Integrand size = 19, antiderivative size = 69 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {1}{3} 2^{3/4} \arctan \left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right )+\frac {1}{3} 2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \left (-x+x^4\right )^{3/4}}{-1+x^3}\right ) \] Output:
1/3*2^(3/4)*arctan(2^(1/4)*(x^4-x)^(3/4)/(x^3-1))+1/3*2^(3/4)*arctanh(2^(1 /4)*(x^4-x)^(3/4)/(x^3-1))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {4 x \sqrt [4]{1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {2 x^3}{1+x^3}\right )}{3 \sqrt [4]{x \left (-1+x^3\right )} \sqrt [4]{1+x^3}} \] Input:
Integrate[1/((1 + x^3)*(-x + x^4)^(1/4)),x]
Output:
(4*x*(1 - x^3)^(1/4)*Hypergeometric2F1[1/4, 1/4, 5/4, (2*x^3)/(1 + x^3)])/ (3*(x*(-1 + x^3))^(1/4)*(1 + x^3)^(1/4))
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2467, 966, 965, 902, 756, 216, 219}
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.
\(\displaystyle \int \frac {1}{\left (x^3+1\right ) \sqrt [4]{x^4-x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{x^3-1} \int \frac {1}{\sqrt [4]{x} \sqrt [4]{x^3-1} \left (x^3+1\right )}dx}{\sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 966 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \int \frac {\sqrt {x}}{\sqrt [4]{x^3-1} \left (x^3+1\right )}d\sqrt [4]{x}}{\sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \int \frac {1}{\sqrt [4]{x-1} (x+1)}dx^{3/4}}{3 \sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 902 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \int \frac {1}{1-2 x}d\frac {x^{3/4}}{\sqrt [4]{x-1}}}{3 \sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{x-1}}+\frac {1}{2} \int \frac {1}{\sqrt {2} \sqrt {x}+1}d\frac {x^{3/4}}{\sqrt [4]{x-1}}\right )}{3 \sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{x-1}}+\frac {\arctan \left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x-1}}\right )}{2 \sqrt [4]{2}}\right )}{3 \sqrt [4]{x^4-x}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{x^3-1} \left (\frac {\arctan \left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x-1}}\right )}{2 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x^{3/4}}{\sqrt [4]{x-1}}\right )}{2 \sqrt [4]{2}}\right )}{3 \sqrt [4]{x^4-x}}\) |
Input:
Int[1/((1 + x^3)*(-x + x^4)^(1/4)),x]
Output:
(4*x^(1/4)*(-1 + x^3)^(1/4)*(ArcTan[(2^(1/4)*x^(3/4))/(-1 + x)^(1/4)]/(2*2 ^(1/4)) + ArcTanh[(2^(1/4)*x^(3/4))/(-1 + x)^(1/4)]/(2*2^(1/4))))/(3*(-x + x^4)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Su bst[Int[1/(c - (b*c - a*d)*x^n), x], x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b , c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*( m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*x)^( 1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && FractionQ[m] && IntegerQ[p]
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]
Time = 4.56 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {3}{4}} \left (-2 \arctan \left (\frac {\left (x^{4}-x \right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right )+\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{4}-x \right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{4}-x \right )^{\frac {1}{4}}}\right )\right )}{6}\) | \(66\) |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \ln \left (-\frac {\sqrt {x^{4}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}-x \right )^{\frac {1}{4}} x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right ) x^{3}+4 \left (x^{4}-x \right )^{\frac {3}{4}}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2}\right )}{\left (1+x \right ) \left (x^{2}-x +1\right )}\right )}{6}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) \ln \left (\frac {\sqrt {x^{4}-x}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{3} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )^{2} \left (x^{4}-x \right )^{\frac {1}{4}} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right ) x^{3}-4 \left (x^{4}-x \right )^{\frac {3}{4}}-\operatorname {RootOf}\left (\textit {\_Z}^{4}-8\right )}{\left (1+x \right ) \left (x^{2}-x +1\right )}\right )}{6}\) | \(229\) |
Input:
int(1/(x^3+1)/(x^4-x)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/6*2^(3/4)*(-2*arctan(1/2*(x^4-x)^(1/4)/x*2^(3/4))+ln((-2^(1/4)*x-(x^4-x) ^(1/4))/(2^(1/4)*x-(x^4-x)^(1/4))))
Leaf count of result is larger than twice the leaf count of optimal. 191 vs. \(2 (53) = 106\).
