Integrand size = 26, antiderivative size = 112 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {3 \left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}+2 \arctan \left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3 \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}+2 \text {arctanh}\left (\frac {x}{\sqrt [4]{x^2+x^4}}\right )-\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}} \]
-3*(x^4+x^2)^(3/4)/x/(x^2+1)+2*arctan(x/(x^4+x^2)^(1/4))-3/4*arctan(2^(1/4 )*x/(x^4+x^2)^(1/4))*2^(3/4)+2*arctanh(x/(x^4+x^2)^(1/4))-3/4*arctanh(2^(1 /4)*x/(x^4+x^2)^(1/4))*2^(3/4)
Time = 0.35 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.39 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {\sqrt {x} \left (-12 \sqrt {x}+8 \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-3\ 2^{3/4} \sqrt [4]{1+x^2} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )+8 \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )-3\ 2^{3/4} \sqrt [4]{1+x^2} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )\right )}{4 \sqrt [4]{x^2+x^4}} \]
(Sqrt[x]*(-12*Sqrt[x] + 8*(1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(1 + x^2)^(1/4)] - 3*2^(3/4)*(1 + x^2)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)] + 8* (1 + x^2)^(1/4)*ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)] - 3*2^(3/4)*(1 + x^2)^(1/ 4)*ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)]))/(4*(x^2 + x^4)^(1/4))
Time = 0.66 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.
| \(\displaystyle \int \frac {2 x^4+1}{\left (x^4-1\right ) \sqrt [4]{x^4+x^2}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt [4]{x^2+1} \int -\frac {2 x^4+1}{\sqrt {x} \sqrt [4]{x^2+1} \left (1-x^4\right )}dx}{\sqrt [4]{x^4+x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^2+1} \int \frac {2 x^4+1}{\sqrt {x} \sqrt [4]{x^2+1} \left (1-x^4\right )}dx}{\sqrt [4]{x^4+x^2}}\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{x^2+1} \int \frac {2 x^4+1}{\sqrt {x} \left (1-x^2\right ) \left (x^2+1\right )^{5/4}}dx}{\sqrt [4]{x^4+x^2}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^2+1} \int \frac {2 x^4+1}{\left (1-x^2\right ) \left (x^2+1\right )^{5/4}}d\sqrt {x}}{\sqrt [4]{x^4+x^2}}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^2+1} \int \left (-\frac {2 x^2}{\left (x^2+1\right )^{5/4}}+\frac {3}{\left (1-x^2\right ) \left (x^2+1\right )^{5/4}}-\frac {2}{\left (x^2+1\right )^{5/4}}\right )d\sqrt {x}}{\sqrt [4]{x^4+x^2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sqrt {x} \sqrt [4]{x^2+1} \left (-\arctan \left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )+\frac {3 \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{4 \sqrt [4]{2}}-\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2+1}}\right )+\frac {3 \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{4 \sqrt [4]{2}}+\frac {3 \sqrt {x}}{2 \sqrt [4]{x^2+1}}\right )}{\sqrt [4]{x^4+x^2}}\) |
(-2*Sqrt[x]*(1 + x^2)^(1/4)*((3*Sqrt[x])/(2*(1 + x^2)^(1/4)) - ArcTan[Sqrt [x]/(1 + x^2)^(1/4)] + (3*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(4*2^ (1/4)) - ArcTanh[Sqrt[x]/(1 + x^2)^(1/4)] + (3*ArcTanh[(2^(1/4)*Sqrt[x])/( 1 + x^2)^(1/4)])/(4*2^(1/4))))/(x^2 + x^4)^(1/4)
3.17.56.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(90)=180\).
Time = 2.36 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.84
| method | result | size |
| pseudoelliptic | \(\frac {6 \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-3 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+8 \ln \left (\frac {x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-8 \ln \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-16 \arctan \left (\frac {\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{x}\right ) \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-24 x}{8 \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\) | \(206\) |
1/8*(6*arctan(1/2*(x^2*(x^2+1))^(1/4)/x*2^(3/4))*2^(3/4)*(x^2*(x^2+1))^(1/ 4)-3*ln((-2^(1/4)*x-(x^2*(x^2+1))^(1/4))/(2^(1/4)*x-(x^2*(x^2+1))^(1/4)))* 2^(3/4)*(x^2*(x^2+1))^(1/4)+8*ln((x+(x^2*(x^2+1))^(1/4))/x)*(x^2*(x^2+1))^ (1/4)-8*ln(((x^2*(x^2+1))^(1/4)-x)/x)*(x^2*(x^2+1))^(1/4)-16*arctan((x^2*( x^2+1))^(1/4)/x)*(x^2*(x^2+1))^(1/4)-24*x)/(x^2*(x^2+1))^(1/4)
Result contains complex when optimal does not.
