x^5+x^3)^(1/4),x, algorithm="giac")
 

Output:

integrate(-(x^4 + x^2 + 1)/((x^5 + x^3)^(1/4)*(x^4 - 1)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=\int -\frac {x^4+x^2+1}{{\left (x^5+x^3\right )}^{1/4}\,\left (x^4-1\right )} \,d x \] Input:

int(-(x^2 + x^4 + 1)/((x^3 + x^5)^(1/4)*(x^4 - 1)),x)
 

Output:

int(-(x^2 + x^4 + 1)/((x^3 + x^5)^(1/4)*(x^4 - 1)), x)
 

Reduce [F]

\[ \int \frac {1+x^2+x^4}{\left (1-x^4\right ) \sqrt [4]{x^3+x^5}} \, dx=-\left (\int \frac {x^{4}}{x^{\frac {19}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}-x^{\frac {3}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right )-\left (\int \frac {x^{2}}{x^{\frac {19}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}-x^{\frac {3}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right )-\left (\int \frac {1}{x^{\frac {19}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}-x^{\frac {3}{4}} \left (x^{2}+1\right )^{\frac {1}{4}}}d x \right ) \] Input:

int((x^4+x^2+1)/(-x^4+1)/(x^5+x^3)^(1/4),x)
 

Output:

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