Integrand size = 29, antiderivative size = 162 \[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}+\frac {3 \arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}-\frac {3 \text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{4\ 2^{3/4}} \] Output:
-1/8*arctan(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3/4)+3/8*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)-3/8*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)
Time = 1.92 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.10 \[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=-\frac {\sqrt {x} \sqrt [4]{1+x^4} \left (\sqrt {2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )-3 \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+3 \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{4\ 2^{3/4} \sqrt [4]{x^2+x^6}} \] Input:
Integrate[(1 - x^2 + x^4)/((-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)] - 3*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)] + 3*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))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.69 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.64, 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, 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.
\(\displaystyle \int \frac {x^4-x^2+1}{\left (x^4-1\right ) \sqrt [4]{x^6+x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt [4]{x^4+1} \int -\frac {x^4-x^2+1}{\sqrt {x} \left (1-x^4\right ) \sqrt [4]{x^4+1}}dx}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4+1} \int \frac {x^4-x^2+1}{\sqrt {x} \left (1-x^4\right ) \sqrt [4]{x^4+1}}dx}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \frac {x^4-x^2+1}{\left (1-x^4\right ) \sqrt [4]{x^4+1}}d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \left (\frac {2-x^2}{\left (1-x^4\right ) \sqrt [4]{x^4+1}}-\frac {1}{\sqrt [4]{x^4+1}}\right )d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \left (2 \sqrt {x} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},x^4,-x^4\right )-\frac {1}{5} x^{5/2} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},x^4,-x^4\right )-\sqrt {x} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )\right )}{\sqrt [4]{x^6+x^2}}\) |
Input:
Int[(1 - x^2 + x^4)/((-1 + x^4)*(x^2 + x^6)^(1/4)),x]
Output:
(-2*Sqrt[x]*(1 + x^4)^(1/4)*(2*Sqrt[x]*AppellF1[1/8, 1, 1/4, 9/8, x^4, -x^ 4] - (x^(5/2)*AppellF1[5/8, 1, 1/4, 13/8, x^4, -x^4])/5 - Sqrt[x]*Hypergeo metric2F1[1/8, 1/4, 9/8, -x^4]))/(x^2 + x^6)^(1/4)
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]
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]
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]
Time = 18.36 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27
method | result | size |
pseudoelliptic | \(-\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) \sqrt {2}-3 \ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right )-6 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right )-6 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right )\right )}{16}\) | \(206\) |
trager | \(\text {Expression too large to display}\) | \(634\) |
Input:
int((x^4-x^2+1)/(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)-3*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)))-6*arctan((2^(1/4) *(x^2*(x^4+1))^(1/4)-x)/x)-6*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4)+x)/x))
Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (123) = 246\).
Time = 67.11 (sec) , antiderivative size = 615, normalized size of antiderivative = 3.80 \[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx =\text {Too large to display} \] Input:
integrate((x^4-x^2+1)/(x^4-1)/(x^6+x^2)^(1/4),x, algorithm="fricas")
Output:
-3/64*8^(3/4)*arctan((x^9 + 4*x^7 + 6*x^5 + 4*x^3 + 8^(3/4)*(3*x^6 - 2*x^4 + 3*x^2)*(x^6 + x^2)^(1/4) + 4*sqrt(2)*sqrt(x^6 + x^2)*(x^5 + 2*x^3 + x) + 2*8^(1/4)*(x^6 + x^2)^(3/4)*(x^4 - 6*x^2 + 1) + x)/(x^9 - 28*x^7 + 6*x^5 - 28*x^3 + x)) - 3/64*8^(3/4)*arctan(-(x^9 + 4*x^7 + 6*x^5 + 4*x^3 - 8^(3 /4)*(3*x^6 - 2*x^4 + 3*x^2)*(x^6 + x^2)^(1/4) + 4*sqrt(2)*sqrt(x^6 + x^2)* (x^5 + 2*x^3 + x) - 2*8^(1/4)*(x^6 + x^2)^(3/4)*(x^4 - 6*x^2 + 1) + x)/(x^ 9 - 28*x^7 + 6*x^5 - 28*x^3 + 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)) - 1/32*2 ^(3/4)*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(x^5 + 2*x^3 + x) + 4*2^(1/4)*sqrt(x^6 + x^2)*x + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 1 /32*2^(3/4)*log(-(4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - 2^(3/4)*(x^5 + 2*x^3 + x) - 4*2^(1/4)*sqrt(x^6 + x^2)*x + 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x) ) - 3/128*8^(3/4)*log(2*(4*8^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 8^(3/4)*(x^6 + x^2)^(3/4) + sqrt(2)*(x^5 + 2*x^3 + x) + 8*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x)) + 3/128*8^(3/4)*log(-2*(4*8^(1/4)*(x^6 + x^2)^(1/4)*x^2 + 8^(3/4)*( x^6 + x^2)^(3/4) - sqrt(2)*(x^5 + 2*x^3 + x) - 8*sqrt(x^6 + x^2)*x)/(x^5 + 2*x^3 + x))
\[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {x^{4} - x^{2} + 1}{\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \] Input:
integrate((x**4-x**2+1)/(x**4-1)/(x**6+x**2)**(1/4),x)
Output:
Integral((x**4 - x**2 + 1)/((x**2*(x**4 + 1))**(1/4)*(x - 1)*(x + 1)*(x**2 + 1)), x)
\[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - x^{2} + 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \] Input:
integrate((x^4-x^2+1)/(x^4-1)/(x^6+x^2)^(1/4),x, algorithm="maxima")
Output:
integrate((x^4 - x^2 + 1)/((x^6 + x^2)^(1/4)*(x^4 - 1)), x)
\[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int { \frac {x^{4} - x^{2} + 1}{{\left (x^{6} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \] Input:
integrate((x^4-x^2+1)/(x^4-1)/(x^6+x^2)^(1/4),x, algorithm="giac")
Output:
integrate((x^4 - x^2 + 1)/((x^6 + x^2)^(1/4)*(x^4 - 1)), x)
Timed out. \[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {x^4-x^2+1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^4-1\right )} \,d x \] Input:
int((x^4 - x^2 + 1)/((x^2 + x^6)^(1/4)*(x^4 - 1)),x)
Output:
int((x^4 - x^2 + 1)/((x^2 + x^6)^(1/4)*(x^4 - 1)), x)
\[ \int \frac {1-x^2+x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^6}} \, dx=\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x^{3}}{x^{8}-1}d x -\left (\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}} x}{x^{8}-1}d x \right )+\int \frac {\sqrt {x}\, \left (x^{4}+1\right )^{\frac {3}{4}}}{x^{9}-x}d x \] Input:
int((x^4-x^2+1)/(x^4-1)/(x^6+x^2)^(1/4),x)
Output:
int((sqrt(x)*(x**4 + 1)**(3/4)*x**3)/(x**8 - 1),x) - int((sqrt(x)*(x**4 + 1)**(3/4)*x)/(x**8 - 1),x) + int((sqrt(x)*(x**4 + 1)**(3/4))/(x**9 - x),x)