Time = 10.41 (sec) , antiderivative size = 434, normalized size of antiderivative = 3.88 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=-\frac {3 \cdot 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 3 \cdot 2^{\frac {3}{4}} {\left (x^{3} + x\right )} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} + x\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x + 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 3 \cdot 2^{\frac {3}{4}} {\left (i \, x^{3} + i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 i \, x^{3} + i \, x\right )} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) + 3 \cdot 2^{\frac {3}{4}} {\left (-i \, x^{3} - i \, x\right )} \log \left (-\frac {4 \, \sqrt {2} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (-3 i \, x^{3} - i \, x\right )} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + x^{2}} x - 4 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x^{3} - x}\right ) - 16 \, {\left (x^{3} + x\right )} \arctan \left (\frac {2 \, {\left ({\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}\right )}}{x}\right ) - 16 \, {\left (x^{3} + x\right )} \log \left (\frac {2 \, x^{3} + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} x^{2} + 2 \, \sqrt {x^{4} + x^{2}} x + x + 2 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{x}\right ) + 48 \, {\left (x^{4} + x^{2}\right )}^{\frac {3}{4}}}{16 \, {\left (x^{3} + x\right )}} \]
-1/16*(3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 + 2^(3/4)* (3*x^3 + x) + 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x) ) - 3*2^(3/4)*(x^3 + x)*log((4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(3* x^3 + x) - 4*2^(1/4)*sqrt(x^4 + x^2)*x + 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 3*2^(3/4)*(I*x^3 + I*x)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*( 3*I*x^3 + I*x) + 4*I*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^(3/4))/(x^3 - x)) + 3*2^(3/4)*(-I*x^3 - I*x)*log(-(4*sqrt(2)*(x^4 + x^2)^(1/4)*x^2 - 2^(3/4)*(-3*I*x^3 - I*x) - 4*I*2^(1/4)*sqrt(x^4 + x^2)*x - 4*(x^4 + x^2)^( 3/4))/(x^3 - x)) - 16*(x^3 + x)*arctan(2*((x^4 + x^2)^(1/4)*x^2 + (x^4 + x ^2)^(3/4))/x) - 16*(x^3 + x)*log((2*x^3 + 2*(x^4 + x^2)^(1/4)*x^2 + 2*sqrt (x^4 + x^2)*x + x + 2*(x^4 + x^2)^(3/4))/x) + 48*(x^4 + x^2)^(3/4))/(x^3 + x)
\[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\int \frac {2 x^{4} + 1}{\sqrt [4]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}\, dx \]
\[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\int { \frac {2 \, x^{4} + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - 1\right )}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\frac {3}{4} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {3}{8} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {3}{8} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) - \frac {3}{{\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}} - 2 \, \arctan \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) - \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
3/4*2^(3/4)*arctan(1/2*2^(3/4)*(1/x^2 + 1)^(1/4)) - 3/8*2^(3/4)*log(2^(1/4 ) + (1/x^2 + 1)^(1/4)) + 3/8*2^(3/4)*log(abs(-2^(1/4) + (1/x^2 + 1)^(1/4)) ) - 3/(1/x^2 + 1)^(1/4) - 2*arctan((1/x^2 + 1)^(1/4)) + log((1/x^2 + 1)^(1 /4) + 1) - log((1/x^2 + 1)^(1/4) - 1)
Timed out. \[ \int \frac {1+2 x^4}{\left (-1+x^4\right ) \sqrt [4]{x^2+x^4}} \, dx=\int \frac {2\,x^4+1}{{\left (x^4+x^2\right )}^{1/4}\,\left (x^4-1\right )} \,d x \]