3.22.46 \(\int \frac {(1+2 x^3)^{4/3} (1+3 x^3)}{x^8 (1+4 x^3)} \, dx\) [2146]

Optimal. Leaf size=157 \[ \frac {\sqrt [3]{1+2 x^3} \left (-4-9 x^3-58 x^6\right )}{28 x^7}-\frac {2 \sqrt [3]{2} \text {ArcTan}\left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{1+2 x^3}}\right )}{\sqrt {3}}+\frac {2}{3} \sqrt [3]{2} \log \left (2 x+2^{2/3} \sqrt [3]{1+2 x^3}\right )-\frac {1}{3} \sqrt [3]{2} \log \left (-2 x^2+2^{2/3} x \sqrt [3]{1+2 x^3}-\sqrt [3]{2} \left (1+2 x^3\right )^{2/3}\right ) \]

[Out]

1/28*(2*x^3+1)^(1/3)*(-58*x^6-9*x^3-4)/x^7-2/3*2^(1/3)*arctan(3^(1/2)*x/(-x+2^(2/3)*(2*x^3+1)^(1/3)))*3^(1/2)+
2/3*2^(1/3)*ln(2*x+2^(2/3)*(2*x^3+1)^(1/3))-1/3*2^(1/3)*ln(-2*x^2+2^(2/3)*x*(2*x^3+1)^(1/3)-2^(1/3)*(2*x^3+1)^
(2/3))

________________________________________________________________________________________

Rubi [A]
time = 0.09, antiderivative size = 141, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {594, 597, 12, 503} \begin {gather*} \frac {2 \sqrt [3]{2} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{2 x^3+1}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {29 \sqrt [3]{2 x^3+1}}{14 x}-\frac {1}{3} \sqrt [3]{2} \log \left (4 x^3+1\right )+\sqrt [3]{2} \log \left (-\sqrt [3]{2 x^3+1}-\sqrt [3]{2} x\right )-\frac {\left (2 x^3+1\right )^{4/3}}{7 x^7}-\frac {\sqrt [3]{2 x^3+1}}{28 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/28*(1 + 2*x^3)^(1/3)/x^4 - (29*(1 + 2*x^3)^(1/3))/(14*x) - (1 + 2*x^3)^(4/3)/(7*x^7) + (2*2^(1/3)*ArcTan[(1
- (2*2^(1/3)*x)/(1 + 2*x^3)^(1/3))/Sqrt[3]])/Sqrt[3] - (2^(1/3)*Log[1 + 4*x^3])/3 + 2^(1/3)*Log[-(2^(1/3)*x)
- (1 + 2*x^3)^(1/3)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 503

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si
mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/
(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 594

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*g*(m + 1))), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 597

