3.4.87 \(\int \frac {(2+x^6) (1-2 x^6+x^8+x^{12})}{x^{10} (-1+x^6)^{3/4}} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 \sqrt [4]{x^6-1} \left (x^{12}+9 x^8-2 x^6+1\right )}{9 x^9} \]

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Rubi [A]  time = 0.11, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {1833, 1584, 449, 1586, 1478} \begin {gather*} \frac {2 \sqrt [4]{x^6-1}}{x}+\frac {2 \left (x^6-1\right )^{9/4}}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + x^6)*(1 - 2*x^6 + x^8 + x^12))/(x^10*(-1 + x^6)^(3/4)),x]

[Out]

(2*(-1 + x^6)^(1/4))/x + (2*(-1 + x^6)^(9/4))/(9*x^9)

Rule 449

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] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 1478

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_.), x_Sym
bol] :> Int[(f*x)^m*(d + e*x^n)^(q + p)*(a/d + (c*x^n)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, q}, x] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1833

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[
Sum[((c*x)^(m + j)*Sum[Coeff[Pq, x, j + (k*n)/2]*x^((k*n)/2), {k, 0, (2*(q - j))/n + 1}]*(a + b*x^n)^p)/c^j, {
j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]

Rubi steps

\begin {align*} \int \frac {\left (2+x^6\right ) \left (1-2 x^6+x^8+x^{12}\right )}{x^{10} \left (-1+x^6\right )^{3/4}} \, dx &=\int \left (\frac {2 x^6+x^{12}}{x^8 \left (-1+x^6\right )^{3/4}}+\frac {2-3 x^6+x^{18}}{x^{10} \left (-1+x^6\right )^{3/4}}\right ) \, dx\\ &=\int \frac {2 x^6+x^{12}}{x^8 \left (-1+x^6\right )^{3/4}} \, dx+\int \frac {2-3 x^6+x^{18}}{x^{10} \left (-1+x^6\right )^{3/4}} \, dx\\ &=\int \frac {2+x^6}{x^2 \left (-1+x^6\right )^{3/4}} \, dx+\int \frac {\sqrt [4]{-1+x^6} \left (-2+x^6+x^{12}\right )}{x^{10}} \, dx\\ &=\frac {2 \sqrt [4]{-1+x^6}}{x}+\int \frac {\left (-1+x^6\right )^{5/4} \left (2+x^6\right )}{x^{10}} \, dx\\ &=\frac {2 \sqrt [4]{-1+x^6}}{x}+\frac {2 \left (-1+x^6\right )^{9/4}}{9 x^9}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 168, normalized size = 5.42 \begin {gather*} \frac {45 \left (1-x^6\right )^{3/4} x^6 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {1}{2};x^6\right )-10 \left (1-x^6\right )^{3/4} \, _2F_1\left (-\frac {3}{2},\frac {3}{4};-\frac {1}{2};x^6\right )-90 \left (1-x^6\right )^{3/4} x^8 \, _2F_1\left (-\frac {1}{6},\frac {3}{4};\frac {5}{6};x^6\right )+x^{12} \left (10 \left (1-x^6\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};x^6\right )+9 \left (1-x^6\right )^{3/4} x^2 \, _2F_1\left (\frac {3}{4},\frac {5}{6};\frac {11}{6};x^6\right )+10 \left (x^6-1\right )\right )}{45 x^9 \left (x^6-1\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + x^6)*(1 - 2*x^6 + x^8 + x^12))/(x^10*(-1 + x^6)^(3/4)),x]

[Out]

(-10*(1 - x^6)^(3/4)*Hypergeometric2F1[-3/2, 3/4, -1/2, x^6] + 45*x^6*(1 - x^6)^(3/4)*Hypergeometric2F1[-1/2,
3/4, 1/2, x^6] - 90*x^8*(1 - x^6)^(3/4)*Hypergeometric2F1[-1/6, 3/4, 5/6, x^6] + x^12*(10*(-1 + x^6) + 10*(1 -
x^6)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, x^6] + 9*x^2*(1 - x^6)^(3/4)*Hypergeometric2F1[3/4, 5/6, 11/6, x^
6]))/(45*x^9*(-1 + x^6)^(3/4))

