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

Optimal. Leaf size=38 \[ \frac {3 \left (x^4+1\right )^{2/3} \left (5 x^8+8 x^7+10 x^4+8 x^3+5\right )}{40 x^8} \]

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Rubi [A]  time = 0.09, antiderivative size = 33, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1833, 1584, 449, 1474, 847, 74} \begin {gather*} \frac {3 \left (x^4+1\right )^{8/3}}{8 x^8}+\frac {3 \left (x^4+1\right )^{5/3}}{5 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + x^4)^(5/3))/(5*x^5) + (3*(1 + x^4)^(8/3))/(8*x^8)

Rule 74

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

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 847

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f -
d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] &&  !IGtQ[n, 0]

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

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 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 (-3+x^4\right ) \left (1+x^4\right )^{2/3} \left (1+x^3+x^4\right )}{x^9} \, dx &=\int \left (\frac {\left (1+x^4\right )^{2/3} \left (-3 x^2+x^6\right )}{x^8}+\frac {\left (1+x^4\right )^{2/3} \left (-3-2 x^4+x^8\right )}{x^9}\right ) \, dx\\ &=\int \frac {\left (1+x^4\right )^{2/3} \left (-3 x^2+x^6\right )}{x^8} \, dx+\int \frac {\left (1+x^4\right )^{2/3} \left (-3-2 x^4+x^8\right )}{x^9} \, dx\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {(1+x)^{2/3} \left (-3-2 x+x^2\right )}{x^3} \, dx,x,x^4\right )+\int \frac {\left (-3+x^4\right ) \left (1+x^4\right )^{2/3}}{x^6} \, dx\\ &=\frac {3 \left (1+x^4\right )^{5/3}}{5 x^5}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {(-3+x) (1+x)^{5/3}}{x^3} \, dx,x,x^4\right )\\ &=\frac {3 \left (1+x^4\right )^{5/3}}{5 x^5}+\frac {3 \left (1+x^4\right )^{8/3}}{8 x^8}\\ \end {align*}

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Mathematica [C]  time = 0.23, size = 158, normalized size = 4.16 \begin {gather*} \frac {24 \, _2F_1\left (-\frac {5}{4},-\frac {2}{3};-\frac {1}{4};-x^4\right )+x^4 \left (x \left (-12 \left (x^4+1\right )^{5/3} \, _2F_1\left (\frac {5}{3},2;\frac {8}{3};x^4+1\right )+18 \left (x^4+1\right )^{5/3} \, _2F_1\left (\frac {5}{3},3;\frac {8}{3};x^4+1\right )+5 \left (3 \left (\left (x^4+1\right )^{2/3}+\log \left (1-\sqrt [3]{x^4+1}\right )\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )-4 \log (x)\right )\right )-40 \, _2F_1\left (-\frac {2}{3},-\frac {1}{4};\frac {3}{4};-x^4\right )\right )}{40 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*Hypergeometric2F1[-5/4, -2/3, -1/4, -x^4] + x^4*(-40*Hypergeometric2F1[-2/3, -1/4, 3/4, -x^4] + x*(-12*(1
+ x^4)^(5/3)*Hypergeometric2F1[5/3, 2, 8/3, 1 + x^4] + 18*(1 + x^4)^(5/3)*Hypergeometric2F1[5/3, 3, 8/3, 1 + x
^4] + 5*(2*Sqrt[3]*ArcTan[(1 + 2*(1 + x^4)^(1/3))/Sqrt[3]] - 4*Log[x] + 3*((1 + x^4)^(2/3) + Log[1 - (1 + x^4)
^(1/3)])))))/(40*x^5)

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IntegrateAlgebraic [A]  time = 0.26, size = 28, normalized size = 0.74 \begin {gather*} \frac {3 \left (1+x^4\right )^{5/3} \left (5+8 x^3+5 x^4\right )}{40 x^8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + x^4)^(5/3)*(5 + 8*x^3 + 5*x^4))/(40*x^8)

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fricas [A]  time = 0.46, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (5 \, x^{8} + 8 \, x^{7} + 10 \, x^{4} + 8 \, x^{3} + 5\right )} {\left (x^{4} + 1\right )}^{\frac {2}{3}}}{40 \, x^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(5*x^8 + 8*x^7 + 10*x^4 + 8*x^3 + 5)*(x^4 + 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 (x^{4} + x^{3} + 1\right )} {\left (x^{4} + 1\right )}^{\frac {2}{3}} {\left (x^{4} - 3\right )}}{x^{9}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.13, size = 25, normalized size = 0.66

