\(\int \frac {(3+x^4) (-1-x^3+x^4)}{x^6 \sqrt [4]{-x+x^5}} \, dx\) [366]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 30 \[ \int \frac {\left (3+x^4\right ) \left (-1-x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \left (-3-7 x^3+3 x^4\right ) \left (-x+x^5\right )^{3/4}}{21 x^6} \]

[Out]

4/21*(3*x^4-7*x^3-3)*(x^5-x)^(3/4)/x^6

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83, number of steps used = 21, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2077, 2050, 2057, 372, 371} \[ \int \frac {\left (3+x^4\right ) \left (-1-x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=-\frac {4 \left (x^5-x\right )^{3/4}}{7 x^6}-\frac {4 \left (x^5-x\right )^{3/4}}{3 x^3}+\frac {4 \left (x^5-x\right )^{3/4}}{7 x^2} \]

[In]

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

[Out]

(-4*(-x + x^5)^(3/4))/(7*x^6) - (4*(-x + x^5)^(3/4))/(3*x^3) + (4*(-x + x^5)^(3/4))/(7*x^2)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2077

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(c*x)
^m*Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !In
tegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {3}{x^6 \sqrt [4]{-x+x^5}}-\frac {3}{x^3 \sqrt [4]{-x+x^5}}+\frac {2}{x^2 \sqrt [4]{-x+x^5}}-\frac {x}{\sqrt [4]{-x+x^5}}+\frac {x^2}{\sqrt [4]{-x+x^5}}\right ) \, dx \\ & = 2 \int \frac {1}{x^2 \sqrt [4]{-x+x^5}} \, dx-3 \int \frac {1}{x^6 \sqrt [4]{-x+x^5}} \, dx-3 \int \frac {1}{x^3 \sqrt [4]{-x+x^5}} \, dx-\int \frac {x}{\sqrt [4]{-x+x^5}} \, dx+\int \frac {x^2}{\sqrt [4]{-x+x^5}} \, dx \\ & = -\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^6}-\frac {4 \left (-x+x^5\right )^{3/4}}{3 x^3}+\frac {8 \left (-x+x^5\right )^{3/4}}{5 x^2}-\frac {9}{7} \int \frac {1}{x^2 \sqrt [4]{-x+x^5}} \, dx-\frac {14}{5} \int \frac {x^2}{\sqrt [4]{-x+x^5}} \, dx-\frac {\left (\sqrt [4]{x} \sqrt [4]{-1+x^4}\right ) \int \frac {x^{3/4}}{\sqrt [4]{-1+x^4}} \, dx}{\sqrt [4]{-x+x^5}}+\frac {\left (\sqrt [4]{x} \sqrt [4]{-1+x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{-1+x^4}} \, dx}{\sqrt [4]{-x+x^5}}+\int \frac {x}{\sqrt [4]{-x+x^5}} \, dx \\ & = -\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^6}-\frac {4 \left (-x+x^5\right )^{3/4}}{3 x^3}+\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^2}+\frac {9}{5} \int \frac {x^2}{\sqrt [4]{-x+x^5}} \, dx-\frac {\left (\sqrt [4]{x} \sqrt [4]{1-x^4}\right ) \int \frac {x^{3/4}}{\sqrt [4]{1-x^4}} \, dx}{\sqrt [4]{-x+x^5}}+\frac {\left (\sqrt [4]{x} \sqrt [4]{1-x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{1-x^4}} \, dx}{\sqrt [4]{-x+x^5}}+\frac {\left (\sqrt [4]{x} \sqrt [4]{-1+x^4}\right ) \int \frac {x^{3/4}}{\sqrt [4]{-1+x^4}} \, dx}{\sqrt [4]{-x+x^5}}-\frac {\left (14 \sqrt [4]{x} \sqrt [4]{-1+x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{-1+x^4}} \, dx}{5 \sqrt [4]{-x+x^5}} \\ & = -\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^6}-\frac {4 \left (-x+x^5\right )^{3/4}}{3 x^3}+\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^2}-\frac {4 x^2 \sqrt [4]{1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{16},\frac {23}{16},x^4\right )}{7 \sqrt [4]{-x+x^5}}+\frac {4 x^3 \sqrt [4]{1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{16},\frac {27}{16},x^4\right )}{11 \sqrt [4]{-x+x^5}}+\frac {\left (\sqrt [4]{x} \sqrt [4]{1-x^4}\right ) \int \frac {x^{3/4}}{\sqrt [4]{1-x^4}} \, dx}{\sqrt [4]{-x+x^5}}-\frac {\left (14 \sqrt [4]{x} \sqrt [4]{1-x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{1-x^4}} \, dx}{5 \sqrt [4]{-x+x^5}}+\frac {\left (9 \sqrt [4]{x} \sqrt [4]{-1+x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{-1+x^4}} \, dx}{5 \sqrt [4]{-x+x^5}} \\ & = -\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^6}-\frac {4 \left (-x+x^5\right )^{3/4}}{3 x^3}+\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^2}-\frac {36 x^3 \sqrt [4]{1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{16},\frac {27}{16},x^4\right )}{55 \sqrt [4]{-x+x^5}}+\frac {\left (9 \sqrt [4]{x} \sqrt [4]{1-x^4}\right ) \int \frac {x^{7/4}}{\sqrt [4]{1-x^4}} \, dx}{5 \sqrt [4]{-x+x^5}} \\ & = -\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^6}-\frac {4 \left (-x+x^5\right )^{3/4}}{3 x^3}+\frac {4 \left (-x+x^5\right )^{3/4}}{7 x^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 2 in optimal.

