\(\int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx\) [261]

Optimal result

Integrand size = 20, antiderivative size = 25 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (3+11 x^3\right ) \left (-x+x^4\right )^{3/4}}{63 x^6} \]

[Out]

4/63*(11*x^3+3)*(x^4-x)^(3/4)/x^6

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2063, 2039} \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (x^4-x\right )^{3/4}}{21 x^6}+\frac {44 \left (x^4-x\right )^{3/4}}{63 x^3} \]

[In]

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

[Out]

(4*(-x + x^4)^(3/4))/(21*x^6) + (44*(-x + x^4)^(3/4))/(63*x^3)

Rule 2039

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*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {4 \left (-x+x^4\right )^{3/4}}{21 x^6}+\frac {11}{7} \int \frac {1}{x^3 \sqrt [4]{-x+x^4}} \, dx \\ & = \frac {4 \left (-x+x^4\right )^{3/4}}{21 x^6}+\frac {44 \left (-x+x^4\right )^{3/4}}{63 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \left (x \left (-1+x^3\right )\right )^{3/4} \left (3+11 x^3\right )}{63 x^6} \]

[In]

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

[Out]

(4*(x*(-1 + x^3))^(3/4)*(3 + 11*x^3))/(63*x^6)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
trager	\(\frac {4 \left (11 x^{3}+3\right ) \left (x^{4}-x \right )^{\frac {3}{4}}}{63 x^{6}}\)	\(22\)
pseudoelliptic	\(\frac {4 \left (11 x^{3}+3\right ) \left (x^{4}-x \right )^{\frac {3}{4}}}{63 x^{6}}\)	\(22\)
risch	\(\frac {\frac {44}{63} x^{6}-\frac {32}{63} x^{3}-\frac {4}{21}}{x^{5} {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{4}}}\)	\(27\)
gosper	\(\frac {4 \left (11 x^{3}+3\right ) \left (x -1\right ) \left (x^{2}+x +1\right )}{63 \left (x^{4}-x \right )^{\frac {1}{4}} x^{5}}\)	\(31\)
meijerg	\(-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (1+\frac {4 x^{3}}{3}\right ) \left (-x^{3}+1\right )^{\frac {3}{4}}}{21 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {21}{4}}}-\frac {4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{4}} \left (-x^{3}+1\right )^{\frac {3}{4}}}{9 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{4}} x^{\frac {9}{4}}}\)	\(73\)

[In]

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

[Out]

4/63*(11*x^3+3)*(x^4-x)^(3/4)/x^6

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \, {\left (x^{4} - x\right )}^{\frac {3}{4}} {\left (11 \, x^{3} + 3\right )}}{63 \, x^{6}} \]

[In]

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

[Out]

4/63*(x^4 - x)^(3/4)*(11*x^3 + 3)/x^6

Sympy [F]

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

[In]

integrate((x**3+1)/x**6/(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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (21) = 42\).

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.32 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {4 \, {\left (x^{4} - x\right )}}{9 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {13}{4}}} + \frac {4 \, {\left (4 \, x^{7} - x^{4} - 3 \, x\right )}}{63 \, {\left (x^{2} + x + 1\right )}^{\frac {1}{4}} {\left (x - 1\right )}^{\frac {1}{4}} x^{\frac {25}{4}}} \]

[In]

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

[Out]

4/9*(x^4 - x)/((x^2 + x + 1)^(1/4)*(x - 1)^(1/4)*x^(13/4)) + 4/63*(4*x^7 - x^4 - 3*x)/((x^2 + x + 1)^(1/4)*(x
- 1)^(1/4)*x^(25/4))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=-\frac {4}{21} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {7}{4}} + \frac {8}{9} \, {\left (-\frac {1}{x^{3}} + 1\right )}^{\frac {3}{4}} \]

[In]

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

[Out]

-4/21*(-1/x^3 + 1)^(7/4) + 8/9*(-1/x^3 + 1)^(3/4)

Mupad [B] (verification not implemented)

Time = 5.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {1+x^3}{x^6 \sqrt [4]{-x+x^4}} \, dx=\frac {12\,{\left (x^4-x\right )}^{3/4}+44\,x^3\,{\left (x^4-x\right )}^{3/4}}{63\,x^6} \]

[In]

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

[Out]

(12*(x^4 - x)^(3/4) + 44*x^3*(x^4 - x)^(3/4))/(63*x^6)