Integrand size = 20, antiderivative size = 121 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{\sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{\sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{2 \sqrt [3]{2}} \] Output:
-1/2*3^(1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)+1/2*ln( -2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/3)-1/4*ln(2*x^2+2^(2/3)*x*(x^4+x^2)^(1/ 3)+2^(1/3)*(x^4+x^2)^(2/3))*2^(2/3)
Time = 0.73 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.21 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {x^{2/3} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )-2 \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+\log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \] Input:
Integrate[(1 + x)/((-1 + x)*(x^2 + x^4)^(1/3)),x]
Output:
-1/2*(x^(2/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3))] - 2*Log[-2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] + Log[2*x^(2/3) + 2^(2/3)*x^(1/3)*(1 + x^2)^(1/3) + 2^(1/3)*(1 + x^2)^(2/ 3)]))/(2^(1/3)*(x^2 + x^4)^(1/3))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.87 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.64, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int \frac {x+1}{(x-1) \sqrt [3]{x^4+x^2}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{2/3} \sqrt [3]{x^2+1} \int -\frac {x+1}{(1-x) x^{2/3} \sqrt [3]{x^2+1}}dx}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {x^{2/3} \sqrt [3]{x^2+1} \int \frac {x+1}{(1-x) x^{2/3} \sqrt [3]{x^2+1}}dx}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \frac {x+1}{(1-x) \sqrt [3]{x^2+1}}d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \left (\frac {2}{(1-x) \sqrt [3]{x^2+1}}-\frac {1}{\sqrt [3]{x^2+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \left (2 \sqrt [3]{x} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )+\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )+\frac {\log \left (\left (1-x^{2/3}\right )^2 \left (x^{2/3}+1\right )\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\frac {2^{2/3} \left (x^{2/3}+1\right )^2}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{3 \sqrt [3]{2}}-\frac {\log \left (x^{2/3}-2^{2/3} \sqrt [3]{x^2+1}+1\right )}{4 \sqrt [3]{2}}\right )}{\sqrt [3]{x^4+x^2}}\) |
Input:
Int[(1 + x)/((-1 + x)*(x^2 + x^4)^(1/3)),x]
Output:
(-3*x^(2/3)*(1 + x^2)^(1/3)*(2*x^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^2, -x^ 2] + ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]]/(2^(1 /3)*Sqrt[3]) + ArcTan[(1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3 ]]/(2*2^(1/3)*Sqrt[3]) - x^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -x^2] + Log[(1 - x^(2/3))^2*(1 + x^(2/3))]/(12*2^(1/3)) + Log[1 + (2^(2/3)*(1 + x^ (2/3))^2)/(1 + x^2)^(2/3) - (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3)]/(6*2^ (1/3)) - Log[1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3)]/(3*2^(1/3)) - Lo g[1 + x^(2/3) - 2^(2/3)*(1 + x^2)^(1/3)]/(4*2^(1/3))))/(x^2 + x^4)^(1/3)
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 = 7.92 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )+2 \ln \left (\frac {-2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{4}\) | \(105\) |
trager | \(\text {Expression too large to display}\) | \(971\) |
Input:
int((1+x)/(-1+x)/(x^4+x^2)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/4*2^(2/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/ x)+2*ln((-2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x)-ln((2^(2/3)*x^2+2^(1/3)*(x^2*( x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2))
Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (94) = 188\).