Time = 10.09 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.90 \[ \int \frac {\left (3+x^4\right ) \left (-1-x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \sqrt [4]{1-x^4} \left (165 \operatorname {Hypergeometric2F1}\left (-\frac {21}{16},\frac {1}{4},-\frac {5}{16},x^4\right )+x^3 \left (385 \operatorname {Hypergeometric2F1}\left (-\frac {9}{16},\frac {1}{4},\frac {7}{16},x^4\right )-462 x \operatorname {Hypergeometric2F1}\left (-\frac {5}{16},\frac {1}{4},\frac {11}{16},x^4\right )-165 x^4 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {7}{16},\frac {23}{16},x^4\right )+105 x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {11}{16},\frac {27}{16},x^4\right )\right )\right )}{1155 x^5 \sqrt [4]{x \left (-1+x^4\right )}} \]

[In]

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

[Out]

(4*(1 - x^4)^(1/4)*(165*Hypergeometric2F1[-21/16, 1/4, -5/16, x^4] + x^3*(385*Hypergeometric2F1[-9/16, 1/4, 7/
16, x^4] - 462*x*Hypergeometric2F1[-5/16, 1/4, 11/16, x^4] - 165*x^4*Hypergeometric2F1[1/4, 7/16, 23/16, x^4]
+ 105*x^5*Hypergeometric2F1[1/4, 11/16, 27/16, x^4])))/(1155*x^5*(x*(-1 + x^4))^(1/4))

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method result size
trager \(\frac {4 \left (3 x^{4}-7 x^{3}-3\right ) \left (x^{5}-x \right )^{\frac {3}{4}}}{21 x^{6}}\) \(27\)
pseudoelliptic \(\frac {4 \left (3 x^{4}-7 x^{3}-3\right ) \left (x^{5}-x \right )^{\frac {3}{4}}}{21 x^{6}}\) \(27\)
risch \(\frac {\frac {4}{7} x^{8}-\frac {8}{7} x^{4}+\frac {4}{7}-\frac {4}{3} x^{7}+\frac {4}{3} x^{3}}{x^{5} {\left (x \left (x^{4}-1\right )\right )}^{\frac {1}{4}}}\) \(37\)
gosper \(\frac {4 \left (x^{2}+1\right ) \left (x -1\right ) \left (1+x \right ) \left (3 x^{4}-7 x^{3}-3\right )}{21 x^{5} \left (x^{5}-x \right )^{\frac {1}{4}}}\) \(38\)
meijerg \(\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {21}{16}, \frac {1}{4}\right ], \left [-\frac {5}{16}\right ], x^{4}\right )}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}}}-\frac {8 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {5}{16}, \frac {1}{4}\right ], \left [\frac {11}{16}\right ], x^{4}\right )}{5 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {5}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} \operatorname {hypergeom}\left (\left [-\frac {9}{16}, \frac {1}{4}\right ], \left [\frac {7}{16}\right ], x^{4}\right )}{3 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}+\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {11}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {11}{16}\right ], \left [\frac {27}{16}\right ], x^{4}\right )}{11 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}-\frac {4 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{4}} x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {7}{16}\right ], \left [\frac {23}{16}\right ], x^{4}\right )}{7 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{4}}}\) \(162\)

[In]

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

[Out]

4/21*(3*x^4-7*x^3-3)*(x^5-x)^(3/4)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (3+x^4\right ) \left (-1-x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=\frac {4 \, {\left (x^{5} - x\right )}^{\frac {3}{4}} {\left (3 \, x^{4} - 7 \, x^{3} - 3\right )}}{21 \, x^{6}} \]

[In]

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

[Out]

4/21*(x^5 - x)^(3/4)*(3*x^4 - 7*x^3 - 3)/x^6

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {\left (3+x^4\right ) \left (-1-x^3+x^4\right )}{x^6 \sqrt [4]{-x+x^5}} \, dx=-\frac {12\,{\left (x^5-x\right )}^{3/4}+28\,x^3\,{\left (x^5-x\right )}^{3/4}-12\,x^4\,{\left (x^5-x\right )}^{3/4}}{21\,x^6} \]

[In]

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

[Out]

-(12*(x^5 - x)^(3/4) + 28*x^3*(x^5 - x)^(3/4) - 12*x^4*(x^5 - x)^(3/4))/(21*x^6)