\(\int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx\) [365]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2081, 79, 37} \[ \int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx=\frac {4}{3 \sqrt [4]{x^4-x^3}}-\frac {28 x}{3 \sqrt [4]{x^4-x^3}} \]

[In]

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

[Out]

4/(3*(-x^3 + x^4)^(1/4)) - (28*x)/(3*(-x^3 + x^4)^(1/4))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx=-\frac {4 (-1+7 x)}{3 \sqrt [4]{(-1+x) x^3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.57

method	result	size
risch	\(-\frac {4 \left (-1+7 x \right )}{3 \left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}\)	\(17\)
pseudoelliptic	\(\frac {-\frac {28 x}{3}+\frac {4}{3}}{\left (x^{3} \left (x -1\right )\right )^{\frac {1}{4}}}\)	\(17\)
gosper	\(-\frac {4 \left (-1+7 x \right )}{3 \left (x^{4}-x^{3}\right )^{\frac {1}{4}}}\)	\(19\)
trager	\(-\frac {4 \left (-1+7 x \right ) \left (x^{4}-x^{3}\right )^{\frac {3}{4}}}{3 \left (x -1\right ) x^{3}}\)	\(27\)
meijerg	\(\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} \left (-4 x +1\right )}{3 \operatorname {signum}\left (x -1\right )^{\frac {1}{4}} x^{\frac {3}{4}} \left (1-x \right )^{\frac {1}{4}}}-\frac {4 \left (-\operatorname {signum}\left (x -1\right )\right )^{\frac {1}{4}} x^{\frac {1}{4}}}{\operatorname {signum}\left (x -1\right )^{\frac {1}{4}} \left (1-x \right )^{\frac {1}{4}}}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60 \[ \int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx=-\frac {4 \, {\left (7 \, x - 1\right )}}{3 \, {\left (x^{4} - x^{3}\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

-4/3*(7*x - 1)/(x^4 - x^3)^(1/4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx=-\frac {4}{3} \, {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{4}} - \frac {8}{{\left (-\frac {1}{x} + 1\right )}^{\frac {1}{4}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.60 \[ \int \frac {1+x}{(-1+x) x \sqrt [4]{-x^3+x^4}} \, dx=-\frac {28\,x-4}{3\,{\left (x^4-x^3\right )}^{1/4}} \]

[In]

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

[Out]

-(28*x - 4)/(3*(x^4 - x^3)^(1/4))