\(\int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx\) [163]

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5 (-1+x) (5+x)}{6 \sqrt [5]{(-1+x)^4}} \]

[Out]

5/6*(-1+x)*(5+x)/((-1+x)^4)^(1/5)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1694, 15, 45} \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5 (x-1)^2}{6 \sqrt [5]{(x-1)^4}}+\frac {5 (x-1)}{\sqrt [5]{(x-1)^4}} \]

[In]

Int[x/(1 - 4*x + 6*x^2 - 4*x^3 + x^4)^(1/5),x]

[Out]

(5*(-1 + x))/((-1 + x)^4)^(1/5) + (5*(-1 + x)^2)/(6*((-1 + x)^4)^(1/5))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1694

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -d/(4*e) + x)*(a + d^4/(256*e^3)
- b*(d/(8*e)) + (c - 3*(d^2/(8*e)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0]
&& NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1+x}{\sqrt [5]{x^4}} \, dx,x,-1+x\right ) \\ & = \frac {(-1+x)^{4/5} \text {Subst}\left (\int \frac {1+x}{x^{4/5}} \, dx,x,-1+x\right )}{\sqrt [5]{(-1+x)^4}} \\ & = \frac {(-1+x)^{4/5} \text {Subst}\left (\int \left (\frac {1}{x^{4/5}}+\sqrt [5]{x}\right ) \, dx,x,-1+x\right )}{\sqrt [5]{(-1+x)^4}} \\ & = -\frac {5 (1-x)}{\sqrt [5]{(-1+x)^4}}+\frac {5 (1-x)^2}{6 \sqrt [5]{(-1+x)^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5 (-1+x) (5+x)}{6 \sqrt [5]{(-1+x)^4}} \]

[In]

Integrate[x/(1 - 4*x + 6*x^2 - 4*x^3 + x^4)^(1/5),x]

[Out]

(5*(-1 + x)*(5 + x))/(6*((-1 + x)^4)^(1/5))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
risch	\(\frac {5 \left (x -1\right ) \left (x +5\right )}{6 \left (\left (x -1\right )^{4}\right )^{\frac {1}{5}}}\)	\(16\)
gosper	\(\frac {5 \left (x -1\right ) \left (x +5\right )}{6 \left (x^{4}-4 x^{3}+6 x^{2}-4 x +1\right )^{\frac {1}{5}}}\)	\(29\)
trager	\(\frac {5 \left (x +5\right ) \left (x^{4}-4 x^{3}+6 x^{2}-4 x +1\right )^{\frac {4}{5}}}{6 \left (x -1\right )^{3}}\)	\(31\)

[In]

int(x/(x^4-4*x^3+6*x^2-4*x+1)^(1/5),x,method=_RETURNVERBOSE)

[Out]

5/6*(x-1)*(x+5)/((x-1)^4)^(1/5)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (15) = 30\).

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.11 \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5 \, {\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}^{\frac {4}{5}} {\left (x + 5\right )}}{6 \, {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}} \]

[In]

integrate(x/(x^4-4*x^3+6*x^2-4*x+1)^(1/5),x, algorithm="fricas")

[Out]

5/6*(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)^(4/5)*(x + 5)/(x^3 - 3*x^2 + 3*x - 1)

Sympy [F]

\[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\int \frac {x}{\sqrt [5]{\left (x - 1\right )^{4}}}\, dx \]

[In]

integrate(x/(x**4-4*x**3+6*x**2-4*x+1)**(1/5),x)

[Out]

Integral(x/((x - 1)**4)**(1/5), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5 \, {\left (x^{2} + 4 \, x - 5\right )}}{6 \, {\left (x - 1\right )}^{\frac {4}{5}}} \]

[In]

integrate(x/(x^4-4*x^3+6*x^2-4*x+1)^(1/5),x, algorithm="maxima")

[Out]

5/6*(x^2 + 4*x - 5)/(x - 1)^(4/5)

Giac [F]

\[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\int { \frac {x}{{\left (x^{4} - 4 \, x^{3} + 6 \, x^{2} - 4 \, x + 1\right )}^{\frac {1}{5}}} \,d x } \]

[In]

integrate(x/(x^4-4*x^3+6*x^2-4*x+1)^(1/5),x, algorithm="giac")

[Out]

integrate(x/(x^4 - 4*x^3 + 6*x^2 - 4*x + 1)^(1/5), x)

Mupad [B] (verification not implemented)

Time = 4.99 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {x}{\sqrt [5]{1-4 x+6 x^2-4 x^3+x^4}} \, dx=\frac {5\,\left (x+5\right )\,{\left (x^4-4\,x^3+6\,x^2-4\,x+1\right )}^{4/5}}{6\,{\left (x-1\right )}^3} \]

[In]

int(x/(6*x^2 - 4*x - 4*x^3 + x^4 + 1)^(1/5),x)

[Out]

(5*(x + 5)*(6*x^2 - 4*x - 4*x^3 + x^4 + 1)^(4/5))/(6*(x - 1)^3)