Time = 0.96 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.77 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {1}{6} \cdot 2^{\frac {3}{4}} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (x^{4} - x\right )}^{\frac {1}{4}} {\left (x^{3} - 1\right )} - 2 \cdot 2^{\frac {1}{4}} {\left (x^{4} - x\right )}^{\frac {3}{4}} x}{4 \, {\left (x^{4} - x\right )}}\right ) + \frac {1}{12} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{3} - 1\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x + 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) - \frac {1}{12} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} - x\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{3} - 1\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} - x} x + 4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}}}{x^{3} + 1}\right ) \] Input:
integrate(1/(x^3+1)/(x^4-x)^(1/4),x, algorithm="fricas")
Output:
1/6*2^(3/4)*arctan(-1/4*(2^(3/4)*(x^4 - x)^(1/4)*(x^3 - 1) - 2*2^(1/4)*(x^ 4 - x)^(3/4)*x)/(x^4 - x)) + 1/12*2^(3/4)*log((4*sqrt(2)*(x^4 - x)^(1/4)*x ^2 + 2^(3/4)*(3*x^3 - 1) + 4*2^(1/4)*sqrt(x^4 - x)*x + 4*(x^4 - x)^(3/4))/ (x^3 + 1)) - 1/12*2^(3/4)*log((4*sqrt(2)*(x^4 - x)^(1/4)*x^2 - 2^(3/4)*(3* x^3 - 1) - 4*2^(1/4)*sqrt(x^4 - x)*x + 4*(x^4 - x)^(3/4))/(x^3 + 1))
\[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int \frac {1}{\sqrt [4]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \] Input:
integrate(1/(x**3+1)/(x**4-x)**(1/4),x)
Output:
Integral(1/((x*(x - 1)*(x**2 + x + 1))**(1/4)*(x + 1)*(x**2 - x + 1)), x)
\[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int { \frac {1}{{\left (x^{4} - x\right )}^{\frac {1}{4}} {\left (x^{3} + 1\right )}} \,d x } \] Input:
integrate(1/(x^3+1)/(x^4-x)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((x^4 - x)^(1/4)*(x^3 + 1)), x)
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {1}{3} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{6} \cdot 2^{\frac {3}{4}} \log \left ({\left | -2^{\frac {1}{4}} + {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}} \right |}\right ) \] Input:
integrate(1/(x^3+1)/(x^4-x)^(1/4),x, algorithm="giac")
Output:
-1/3*2^(3/4)*arctan(1/2*2^(3/4)*(-1/x^3 + 1)^(1/4)) + 1/6*2^(3/4)*log(2^(1 /4) + (-1/x^3 + 1)^(1/4)) - 1/6*2^(3/4)*log(abs(-2^(1/4) + (-1/x^3 + 1)^(1 /4)))
Timed out. \[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int \frac {1}{{\left (x^4-x\right )}^{1/4}\,\left (x^3+1\right )} \,d x \] Input:
int(1/((x^4 - x)^(1/4)*(x^3 + 1)),x)
Output:
int(1/((x^4 - x)^(1/4)*(x^3 + 1)), x)
\[ \int \frac {1}{\left (1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\int \frac {1}{x^{\frac {13}{4}} \left (x^{3}-1\right )^{\frac {1}{4}}+x^{\frac {1}{4}} \left (x^{3}-1\right )^{\frac {1}{4}}}d x \] Input:
int(1/(x^3+1)/(x^4-x)^(1/4),x)
Output:
int(1/(x**(1/4)*(x**3 - 1)**(1/4)*x**3 + x**(1/4)*(x**3 - 1)**(1/4)),x)