Integrand size = 29, antiderivative size = 187 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {3 \left (x^2+x^6\right )^{3/4}}{2 x \left (1+x^4\right )}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{8 \sqrt [4]{2}}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{8\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{8 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{8\ 2^{3/4}} \]
-3/2*(x^6+x^2)^(3/4)/x/(x^4+1)-1/16*arctan(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3 /4)+1/16*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)-1/16*arctanh(2^(1/4)*x/(x^6+x^2)^(1/4))*2^(3/4)-1/16*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 = 1.32 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {\sqrt {x} \left (24 \sqrt {x}+2^{3/4} \sqrt [4]{1+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )-\sqrt [4]{2} \sqrt [4]{1+x^4} \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+2^{3/4} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\sqrt [4]{2} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{16 \sqrt [4]{x^2+x^6}} \]
-1/16*(Sqrt[x]*(24*Sqrt[x] + 2^(3/4)*(1 + x^4)^(1/4)*ArcTan[(2^(1/4)*Sqrt[ x])/(1 + x^4)^(1/4)] - 2^(1/4)*(1 + x^4)^(1/4)*ArcTan[(2^(3/4)*Sqrt[x]*(1 + x^4)^(1/4))/(Sqrt[2]*x - Sqrt[1 + x^4])] + 2^(3/4)*(1 + x^4)^(1/4)*ArcTa nh[(2^(1/4)*Sqrt[x])/(1 + x^4)^(1/4)] + 2^(1/4)*(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])]))/(x^2 + x^6)^(1/4)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.56 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.41, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2467, 25, 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.
\(\displaystyle \int \frac {x^8-x^4+1}{\sqrt [4]{x^6+x^2} \left (x^8-1\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x} \sqrt [4]{x^4+1} \int -\frac {x^8-x^4+1}{\sqrt {x} \sqrt [4]{x^4+1} \left (1-x^8\right )}dx}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4+1} \int \frac {x^8-x^4+1}{\sqrt {x} \sqrt [4]{x^4+1} \left (1-x^8\right )}dx}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^4+1} \int \frac {x^8-x^4+1}{\sqrt {x} \left (1-x^4\right ) \left (x^4+1\right )^{5/4}}dx}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \frac {x^8-x^4+1}{\left (1-x^4\right ) \left (x^4+1\right )^{5/4}}d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \int \left (\frac {1}{\left (1-x^4\right ) \left (x^4+1\right )^{5/4}}-\frac {x^4}{\left (x^4+1\right )^{5/4}}\right )d\sqrt {x}}{\sqrt [4]{x^6+x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^4+1} \left (\sqrt {x} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^4,-x^4\right )-\frac {1}{9} x^{9/2} \operatorname {Hypergeometric2F1}\left (\frac {9}{8},\frac {5}{4},\frac {17}{8},-x^4\right )\right )}{\sqrt [4]{x^6+x^2}}\) |
(-2*Sqrt[x]*(1 + x^4)^(1/4)*(Sqrt[x]*AppellF1[1/8, 1, 5/4, 9/8, x^4, -x^4] - (x^(9/2)*Hypergeometric2F1[9/8, 5/4, 17/8, -x^4])/9))/(x^2 + x^6)^(1/4)
3.24.64.3.1 Defintions of rubi rules used
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]))
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 = 55.08 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.50
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-48 x}{32 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\) | \(281\) |
risch | \(\text {Expression too large to display}\) | \(648\) |
trager | \(\text {Expression too large to display}\) | \(656\) |
1/32*(2*arctan(1/2*(x^2*(x^4+1))^(1/4)/x*2^(3/4))*2^(3/4)*(x^2*(x^4+1))^(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 ^(3/4)*(x^2*(x^4+1))^(1/4)+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^(1/4)*(x^2*(x^4+1))^(1/4)+2*arctan((2^(1/4)*(x^2*(x^4+1))^(1 /4)+x)/x)*2^(1/4)*(x^2*(x^4+1))^(1/4)+2*arctan((2^(1/4)*(x^2*(x^4+1))^(1/4 )-x)/x)*2^(1/4)*(x^2*(x^4+1))^(1/4)-48*x)/(x^2*(x^4+1))^(1/4)
Result contains complex when optimal does not.
Time = 32.53 (sec) , antiderivative size = 733, normalized size of antiderivative = 3.92 \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\text {Too large to display} \]
-1/64*(2^(3/4)*(x^5 + x)*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)) - 2^(3/4)*(x^5 + x)*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)) + 2^(3/4)*(-I*x^5 - I*x)*log((4*sqrt(2)*(x^6 + x^ 2)^(1/4)*x^2 + 2^(3/4)*(I*x^5 + 2*I*x^3 + I*x) - 4*I*2^(1/4)*sqrt(x^6 + x^ 2)*x - 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) + 2^(3/4)*(I*x^5 + I*x)*log ((4*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + 2^(3/4)*(-I*x^5 - 2*I*x^3 - I*x) + 4*I *2^(1/4)*sqrt(x^6 + x^2)*x - 4*(x^6 + x^2)^(3/4))/(x^5 - 2*x^3 + x)) - 2^( 1/4)*((I + 1)*x^5 + (I + 1)*x)*log(-2*(4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - (2*I - 2)*2^(3/4)*sqrt(x^6 + x^2)*x - 2^(1/4)*(-(I + 1)*x^5 + (2*I + 2)*x ^3 - (I + 1)*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4)*(-(I + 1)*x^5 - (I + 1)*x)*log(-2*(4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + (2*I - 2) *2^(3/4)*sqrt(x^6 + x^2)*x - 2^(1/4)*((I + 1)*x^5 - (2*I + 2)*x^3 + (I + 1 )*x) - 4*(x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4)*(-(I - 1)*x^5 - ( I - 1)*x)*log(-2*(-4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 + (2*I + 2)*2^(3/4)*s qrt(x^6 + x^2)*x - 2^(1/4)*((I - 1)*x^5 - (2*I - 2)*x^3 + (I - 1)*x) - 4*( x^6 + x^2)^(3/4))/(x^5 + 2*x^3 + x)) - 2^(1/4)*((I - 1)*x^5 + (I - 1)*x)*l og(-2*(-4*I*sqrt(2)*(x^6 + x^2)^(1/4)*x^2 - (2*I + 2)*2^(3/4)*sqrt(x^6 + x ^2)*x - 2^(1/4)*(-(I - 1)*x^5 + (2*I - 2)*x^3 - (I - 1)*x) - 4*(x^6 + x...
\[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int \frac {x^{8} - x^{4} + 1}{\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \]
Integral((x**8 - x**4 + 1)/((x**2*(x**4 + 1))**(1/4)*(x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)), x)
\[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} - x^{4} + 1}{{\left (x^{8} - 1\right )} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} - x^{4} + 1}{{\left (x^{8} - 1\right )} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {1-x^4+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int \frac {x^8-x^4+1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^8-1\right )} \,d x \]