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IntegrateAlgebraic [A]  time = 7.46, size = 31, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt [4]{-1+x^6} \left (1-2 x^6+9 x^8+x^{12}\right )}{9 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 + x^6)*(1 - 2*x^6 + x^8 + x^12))/(x^10*(-1 + x^6)^(3/4)),x]

[Out]

(2*(-1 + x^6)^(1/4)*(1 - 2*x^6 + 9*x^8 + x^12))/(9*x^9)

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fricas [A]  time = 0.47, size = 27, normalized size = 0.87 \begin {gather*} \frac {2 \, {\left (x^{12} + 9 \, x^{8} - 2 \, x^{6} + 1\right )} {\left (x^{6} - 1\right )}^{\frac {1}{4}}}{9 \, x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+2)*(x^12+x^8-2*x^6+1)/x^10/(x^6-1)^(3/4),x, algorithm="fricas")

[Out]

2/9*(x^12 + 9*x^8 - 2*x^6 + 1)*(x^6 - 1)^(1/4)/x^9

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{12} + x^{8} - 2 \, x^{6} + 1\right )} {\left (x^{6} + 2\right )}}{{\left (x^{6} - 1\right )}^{\frac {3}{4}} x^{10}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+2)*(x^12+x^8-2*x^6+1)/x^10/(x^6-1)^(3/4),x, algorithm="giac")

[Out]

integrate((x^12 + x^8 - 2*x^6 + 1)*(x^6 + 2)/((x^6 - 1)^(3/4)*x^10), x)

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maple [A]  time = 0.14, size = 28, normalized size = 0.90

method	result	size
trager	\(\frac {2 \left (x^{6}-1\right )^{\frac {1}{4}} \left (x^{12}+9 x^{8}-2 x^{6}+1\right )}{9 x^{9}}\)	\(28\)
risch	\(\frac {\frac {2}{9} x^{18}-\frac {2}{3} x^{12}+2 x^{14}-2 x^{8}+\frac {2}{3} x^{6}-\frac {2}{9}}{\left (x^{6}-1\right )^{\frac {3}{4}} x^{9}}\)	\(38\)
gosper	\(\frac {2 \left (-1+x \right ) \left (x^{2}+x +1\right ) \left (1+x \right ) \left (x^{2}-x +1\right ) \left (x^{12}+9 x^{8}-2 x^{6}+1\right )}{9 x^{9} \left (x^{6}-1\right )^{\frac {3}{4}}}\)	\(48\)
meijerg	\(\frac {\left (-\mathrm {signum}\left (x^{6}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {5}{2}\right ], x^{6}\right ) x^{9}}{9 \mathrm {signum}\left (x^{6}-1\right )^{\frac {3}{4}}}+\frac {\left (-\mathrm {signum}\left (x^{6}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [\frac {3}{4}, \frac {5}{6}\right ], \left [\frac {11}{6}\right ], x^{6}\right ) x^{5}}{5 \mathrm {signum}\left (x^{6}-1\right )^{\frac {3}{4}}}+\frac {\left (-\mathrm {signum}\left (x^{6}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}\right ], x^{6}\right )}{\mathrm {signum}\left (x^{6}-1\right )^{\frac {3}{4}} x^{3}}-\frac {2 \left (-\mathrm {signum}\left (x^{6}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [-\frac {1}{2}\right ], x^{6}\right )}{9 \mathrm {signum}\left (x^{6}-1\right )^{\frac {3}{4}} x^{9}}-\frac {2 \left (-\mathrm {signum}\left (x^{6}-1\right )\right )^{\frac {3}{4}} \hypergeom \left (\left [-\frac {1}{6}, \frac {3}{4}\right ], \left [\frac {5}{6}\right ], x^{6}\right )}{\mathrm {signum}\left (x^{6}-1\right )^{\frac {3}{4}} x}\)	\(161\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6+2)*(x^12+x^8-2*x^6+1)/x^10/(x^6-1)^(3/4),x,method=_RETURNVERBOSE)

