3.5.89 \(\int \frac {(1+x^5)^{2/3} (-3+2 x^5) (4+3 x^3+4 x^5)}{x^9} \, dx\)

Optimal. Leaf size=38 \[ \frac {3 \left (x^5+1\right )^{2/3} \left (5 x^{10}+6 x^8+10 x^5+6 x^3+5\right )}{10 x^8} \]

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Rubi [B]  time = 0.17, antiderivative size = 87, normalized size of antiderivative = 2.29, number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1826, 1835, 1586, 1584, 449} \begin {gather*} -\frac {18 \left (x^5+1\right )^{2/3}}{5 x^5}-\frac {15 \left (x^5+1\right )^{2/3}}{14 x^8}+\frac {15 \left (x^5+1\right )^{2/3}}{x^3}+\frac {3 \left (x^5+1\right )^{2/3} \left (35 x^{11}+42 x^9-280 x^6+126 x^4+60 x\right )}{70 x^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^5)^(2/3)*(-3 + 2*x^5)*(4 + 3*x^3 + 4*x^5))/x^9,x]

[Out]

(-15*(1 + x^5)^(2/3))/(14*x^8) - (18*(1 + x^5)^(2/3))/(5*x^5) + (15*(1 + x^5)^(2/3))/x^3 + (3*(1 + x^5)^(2/3)*
(60*x + 126*x^4 - 280*x^6 + 42*x^9 + 35*x^11))/(70*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 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 1826

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

Rule 1835

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{Pq0 = Coeff[Pq, x, 0]}, Simp[(Pq
0*(c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(2*a*c*(m + 1)), Int[(c*x)^(m + 1)*ExpandToSum
[(2*a*(m + 1)*(Pq - Pq0))/x - 2*b*Pq0*(m + n*(p + 1) + 1)*x^(n - 1), x]*(a + b*x^n)^p, x], x] /; NeQ[Pq0, 0]]
/; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[m, -1] && LeQ[n - 1, Expon[Pq, x]]

Rubi steps

\begin {align*} \int \frac {\left (1+x^5\right )^{2/3} \left (-3+2 x^5\right ) \left (4+3 x^3+4 x^5\right )}{x^9} \, dx &=\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}+\frac {10}{3} \int \frac {\frac {18}{7}+\frac {27 x^3}{5}-12 x^5+\frac {9 x^8}{5}+\frac {3 x^{10}}{2}}{x^9 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}-\frac {5}{24} \int \frac {-\frac {432 x^2}{5}+216 x^4-\frac {144 x^7}{5}-24 x^9}{x^8 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}-\frac {5}{24} \int \frac {-\frac {432 x}{5}+216 x^3-\frac {144 x^6}{5}-24 x^8}{x^7 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}-\frac {5}{24} \int \frac {-\frac {432}{5}+216 x^2-\frac {144 x^5}{5}-24 x^7}{x^6 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}-\frac {18 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}+\frac {1}{48} \int \frac {-2160 x+240 x^6}{x^5 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}-\frac {18 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}+\frac {1}{48} \int \frac {-2160+240 x^5}{x^4 \sqrt [3]{1+x^5}} \, dx\\ &=-\frac {15 \left (1+x^5\right )^{2/3}}{14 x^8}-\frac {18 \left (1+x^5\right )^{2/3}}{5 x^5}+\frac {15 \left (1+x^5\right )^{2/3}}{x^3}+\frac {3 \left (1+x^5\right )^{2/3} \left (60 x+126 x^4-280 x^6+42 x^9+35 x^{11}\right )}{70 x^9}\\ \end {align*}

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Mathematica [C]  time = 0.20, size = 154, normalized size = 4.05 \begin {gather*} -\frac {27}{25} \left (x^5+1\right )^{5/3} \, _2F_1\left (\frac {5}{3},2;\frac {8}{3};x^5+1\right )+\frac {3 \, _2F_1\left (-\frac {8}{5},-\frac {2}{3};-\frac {3}{5};-x^5\right )}{2 x^8}+\frac {4 \, _2F_1\left (-\frac {2}{3},-\frac {3}{5};\frac {2}{5};-x^5\right )}{3 x^3}+4 x^2 \, _2F_1\left (-\frac {2}{3},\frac {2}{5};\frac {7}{5};-x^5\right )+\frac {3}{5} \left (3 \left (x^5+1\right )^{2/3}+3 \log \left (1-\sqrt [3]{x^5+1}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^5+1}+1}{\sqrt {3}}\right )-5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^5)^(2/3)*(-3 + 2*x^5)*(4 + 3*x^3 + 4*x^5))/x^9,x]

