\(\int \frac {-1+x^3}{x^3 (1+x^3) \sqrt [4]{x+x^4}} \, dx\) [362]

Optimal result

Integrand size = 25, antiderivative size = 30 \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {4 \left (1+7 x^3\right ) \left (x+x^4\right )^{3/4}}{9 x^3 \left (1+x^3\right )} \]

[Out]

4/9*(7*x^3+1)*(x^4+x)^(3/4)/x^3/(x^3+1)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2081, 464, 270} \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {28 x}{9 \sqrt [4]{x^4+x}}+\frac {4}{9 \sqrt [4]{x^4+x} x^2} \]

[In]

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

[Out]

4/(9*x^2*(x + x^4)^(1/4)) + (28*x)/(9*(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 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^{13/4} \left (1+x^3\right )^{5/4}} \, dx}{\sqrt [4]{x+x^4}} \\ & = \frac {4}{9 x^2 \sqrt [4]{x+x^4}}+\frac {\left (7 \sqrt [4]{x} \sqrt [4]{1+x^3}\right ) \int \frac {1}{\sqrt [4]{x} \left (1+x^3\right )^{5/4}} \, dx}{3 \sqrt [4]{x+x^4}} \\ & = \frac {4}{9 x^2 \sqrt [4]{x+x^4}}+\frac {28 x}{9 \sqrt [4]{x+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {4+28 x^3}{9 x^2 \sqrt [4]{x+x^4}} \]

[In]

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

[Out]

(4 + 28*x^3)/(9*x^2*(x + x^4)^(1/4))

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {\frac {28 x^{3}}{9}+\frac {4}{9}}{\left (x^{4}+x \right )^{\frac {1}{4}} x^{2}}\)	\(20\)
pseudoelliptic	\(\frac {\frac {28 x^{3}}{9}+\frac {4}{9}}{\left (x^{4}+x \right )^{\frac {1}{4}} x^{2}}\)	\(20\)
risch	\(\frac {\frac {28 x^{3}}{9}+\frac {4}{9}}{x^{2} {\left (x \left (x^{3}+1\right )\right )}^{\frac {1}{4}}}\)	\(22\)
trager	\(\frac {4 \left (7 x^{3}+1\right ) \left (x^{4}+x \right )^{\frac {3}{4}}}{9 x^{3} \left (x^{3}+1\right )}\)	\(27\)
meijerg	\(\frac {\frac {16 x^{3}}{9}+\frac {4}{9}}{x^{\frac {9}{4}} \left (x^{3}+1\right )^{\frac {1}{4}}}+\frac {4 x^{\frac {3}{4}}}{3 \left (x^{3}+1\right )^{\frac {1}{4}}}\)	\(33\)

[In]

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

[Out]

4/9*(7*x^3+1)/(x^4+x)^(1/4)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {4 \, {\left (x^{4} + x\right )}^{\frac {3}{4}} {\left (7 \, x^{3} + 1\right )}}{9 \, {\left (x^{6} + x^{3}\right )}} \]

[In]

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

[Out]

4/9*(x^4 + x)^(3/4)*(7*x^3 + 1)/(x^6 + x^3)

Sympy [F]

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

[In]

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

[Out]

Integral((x - 1)*(x**2 + x + 1)/(x**3*(x*(x + 1)*(x**2 - x + 1))**(1/4)*(x + 1)*(x**2 - x + 1)), x)

Maxima [F]

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

[In]

integrate((x^3-1)/x^3/(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^3), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.63 \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {4}{9} \, {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {3}{4}} + \frac {8}{3 \, {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

4/9*(1/x^3 + 1)^(3/4) + 8/3/(1/x^3 + 1)^(1/4)

Mupad [B] (verification not implemented)

Time = 4.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {-1+x^3}{x^3 \left (1+x^3\right ) \sqrt [4]{x+x^4}} \, dx=\frac {4\,\left (7\,x^3+1\right )\,{\left (x^4+x\right )}^{3/4}}{9\,x^3\,\left (x^3+1\right )} \]

[In]

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

[Out]

(4*(7*x^3 + 1)*(x + x^4)^(3/4))/(9*x^3*(x^3 + 1))