[Out]

2/9*(x^6-1)^(1/4)*(x^12+9*x^8-2*x^6+1)/x^9

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maxima [B]  time = 0.90, size = 58, normalized size = 1.87 \begin {gather*} \frac {2 \, {\left (x^{18} + 9 \, x^{14} - 3 \, x^{12} - 9 \, x^{8} + 3 \, x^{6} - 1\right )}}{9 \, {\left (x^{2} + x + 1\right )}^{\frac {3}{4}} {\left (x^{2} - x + 1\right )}^{\frac {3}{4}} {\left (x + 1\right )}^{\frac {3}{4}} {\left (x - 1\right )}^{\frac {3}{4}} x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^6+2)*(x^12+x^8-2*x^6+1)/x^10/(x^6-1)^(3/4),x, algorithm="maxima")

[Out]

2/9*(x^18 + 9*x^14 - 3*x^12 - 9*x^8 + 3*x^6 - 1)/((x^2 + x + 1)^(3/4)*(x^2 - x + 1)^(3/4)*(x + 1)^(3/4)*(x - 1
)^(3/4)*x^9)

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mupad [B]  time = 0.40, size = 49, normalized size = 1.58 \begin {gather*} \frac {2\,{\left (x^6-1\right )}^{1/4}}{x}-\frac {4\,{\left (x^6-1\right )}^{1/4}}{9\,x^3}+\frac {2\,x^3\,{\left (x^6-1\right )}^{1/4}}{9}+\frac {2\,{\left (x^6-1\right )}^{1/4}}{9\,x^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^6 + 2)*(x^8 - 2*x^6 + x^12 + 1))/(x^10*(x^6 - 1)^(3/4)),x)

[Out]

(2*(x^6 - 1)^(1/4))/x - (4*(x^6 - 1)^(1/4))/(9*x^3) + (2*x^3*(x^6 - 1)^(1/4))/9 + (2*(x^6 - 1)^(1/4))/(9*x^9)

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sympy [C]  time = 6.02, size = 139, normalized size = 4.48 \begin {gather*} \frac {x^{9} e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {x^{6}} \right )}}{9} + \frac {x^{5} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {5}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{6} \\ \frac {11}{6} \end {matrix}\middle | {x^{6}} \right )}}{6 \Gamma \left (\frac {11}{6}\right )} - \frac {e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{6}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{6}, \frac {3}{4} \\ \frac {5}{6} \end {matrix}\middle | {x^{6}} \right )}}{3 x \Gamma \left (\frac {5}{6}\right )} - \frac {e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2} \end {matrix}\middle | {x^{6}} \right )}}{x^{3}} + \frac {2 e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {x^{6}} \right )}}{9 x^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**6+2)*(x**12+x**8-2*x**6+1)/x**10/(x**6-1)**(3/4),x)

[Out]

x**9*exp(-3*I*pi/4)*hyper((3/4, 3/2), (5/2,), x**6)/9 + x**5*exp(-3*I*pi/4)*gamma(5/6)*hyper((3/4, 5/6), (11/6
,), x**6)/(6*gamma(11/6)) - exp(I*pi/4)*gamma(-1/6)*hyper((-1/6, 3/4), (5/6,), x**6)/(3*x*gamma(5/6)) - exp(I*
pi/4)*hyper((-1/2, 3/4), (1/2,), x**6)/x**3 + 2*exp(I*pi/4)*hyper((-3/2, 3/4), (-1/2,), x**6)/(9*x**9)

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