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (1+2 x^3\right )^{4/3} \left (1+3 x^3\right )}{x^8 \left (1+4 x^3\right )} \, dx &=-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}+\frac {1}{7} \int \frac {\sqrt [3]{1+2 x^3} \left (1+18 x^3\right )}{x^5 \left (1+4 x^3\right )} \, dx\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}+\frac {1}{28} \int \frac {58+120 x^3}{x^2 \left (1+2 x^3\right )^{2/3} \left (1+4 x^3\right )} \, dx\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}-\frac {1}{28} \int \frac {112 x}{\left (1+2 x^3\right )^{2/3} \left (1+4 x^3\right )} \, dx\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}-4 \int \frac {x}{\left (1+2 x^3\right )^{2/3} \left (1+4 x^3\right )} \, dx\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}-4 \text {Subst}\left (\int \frac {x}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1+2 x^3}}\right )\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}+\frac {1}{3} \left (2\ 2^{2/3}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{2} x} \, dx,x,\frac {x}{\sqrt [3]{1+2 x^3}}\right )-\frac {1}{3} \left (2\ 2^{2/3}\right ) \text {Subst}\left (\int \frac {1+\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1+2 x^3}}\right )\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}+\frac {2}{3} \sqrt [3]{2} \log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}\right )-\frac {1}{3} \sqrt [3]{2} \text {Subst}\left (\int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1+2 x^3}}\right )-2^{2/3} \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1+2 x^3}}\right )\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}-\frac {1}{3} \sqrt [3]{2} \log \left (1+\frac {2^{2/3} x^2}{\left (1+2 x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}\right )+\frac {2}{3} \sqrt [3]{2} \log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}\right )-\left (2 \sqrt [3]{2}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 x}{\sqrt [3]{\frac {1}{2}+x^3}}\right )\\ &=-\frac {\sqrt [3]{1+2 x^3}}{28 x^4}-\frac {29 \sqrt [3]{1+2 x^3}}{14 x}-\frac {\left (1+2 x^3\right )^{4/3}}{7 x^7}+\frac {2 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \sqrt [3]{2} \log \left (1+\frac {2^{2/3} x^2}{\left (1+2 x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}\right )+\frac {2}{3} \sqrt [3]{2} \log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1+2 x^3}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.29, size = 157, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{1+2 x^3} \left (-4-9 x^3-58 x^6\right )}{28 x^7}-\frac {2 \sqrt [3]{2} \text {ArcTan}\left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{1+2 x^3}}\right )}{\sqrt {3}}+\frac {2}{3} \sqrt [3]{2} \log \left (2 x+2^{2/3} \sqrt [3]{1+2 x^3}\right )-\frac {1}{3} \sqrt [3]{2} \log \left (-2 x^2+2^{2/3} x \sqrt [3]{1+2 x^3}-\sqrt [3]{2} \left (1+2 x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*x^3)^(1/3)*(-4 - 9*x^3 - 58*x^6))/(28*x^7) - (2*2^(1/3)*ArcTan[(Sqrt[3]*x)/(-x + 2^(2/3)*(1 + 2*x^3)^(
1/3))])/Sqrt[3] + (2*2^(1/3)*Log[2*x + 2^(2/3)*(1 + 2*x^3)^(1/3)])/3 - (2^(1/3)*Log[-2*x^2 + 2^(2/3)*x*(1 + 2*
x^3)^(1/3) - 2^(1/3)*(1 + 2*x^3)^(2/3)])/3

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 9.76, size = 1176, normalized size = 7.49