[Out]

(3*Hypergeometric2F1[-8/5, -2/3, -3/5, -x^5])/(2*x^8) + (4*Hypergeometric2F1[-2/3, -3/5, 2/5, -x^5])/(3*x^3) +
4*x^2*Hypergeometric2F1[-2/3, 2/5, 7/5, -x^5] - (27*(1 + x^5)^(5/3)*Hypergeometric2F1[5/3, 2, 8/3, 1 + x^5])/
25 + (3*(3*(1 + x^5)^(2/3) + 2*Sqrt[3]*ArcTan[(1 + 2*(1 + x^5)^(1/3))/Sqrt[3]] - 5*Log[x] + 3*Log[1 - (1 + x^5
)^(1/3)]))/5

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IntegrateAlgebraic [A]  time = 0.95, size = 28, normalized size = 0.74 \begin {gather*} \frac {3 \left (1+x^5\right )^{5/3} \left (5+6 x^3+5 x^5\right )}{10 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^5)^(2/3)*(-3 + 2*x^5)*(4 + 3*x^3 + 4*x^5))/x^9,x]

[Out]

(3*(1 + x^5)^(5/3)*(5 + 6*x^3 + 5*x^5))/(10*x^8)

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fricas [A]  time = 0.47, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (5 \, x^{10} + 6 \, x^{8} + 10 \, x^{5} + 6 \, x^{3} + 5\right )} {\left (x^{5} + 1\right )}^{\frac {2}{3}}}{10 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(2/3)*(2*x^5-3)*(4*x^5+3*x^3+4)/x^9,x, algorithm="fricas")

[Out]

3/10*(5*x^10 + 6*x^8 + 10*x^5 + 6*x^3 + 5)*(x^5 + 1)^(2/3)/x^8

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x^{5} + 3 \, x^{3} + 4\right )} {\left (2 \, x^{5} - 3\right )} {\left (x^{5} + 1\right )}^{\frac {2}{3}}}{x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(2/3)*(2*x^5-3)*(4*x^5+3*x^3+4)/x^9,x, algorithm="giac")

[Out]

integrate((4*x^5 + 3*x^3 + 4)*(2*x^5 - 3)*(x^5 + 1)^(2/3)/x^9, x)

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maple [A]  time = 0.10, size = 35, normalized size = 0.92

method	result	size
trager	\(\frac {3 \left (x^{5}+1\right )^{\frac {2}{3}} \left (5 x^{10}+6 x^{8}+10 x^{5}+6 x^{3}+5\right )}{10 x^{8}}\)	\(35\)
gosper	\(\frac {3 \left (x^{4}-x^{3}+x^{2}-x +1\right ) \left (1+x \right ) \left (5 x^{5}+6 x^{3}+5\right ) \left (x^{5}+1\right )^{\frac {2}{3}}}{10 x^{8}}\)	\(44\)
risch	\(\frac {\frac {18}{5} x^{8}+\frac {9}{5} x^{3}+\frac {9}{2} x^{10}+\frac {9}{2} x^{5}+\frac {3}{2}+\frac {3}{2} x^{15}+\frac {9}{5} x^{13}}{x^{8} \left (x^{5}+1\right )^{\frac {1}{3}}}\)	\(45\)
meijerg	\(4 \hypergeom \left (\left [-\frac {2}{3}, \frac {2}{5}\right ], \left [\frac {7}{5}\right ], -x^{5}\right ) x^{2}-\frac {2 \Gamma \left (\frac {2}{3}\right ) \sqrt {3}\, \left (-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+5 \ln \relax (x )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}-\frac {2 \hypergeom \left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{5}\right ) \pi \sqrt {3}\, x^{5}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{5 \pi }+\frac {4 \hypergeom \left (\left [-\frac {2}{3}, -\frac {3}{5}\right ], \left [\frac {2}{5}\right ], -x^{5}\right )}{3 x^{3}}+\frac {3 \Gamma \left (\frac {2}{3}\right ) \sqrt {3}\, \left (\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{5}}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}-1+5 \ln \relax (x )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{5}\right ) \pi \sqrt {3}\, x^{5}}{9 \Gamma \left (\frac {2}{3}\right )}\right )}{5 \pi }+\frac {3 \hypergeom \left (\left [-\frac {8}{5}, -\frac {2}{3}\right ], \left [-\frac {3}{5}\right ], -x^{5}\right )}{2 x^{8}}\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5+1)^(2/3)*(2*x^5-3)*(4*x^5+3*x^3+4)/x^9,x,method=_RETURNVERBOSE)

