\(\int \frac {1+x^3}{x^6 (-1+x^3) \sqrt [4]{-x+x^4}} \, dx\) [458]

Optimal result

Integrand size = 27, antiderivative size = 37 \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 \left (-x+x^4\right )^{3/4} \left (-1-5 x^3+20 x^6\right )}{21 x^6 \left (-1+x^3\right )} \]

[Out]

-4/21*(x^4-x)^(3/4)*(20*x^6-5*x^3-1)/x^6/(x^3-1)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2081, 464, 277, 270} \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {80 x}{21 \sqrt [4]{x^4-x}}+\frac {4}{21 \sqrt [4]{x^4-x} x^5}+\frac {20}{21 \sqrt [4]{x^4-x} x^2} \]

[In]

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

[Out]

4/(21*x^5*(-x + x^4)^(1/4)) + 20/(21*x^2*(-x + x^4)^(1/4)) - (80*x)/(21*(-x + x^4)^(1/4))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \int \frac {1+x^3}{x^{25/4} \left (-1+x^3\right )^{5/4}} \, dx}{\sqrt [4]{-x+x^4}} \\ & = \frac {4}{21 x^5 \sqrt [4]{-x+x^4}}+\frac {\left (15 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \int \frac {1}{x^{13/4} \left (-1+x^3\right )^{5/4}} \, dx}{7 \sqrt [4]{-x+x^4}} \\ & = \frac {4}{21 x^5 \sqrt [4]{-x+x^4}}+\frac {20}{21 x^2 \sqrt [4]{-x+x^4}}+\frac {\left (20 \sqrt [4]{x} \sqrt [4]{-1+x^3}\right ) \int \frac {1}{\sqrt [4]{x} \left (-1+x^3\right )^{5/4}} \, dx}{7 \sqrt [4]{-x+x^4}} \\ & = \frac {4}{21 x^5 \sqrt [4]{-x+x^4}}+\frac {20}{21 x^2 \sqrt [4]{-x+x^4}}-\frac {80 x}{21 \sqrt [4]{-x+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {4+20 x^3-80 x^6}{21 x^5 \sqrt [4]{x \left (-1+x^3\right )}} \]

[In]

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

[Out]

(4 + 20*x^3 - 80*x^6)/(21*x^5*(x*(-1 + x^3))^(1/4))

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73

method	result	size
gosper	\(-\frac {4 \left (20 x^{6}-5 x^{3}-1\right )}{21 \left (x^{4}-x \right )^{\frac {1}{4}} x^{5}}\)	\(27\)
risch	\(-\frac {4 \left (20 x^{6}-5 x^{3}-1\right )}{21 x^{5} {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{4}}}\)	\(27\)
pseudoelliptic	\(\frac {-\frac {80}{21} x^{6}+\frac {20}{21} x^{3}+\frac {4}{21}}{\left (x^{4}-x \right )^{\frac {1}{4}} x^{5}}\)	\(27\)
trager	\(-\frac {4 \left (x^{4}-x \right )^{\frac {3}{4}} \left (20 x^{6}-5 x^{3}-1\right )}{21 x^{6} \left (x^{3}-1\right )}\)	\(34\)
meijerg	\(\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (-32 x^{6}+8 x^{3}+3\right ) \left (-x^{3}+1\right )^{\frac {3}{4}}}{21 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}} \left (-3 x^{3}+3\right )}+\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (-4 x^{3}+1\right )}{9 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}} \left (-x^{3}+1\right )^{\frac {1}{4}}}\)	\(94\)

[In]

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

[Out]

-4/21*(20*x^6-5*x^3-1)/(x^4-x)^(1/4)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=-\frac {4 \, {\left (20 \, x^{6} - 5 \, x^{3} - 1\right )} {\left (x^{4} - x\right )}^{\frac {3}{4}}}{21 \, {\left (x^{9} - x^{6}\right )}} \]

[In]

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

[Out]

-4/21*(20*x^6 - 5*x^3 - 1)*(x^4 - x)^(3/4)/(x^9 - x^6)

Sympy [F]

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

[In]

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

[Out]

Integral((x + 1)*(x**2 - x + 1)/(x**6*(x*(x - 1)*(x**2 + x + 1))**(1/4)*(x - 1)*(x**2 + x + 1)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {4}{21} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {7}{4}} - \frac {4}{3} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {3}{4}} - \frac {8}{3 \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

4/21*(-1/x^3 + 1)^(7/4) - 4/3*(-1/x^3 + 1)^(3/4) - 8/3/(-1/x^3 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 5.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {1+x^3}{x^6 \left (-1+x^3\right ) \sqrt [4]{-x+x^4}} \, dx=\frac {4\,{\left (x^4-x\right )}^{3/4}\,\left (-20\,x^6+5\,x^3+1\right )}{21\,x^6\,\left (x^3-1\right )} \]

[In]

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

[Out]

(4*(x^4 - x)^(3/4)*(5*x^3 - 20*x^6 + 1))/(21*x^6*(x^3 - 1))