method	result	size
gosper	\(\frac {3 \left (x^{4}+1\right )^{\frac {5}{3}} \left (5 x^{4}+8 x^{3}+5\right )}{40 x^{8}}\)	\(25\)
trager	\(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}} \left (5 x^{8}+8 x^{7}+10 x^{4}+8 x^{3}+5\right )}{40 x^{8}}\)	\(35\)
risch	\(\frac {\frac {3}{5} x^{11}+\frac {6}{5} x^{7}+\frac {3}{5} x^{3}+\frac {9}{8} x^{8}+\frac {9}{8} x^{4}+\frac {3}{8}+\frac {3}{8} x^{12}}{x^{8} \left (x^{4}+1\right )^{\frac {1}{3}}}\)	\(45\)
meijerg	\(\frac {\Gamma \left (\frac {2}{3}\right ) \sqrt {3}\, \left (\frac {\pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right ) x^{4}}-\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}-1+4 \ln \relax (x )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}+\frac {\hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 3\right ], -x^{4}\right ) \pi \sqrt {3}\, x^{4}}{9 \Gamma \left (\frac {2}{3}\right )}\right )}{6 \pi }+\frac {3 \hypergeom \left (\left [-\frac {5}{4}, -\frac {2}{3}\right ], \left [-\frac {1}{4}\right ], -x^{4}\right )}{5 x^{5}}+\frac {\Gamma \left (\frac {2}{3}\right ) \sqrt {3}\, \left (\frac {\pi \sqrt {3}}{2 \Gamma \left (\frac {2}{3}\right ) x^{8}}+\frac {2 \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right ) x^{4}}+\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \pi \sqrt {3}}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {4 \hypergeom \left (\left [1, 1, \frac {7}{3}\right ], \left [2, 4\right ], -x^{4}\right ) \pi \sqrt {3}\, x^{4}}{81 \Gamma \left (\frac {2}{3}\right )}\right )}{4 \pi }-\frac {\Gamma \left (\frac {2}{3}\right ) \sqrt {3}\, \left (-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}-\frac {2 \hypergeom \left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right ) \pi \sqrt {3}\, x^{4}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{12 \pi }-\frac {\hypergeom \left (\left [-\frac {2}{3}, -\frac {1}{4}\right ], \left [\frac {3}{4}\right ], -x^{4}\right )}{x}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/40*(x^4+1)^(5/3)*(5*x^4+8*x^3+5)/x^8

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {2 \, {\left (x^{4} + 1\right )}^{\frac {5}{3}} + {\left (x^{4} + 1\right )}^{\frac {2}{3}}}{8 \, {\left (2 \, x^{4} - {\left (x^{4} + 1\right )}^{2} + 1\right )}} + \int \frac {{\left (x^{5} + x^{4} - 2 \, x - 3\right )} {\left (x^{4} + 1\right )}^{\frac {2}{3}}}{x^{6}}\,{d x} - \frac {1}{24} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{12} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^4 + 1)^(1/3) + 1)) - 1/8*(2*(x^4 + 1)^(5/3) + (x^4 + 1)^(2/3))/(2*x^4 -
(x^4 + 1)^2 + 1) + integrate((x^5 + x^4 - 2*x - 3)*(x^4 + 1)^(2/3)/x^6, x) - 1/24*log((x^4 + 1)^(2/3) + (x^4 +
1)^(1/3) + 1) + 1/12*log((x^4 + 1)^(1/3) - 1)

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mupad [B]  time = 0.47, size = 58, normalized size = 1.53 \begin {gather*} \frac {3\,{\left (x^4+1\right )}^{2/3}}{8}+\frac {3\,{\left (x^4+1\right )}^{2/3}}{5\,x}+\frac {3\,{\left (x^4+1\right )}^{2/3}}{4\,x^4}+\frac {3\,{\left (x^4+1\right )}^{2/3}}{5\,x^5}+\frac {3\,{\left (x^4+1\right )}^{2/3}}{8\,x^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 5.46, size = 180, normalized size = 4.74 \begin {gather*} - \frac {x^{\frac {8}{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^{4}}} \right )}}{4 \Gamma \left (\frac {1}{3}\right )} + \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} - \frac {3 \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {2}{3} \\ - \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{5} \Gamma \left (- \frac {1}{4}\right )} + \frac {\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^{4}}} \right )}}{2 x^{\frac {4}{3}} \Gamma \left (\frac {4}{3}\right )} + \frac {3 \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 x^{\frac {16}{3}} \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**(8/3)*gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), exp_polar(I*pi)/x**4)/(4*gamma(1/3)) + gamma(-1/4)*hyper((-2
/3, -1/4), (3/4,), x**4*exp_polar(I*pi))/(4*x*gamma(3/4)) - 3*gamma(-5/4)*hyper((-5/4, -2/3), (-1/4,), x**4*ex
p_polar(I*pi))/(4*x**5*gamma(-1/4)) + gamma(1/3)*hyper((-2/3, 1/3), (4/3,), exp_polar(I*pi)/x**4)/(2*x**(4/3)*
gamma(4/3)) + 3*gamma(4/3)*hyper((-2/3, 4/3), (7/3,), exp_polar(I*pi)/x**4)/(4*x**(16/3)*gamma(7/3))

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