[Out]

3/10*(x^5+1)^(2/3)*(5*x^10+6*x^8+10*x^5+6*x^3+5)/x^8

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maxima [A]  time = 0.50, size = 50, normalized size = 1.32 \begin {gather*} \frac {3 \, {\left (5 \, x^{10} + 6 \, x^{8} + 10 \, x^{5} + 6 \, x^{3} + 5\right )} {\left (x^{4} - x^{3} + x^{2} - x + 1\right )}^{\frac {2}{3}} {\left (x + 1\right )}^{\frac {2}{3}}}{10 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^5+1)^(2/3)*(2*x^5-3)*(4*x^5+3*x^3+4)/x^9,x, algorithm="maxima")

[Out]

3/10*(5*x^10 + 6*x^8 + 10*x^5 + 6*x^3 + 5)*(x^4 - x^3 + x^2 - x + 1)^(2/3)*(x + 1)^(2/3)/x^8

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mupad [B]  time = 0.49, size = 52, normalized size = 1.37 \begin {gather*} {\left (x^5+1\right )}^{2/3}\,\left (\frac {3\,x^2}{2}+\frac {9}{5}\right )+\frac {3\,{\left (x^5+1\right )}^{2/3}}{x^3}+\frac {9\,{\left (x^5+1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^5+1\right )}^{2/3}}{2\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^5 + 1)^(2/3)*(2*x^5 - 3)*(3*x^3 + 4*x^5 + 4))/x^9,x)

[Out]

(x^5 + 1)^(2/3)*((3*x^2)/2 + 9/5) + (3*(x^5 + 1)^(2/3))/x^3 + (9*(x^5 + 1)^(2/3))/(5*x^5) + (3*(x^5 + 1)^(2/3)
)/(2*x^8)

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sympy [C]  time = 5.64, size = 185, normalized size = 4.87 \begin {gather*} - \frac {6 x^{\frac {10}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 \Gamma \left (\frac {1}{3}\right )} + \frac {8 x^{2} \Gamma \left (\frac {2}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {2}{5} \\ \frac {7}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 \Gamma \left (\frac {7}{5}\right )} - \frac {4 \Gamma \left (- \frac {3}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {3}{5} \\ \frac {2}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{3} \Gamma \left (\frac {2}{5}\right )} - \frac {12 \Gamma \left (- \frac {8}{5}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{5}, - \frac {2}{3} \\ - \frac {3}{5} \end {matrix}\middle | {x^{5} e^{i \pi }} \right )}}{5 x^{8} \Gamma \left (- \frac {3}{5}\right )} + \frac {9 \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{5}}} \right )}}{5 x^{\frac {5}{3}} \Gamma \left (\frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**5+1)**(2/3)*(2*x**5-3)*(4*x**5+3*x**3+4)/x**9,x)

[Out]

-6*x**(10/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(I*pi)/x**5)/(5*gamma(1/3)) + 8*x**2*gamma(2/5)*
hyper((-2/3, 2/5), (7/5,), x**5*exp_polar(I*pi))/(5*gamma(7/5)) - 4*gamma(-3/5)*hyper((-2/3, -3/5), (2/5,), x*
*5*exp_polar(I*pi))/(5*x**3*gamma(2/5)) - 12*gamma(-8/5)*hyper((-8/5, -2/3), (-3/5,), x**5*exp_polar(I*pi))/(5
*x**8*gamma(-3/5)) + 9*gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(I*pi)/x**5)/(5*x**(5/3)*gamma(4/3))

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