3.5.82 \(\int \frac {(-3+x^4) \sqrt [3]{1+x^4} (1+x^3+x^4)}{x^8} \, dx\)

Optimal. Leaf size=38 \[ \frac {3 \sqrt [3]{x^4+1} \left (4 x^8+7 x^7+8 x^4+7 x^3+4\right )}{28 x^7} \]

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Rubi [A]  time = 0.08, 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, 446, 74, 1478, 449} \begin {gather*} \frac {3 \left (x^4+1\right )^{4/3}}{4 x^4}+\frac {3 \left (x^4+1\right )^{7/3}}{7 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + x^4)^(4/3))/(4*x^4) + (3*(1 + x^4)^(7/3))/(7*x^7)

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 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 ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx &=\int \left (\frac {\sqrt [3]{1+x^4} \left (-3 x^2+x^6\right )}{x^7}+\frac {\sqrt [3]{1+x^4} \left (-3-2 x^4+x^8\right )}{x^8}\right ) \, dx\\ &=\int \frac {\sqrt [3]{1+x^4} \left (-3 x^2+x^6\right )}{x^7} \, dx+\int \frac {\sqrt [3]{1+x^4} \left (-3-2 x^4+x^8\right )}{x^8} \, dx\\ &=\int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4}}{x^5} \, dx+\int \frac {\left (-3+x^4\right ) \left (1+x^4\right )^{4/3}}{x^8} \, dx\\ &=\frac {3 \left (1+x^4\right )^{7/3}}{7 x^7}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {(-3+x) \sqrt [3]{1+x}}{x^2} \, dx,x,x^4\right )\\ &=\frac {3 \left (1+x^4\right )^{4/3}}{4 x^4}+\frac {3 \left (1+x^4\right )^{7/3}}{7 x^7}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 169, normalized size = 4.45 \begin {gather*} \frac {1}{16} \left (-9 \left (x^4+1\right )^{4/3} \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};x^4+1\right )+16 x \, _2F_1\left (-\frac {1}{3},\frac {1}{4};\frac {5}{4};-x^4\right )+12 \sqrt [3]{x^4+1}+4 \log \left (1-\sqrt [3]{x^4+1}\right )-2 \log \left (\left (x^4+1\right )^{2/3}+\sqrt [3]{x^4+1}+1\right )-4 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )\right )+\frac {3 \, _2F_1\left (-\frac {7}{4},-\frac {1}{3};-\frac {3}{4};-x^4\right )}{7 x^7}+\frac {2 \, _2F_1\left (-\frac {3}{4},-\frac {1}{3};\frac {1}{4};-x^4\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Hypergeometric2F1[-7/4, -1/3, -3/4, -x^4])/(7*x^7) + (2*Hypergeometric2F1[-3/4, -1/3, 1/4, -x^4])/(3*x^3) +
(12*(1 + x^4)^(1/3) - 4*Sqrt[3]*ArcTan[(1 + 2*(1 + x^4)^(1/3))/Sqrt[3]] + 16*x*Hypergeometric2F1[-1/3, 1/4, 5
/4, -x^4] - 9*(1 + x^4)^(4/3)*Hypergeometric2F1[4/3, 2, 7/3, 1 + x^4] + 4*Log[1 - (1 + x^4)^(1/3)] - 2*Log[1 +
(1 + x^4)^(1/3) + (1 + x^4)^(2/3)])/16

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IntegrateAlgebraic [A]  time = 0.27, size = 28, normalized size = 0.74 \begin {gather*} \frac {3 \left (1+x^4\right )^{4/3} \left (4+7 x^3+4 x^4\right )}{28 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + x^4)^(4/3)*(4 + 7*x^3 + 4*x^4))/(28*x^7)

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fricas [A]  time = 0.46, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (4 \, x^{8} + 7 \, x^{7} + 8 \, x^{4} + 7 \, x^{3} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{28 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/28*(4*x^8 + 7*x^7 + 8*x^4 + 7*x^3 + 4)*(x^4 + 1)^(1/3)/x^7