method	result	size
trager	\(\text {Expression too large to display}\)	\(1176\)
risch	\(\text {Expression too large to display}\)	\(1403\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/28*(58*x^6+9*x^3+4)/x^7*(2*x^3+1)^(1/3)+4*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*ln((516*Root
Of(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*x^3+1710*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf
(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+1440*(2*x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_
Z^2)*RootOf(_Z^3-2)^2*x-516*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4-1710*RootOf(
RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3+516*RootOf(_Z^3-2)^2*x^3+1710*RootOf(RootOf(_
Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3+303*(2*x^3+1)^(1/3)*RootOf(_Z^3-2)*x^2-2880*(2*x^3+1)
^(1/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^2-303*(2*x^3+1)^(2/3)*x+86*RootOf(_Z^3-2)^2+285*
RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2))/(4*x^3+1))-2/3*ln(-(462*RootOf(RootOf(_Z^
3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*x^3-3420*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_
Z^2)^2*RootOf(_Z^3-2)^3*x^3+2880*(2*x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_
Z^3-2)^2*x-462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4+3420*RootOf(RootOf(_Z^3-2
)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3-308*RootOf(_Z^3-2)^2*x^3+2280*RootOf(RootOf(_Z^3-2)^2+6*_Z
*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3-1566*(2*x^3+1)^(1/3)*RootOf(_Z^3-2)*x^2-5760*(2*x^3+1)^(1/3)*RootO
f(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^2+1566*(2*x^3+1)^(2/3)*x-231*RootOf(_Z^3-2)^2+1710*RootOf(Ro
otOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2))/(4*x^3+1))*RootOf(_Z^3-2)-4*ln(-(462*RootOf(RootOf
(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*x^3-3420*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+
36*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+2880*(2*x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*Root
Of(_Z^3-2)^2*x-462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4+3420*RootOf(RootOf(_Z
^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3-308*RootOf(_Z^3-2)^2*x^3+2280*RootOf(RootOf(_Z^3-2)^2+
6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3-1566*(2*x^3+1)^(1/3)*RootOf(_Z^3-2)*x^2-5760*(2*x^3+1)^(1/3)*R
ootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^2+1566*(2*x^3+1)^(2/3)*x-231*RootOf(_Z^3-2)^2+1710*RootO
f(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2))/(4*x^3+1))*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf
(_Z^3-2)+36*_Z^2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (123) = 246\).
time = 1.62, size = 291, normalized size = 1.85 \begin {gather*} -\frac {56 \, \sqrt {3} 2^{\frac {1}{3}} x^{7} \arctan \left (-\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} {\left (20 \, x^{8} + 10 \, x^{5} - x^{2}\right )} {\left (2 \, x^{3} + 1\right )}^{\frac {1}{3}} - 6 \, \sqrt {3} 2^{\frac {1}{3}} {\left (8 \, x^{7} - 2 \, x^{4} - x\right )} {\left (2 \, x^{3} + 1\right )}^{\frac {2}{3}} + \sqrt {3} {\left (8 \, x^{9} + 60 \, x^{6} + 24 \, x^{3} - 1\right )}}{3 \, {\left (152 \, x^{9} + 60 \, x^{6} - 12 \, x^{3} - 1\right )}}\right ) - 56 \cdot 2^{\frac {1}{3}} x^{7} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (2 \, x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (4 \, x^{3} + 1\right )} + 6 \, {\left (2 \, x^{3} + 1\right )}^{\frac {2}{3}} x}{4 \, x^{3} + 1}\right ) + 28 \cdot 2^{\frac {1}{3}} x^{7} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (2 \, x^{4} - x\right )} {\left (2 \, x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (20 \, x^{6} + 10 \, x^{3} - 1\right )} + 12 \, {\left (x^{5} + x^{2}\right )} {\left (2 \, x^{3} + 1\right )}^{\frac {1}{3}}}{16 \, x^{6} + 8 \, x^{3} + 1}\right ) + 9 \, {\left (58 \, x^{6} + 9 \, x^{3} + 4\right )} {\left (2 \, x^{3} + 1\right )}^{\frac {1}{3}}}{252 \, x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/252*(56*sqrt(3)*2^(1/3)*x^7*arctan(-1/3*(6*sqrt(3)*2^(2/3)*(20*x^8 + 10*x^5 - x^2)*(2*x^3 + 1)^(1/3) - 6*sq
rt(3)*2^(1/3)*(8*x^7 - 2*x^4 - x)*(2*x^3 + 1)^(2/3) + sqrt(3)*(8*x^9 + 60*x^6 + 24*x^3 - 1))/(152*x^9 + 60*x^6
- 12*x^3 - 1)) - 56*2^(1/3)*x^7*log((6*2^(1/3)*(2*x^3 + 1)^(1/3)*x^2 + 2^(2/3)*(4*x^3 + 1) + 6*(2*x^3 + 1)^(2
/3)*x)/(4*x^3 + 1)) + 28*2^(1/3)*x^7*log((3*2^(2/3)*(2*x^4 - x)*(2*x^3 + 1)^(2/3) - 2^(1/3)*(20*x^6 + 10*x^3 -
1) + 12*(x^5 + x^2)*(2*x^3 + 1)^(1/3))/(16*x^6 + 8*x^3 + 1)) + 9*(58*x^6 + 9*x^3 + 4)*(2*x^3 + 1)^(1/3))/x^7

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (2 x^{3} + 1\right )^{\frac {4}{3}} \cdot \left (3 x^{3} + 1\right )}{x^{8} \cdot \left (4 x^{3} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3+1)**(4/3)*(3*x**3+1)/x**8/(4*x**3+1),x)

[Out]

Integral((2*x**3 + 1)**(4/3)*(3*x**3 + 1)/(x**8*(4*x**3 + 1)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (2\,x^3+1\right )}^{4/3}\,\left (3\,x^3+1\right )}{x^8\,\left (4\,x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________