Time = 2.26 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.54 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {1}{6} \cdot 4^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {3 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (x^{4} + 2 \, x^{3} - 6 \, x^{2} + 2 \, x + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} + 6 \cdot 4^{\frac {1}{3}} \sqrt {3} {\left (x^{5} + 14 \, x^{4} + 6 \, x^{3} + 14 \, x^{2} + x\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{7} + 30 \, x^{6} + 51 \, x^{5} + 52 \, x^{4} + 51 \, x^{3} + 30 \, x^{2} + x\right )}}{3 \, {\left (x^{7} - 6 \, x^{6} - 93 \, x^{5} - 20 \, x^{4} - 93 \, x^{3} - 6 \, x^{2} + x\right )}}\right ) - \frac {1}{12} \cdot 4^{\frac {1}{3}} \log \left (\frac {6 \cdot 4^{\frac {1}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} {\left (x^{2} + 4 \, x + 1\right )} + 4^{\frac {2}{3}} {\left (x^{5} + 14 \, x^{4} + 6 \, x^{3} + 14 \, x^{2} + x\right )} + 24 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{3} + x^{2} + x\right )}}{x^{5} - 4 \, x^{4} + 6 \, x^{3} - 4 \, x^{2} + x}\right ) + \frac {1}{6} \cdot 4^{\frac {1}{3}} \log \left (-\frac {3 \cdot 4^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x + 4^{\frac {1}{3}} {\left (x^{3} - 2 \, x^{2} + x\right )} - 6 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{x^{3} - 2 \, x^{2} + x}\right ) \] Input:
integrate((1+x)/(-1+x)/(x^4+x^2)^(1/3),x, algorithm="fricas")
Output:
-1/6*4^(1/3)*sqrt(3)*arctan(1/3*(3*4^(2/3)*sqrt(3)*(x^4 + 2*x^3 - 6*x^2 + 2*x + 1)*(x^4 + x^2)^(2/3) + 6*4^(1/3)*sqrt(3)*(x^5 + 14*x^4 + 6*x^3 + 14* x^2 + x)*(x^4 + x^2)^(1/3) + sqrt(3)*(x^7 + 30*x^6 + 51*x^5 + 52*x^4 + 51* x^3 + 30*x^2 + x))/(x^7 - 6*x^6 - 93*x^5 - 20*x^4 - 93*x^3 - 6*x^2 + x)) - 1/12*4^(1/3)*log((6*4^(1/3)*(x^4 + x^2)^(2/3)*(x^2 + 4*x + 1) + 4^(2/3)*( x^5 + 14*x^4 + 6*x^3 + 14*x^2 + x) + 24*(x^4 + x^2)^(1/3)*(x^3 + x^2 + x)) /(x^5 - 4*x^4 + 6*x^3 - 4*x^2 + x)) + 1/6*4^(1/3)*log(-(3*4^(2/3)*(x^4 + x ^2)^(1/3)*x + 4^(1/3)*(x^3 - 2*x^2 + x) - 6*(x^4 + x^2)^(2/3))/(x^3 - 2*x^ 2 + x))
\[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x + 1}{\sqrt [3]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right )}\, dx \] Input:
integrate((1+x)/(-1+x)/(x**4+x**2)**(1/3),x)
Output:
Integral((x + 1)/((x**2*(x**2 + 1))**(1/3)*(x - 1)), x)
\[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \] Input:
integrate((1+x)/(-1+x)/(x^4+x^2)^(1/3),x, algorithm="maxima")
Output:
integrate((x + 1)/((x^4 + x^2)^(1/3)*(x - 1)), x)
\[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int { \frac {x + 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \] Input:
integrate((1+x)/(-1+x)/(x^4+x^2)^(1/3),x, algorithm="giac")
Output:
integrate((x + 1)/((x^4 + x^2)^(1/3)*(x - 1)), x)
Timed out. \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x+1}{{\left (x^4+x^2\right )}^{1/3}\,\left (x-1\right )} \,d x \] Input:
int((x + 1)/((x^2 + x^4)^(1/3)*(x - 1)),x)
Output:
int((x + 1)/((x^2 + x^4)^(1/3)*(x - 1)), x)
\[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x}{x^{\frac {5}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}-x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x +\int \frac {1}{x^{\frac {5}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}-x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x \] Input:
int((1+x)/(-1+x)/(x^4+x^2)^(1/3),x)
Output:
int(x/(x**(2/3)*(x**2 + 1)**(1/3)*x - x**(2/3)*(x**2 + 1)**(1/3)),x) + int (1/(x**(2/3)*(x**2 + 1)**(1/3)*x - x**(2/3)*(x**2 + 1)**(1/3)),x)