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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 {1}{3}} {\left (x^{4} - 3\right )}}{x^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + x^3 + 1)*(x^4 + 1)^(1/3)*(x^4 - 3)/x^8, x)

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maple [A]  time = 0.12, size = 25, normalized size = 0.66

method	result	size
gosper	\(\frac {3 \left (x^{4}+1\right )^{\frac {4}{3}} \left (4 x^{4}+7 x^{3}+4\right )}{28 x^{7}}\)	\(25\)
trager	\(\frac {3 \left (x^{4}+1\right )^{\frac {1}{3}} \left (4 x^{8}+7 x^{7}+8 x^{4}+7 x^{3}+4\right )}{28 x^{7}}\)	\(35\)
risch	\(\frac {\frac {9}{7} x^{8}+\frac {9}{7} x^{4}+\frac {3}{7}+\frac {3}{2} x^{7}+\frac {3}{4} x^{3}+\frac {3}{7} x^{12}+\frac {3}{4} x^{11}}{x^{7} \left (x^{4}+1\right )^{\frac {2}{3}}}\)	\(45\)
meijerg	\(\frac {2 \hypergeom \left (\left [-\frac {3}{4}, -\frac {1}{3}\right ], \left [\frac {1}{4}\right ], -x^{4}\right )}{3 x^{3}}+\frac {\frac {3 \Gamma \left (\frac {2}{3}\right )}{x^{4}}-\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}-1+4 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )+\frac {\hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 3\right ], -x^{4}\right ) \Gamma \left (\frac {2}{3}\right ) x^{4}}{3}}{4 \Gamma \left (\frac {2}{3}\right )}+\frac {3 \hypergeom \left (\left [-\frac {7}{4}, -\frac {1}{3}\right ], \left [-\frac {3}{4}\right ], -x^{4}\right )}{7 x^{7}}+\hypergeom \left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -x^{4}\right ) x -\frac {-3 \left (3+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+4 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )-\hypergeom \left (\left [\frac {2}{3}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right ) \Gamma \left (\frac {2}{3}\right ) x^{4}}{12 \Gamma \left (\frac {2}{3}\right )}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/28*(x^4+1)^(4/3)*(4*x^4+7*x^3+4)/x^7

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maxima [A]  time = 0.46, size = 34, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (4 \, x^{8} + 7 \, x^{7} + 8 \, x^{4} + 7 \, x^{3} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{28 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/28*(4*x^8 + 7*x^7 + 8*x^4 + 7*x^3 + 4)*(x^4 + 1)^(1/3)/x^7

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mupad [B]  time = 0.45, size = 50, normalized size = 1.32 \begin {gather*} \left (\frac {3\,x}{7}+\frac {3}{4}\right )\,{\left (x^4+1\right )}^{1/3}+\frac {6\,{\left (x^4+1\right )}^{1/3}}{7\,x^3}+\frac {3\,{\left (x^4+1\right )}^{1/3}}{4\,x^4}+\frac {3\,{\left (x^4+1\right )}^{1/3}}{7\,x^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 + 1)^(1/3)*(x^4 - 3)*(x^3 + x^4 + 1))/x^8,x)

[Out]

((3*x)/7 + 3/4)*(x^4 + 1)^(1/3) + (6*(x^4 + 1)^(1/3))/(7*x^3) + (3*(x^4 + 1)^(1/3))/(4*x^4) + (3*(x^4 + 1)^(1/
3))/(7*x^7)

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sympy [C]  time = 4.81, size = 178, normalized size = 4.68 \begin {gather*} - \frac {x^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {2}{3}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{3} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{2 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {3 \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {1}{3} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} + \frac {3 \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 x^{\frac {8}{3}} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-3)*(x**4+1)**(1/3)*(x**4+x**3+1)/x**8,x)

[Out]

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

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