\(\int \frac {(-3+x^4) \sqrt [3]{1+x^4} (1+x^3+x^4)}{x^8} \, dx\) [482]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\frac {3 \sqrt [3]{1+x^4} \left (4+7 x^3+8 x^4+7 x^7+4 x^8\right )}{28 x^7} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1847, 1598, 457, 75, 1492, 460} \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\frac {3 \left (x^4+1\right )^{4/3}}{4 x^4}+\frac {3 \left (x^4+1\right )^{7/3}}{7 x^7} \]

[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 75

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 457

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 460

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 1492

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/e)*x^n)^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 1598

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 1847

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)/c^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, {
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*} \text {integral}& = \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} \text {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*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\frac {3 \left (1+x^4\right )^{4/3} \left (4+7 x^3+4 x^4\right )}{28 x^7} \]

[In]

Integrate[((-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)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66

method	result	size
gosper	\(\frac {3 \left (x^{4}+1\right )^{\frac {4}{3}} \left (4 x^{4}+7 x^{3}+4\right )}{28 x^{7}}\)	\(25\)
pseudoelliptic	\(\frac {3 \left (x^{4}+1\right )^{\frac {4}{3}} \left (4 x^{4}+7 x^{3}+4\right )}{28 x^{7}}\)	\(25\)
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	\(\frac {2 \operatorname {hypergeom}\left (\left [-\frac {3}{4}, -\frac {1}{3}\right ], \left [\frac {1}{4}\right ], -x^{4}\right )}{3 x^{3}}+\frac {\frac {\Gamma \left (\frac {2}{3}\right ) x^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{3}\right ], \left [2, 3\right ], -x^{4}\right )}{3}-\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}-1+4 \ln \left (x \right )\right ) \Gamma \left (\frac {2}{3}\right )+\frac {3 \Gamma \left (\frac {2}{3}\right )}{x^{4}}}{4 \Gamma \left (\frac {2}{3}\right )}+\frac {3 \operatorname {hypergeom}\left (\left [-\frac {7}{4}, -\frac {1}{3}\right ], \left [-\frac {3}{4}\right ], -x^{4}\right )}{7 x^{7}}+x \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -x^{4}\right )-\frac {-\Gamma \left (\frac {2}{3}\right ) x^{4} \operatorname {hypergeom}\left (\left [\frac {2}{3}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right )-3 \left (3+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )\right ) \Gamma \left (\frac {2}{3}\right )}{12 \Gamma \left (\frac {2}{3}\right )}\)	\(148\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\frac {3 \, {\left (4 \, x^{8} + 7 \, x^{7} + 8 \, x^{4} + 7 \, x^{3} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{28 \, x^{7}} \]

[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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.57 (sec) , antiderivative size = 178, normalized size of antiderivative = 4.68 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=- \frac {x^{\frac {4}{3}} \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {2}{3}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} - \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{3} \\ \frac {1}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{2 x^{3} \Gamma \left (\frac {1}{4}\right )} - \frac {3 \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {1}{3} \\ - \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x^{7} \Gamma \left (- \frac {3}{4}\right )} + \frac {3 \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 x^{\frac {8}{3}} \Gamma \left (\frac {5}{3}\right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\frac {3 \, {\left (4 \, x^{8} + 7 \, x^{7} + 8 \, x^{4} + 7 \, x^{3} + 4\right )} {\left (x^{4} + 1\right )}^{\frac {1}{3}}}{28 \, x^{7}} \]

[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

Giac [F]

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

[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)

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {\left (-3+x^4\right ) \sqrt [3]{1+x^4} \left (1+x^3+x^4\right )}{x^8} \, dx=\left (\frac {3\,x}{7}+\frac {3}{4}\right )\,{\left (x^4+1\right )}^{1/3}+\frac {6\,{\left (x^4+1\right )}^{1/3}}{7\,x^3}+\frac {3\,{\left (x^4+1\right )}^{1/3}}{4\,x^4}+\frac {3\,{\left (x^4+1\right )}^{1/3}}{7\,x^7} \]

[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)