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

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

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 176, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {580, 583, 12, 494, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {29 \sqrt [3]{2 x^3+1}}{14 x}+\frac {2}{3} \sqrt [3]{2} \log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{2 x^3+1}}+1\right )+\frac {2 \sqrt [3]{2} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{2 x^3+1}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\left (2 x^3+1\right )^{4/3}}{7 x^7}-\frac {\sqrt [3]{2 x^3+1}}{28 x^4}-\frac {1}{3} \sqrt [3]{2} \log \left (-\frac {\sqrt [3]{2} x}{\sqrt [3]{2 x^3+1}}+\frac {2^{2/3} x^2}{\left (2 x^3+1\right )^{2/3}}+1\right ) \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 + (2^(2/3)*x^2)/(1 + 2*x^3)^(2/3) - (2^
(1/3)*x)/(1 + 2*x^3)^(1/3)])/3 + (2*2^(1/3)*Log[1 + (2^(1/3)*x)/(1 + 2*x^3)^(1/3)])/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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 580

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 583

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]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 \operatorname {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 ) \operatorname {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 ) \operatorname {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} \operatorname {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} \operatorname {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 ) \operatorname {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 [C]  time = 0.10, size = 71, normalized size = 0.45 \begin {gather*} -\frac {2 x^2 \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {2 x^3}{4 x^3+1}\right )}{\left (4 x^3+1\right )^{2/3}}-\frac {\sqrt [3]{2 x^3+1} \left (58 x^6+9 x^3+4\right )}{28 x^7} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((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)*(4 + 9*x^3 + 58*x^6))/x^7 - (2*x^2*Hypergeometric2F1[2/3, 2/3, 5/3, (2*x^3)/(1 + 4*x^
3)])/(1 + 4*x^3)^(2/3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.36, 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} \tan ^{-1}\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]

IntegrateAlgebraic[((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

________________________________________________________________________________________

fricas [B]  time = 2.13, 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

________________________________________________________________________________________

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

________________________________________________________________________________________

maple [C]  time = 16.63, size = 1174, normalized size = 7.48

method	result	size
trager	\(\text {Expression too large to display}\)	\(1174\)
risch	\(\text {Expression too large to display}\)	\(1409\)

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)-2/3*ln((3420*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2
*RootOf(_Z^3-2)^3*x^3-462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*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-3420*RootOf(RootOf(_Z^3-2)^
2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3+462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*Roo
tOf(_Z^3-2)^4-2280*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3+308*RootOf(_Z^3-2)^
2*x^3+5760*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*(2*x^3+1)^(1/3)*x^2+1566*(2*x^3+1)^(1/3)*RootO
f(_Z^3-2)*x^2-1566*(2*x^3+1)^(2/3)*x-1710*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)+
231*RootOf(_Z^3-2)^2)/(4*x^3+1))*RootOf(_Z^3-2)-4*ln((3420*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2
)^2*RootOf(_Z^3-2)^3*x^3-462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*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-3420*RootOf(RootOf(_Z^3-
2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3+462*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*
RootOf(_Z^3-2)^4-2280*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3+308*RootOf(_Z^3-
2)^2*x^3+5760*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*(2*x^3+1)^(1/3)*x^2+1566*(2*x^3+1)^(1/3)*Ro
otOf(_Z^3-2)*x^2-1566*(2*x^3+1)^(2/3)*x-1710*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-
2)+231*RootOf(_Z^3-2)^2)/(4*x^3+1))*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)+4*RootOf(RootOf(_Z^3-
2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*ln((306*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2
)^3*x^3+48*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*x^3+36*(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-306*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3
-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3-48*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4+306*R
ootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)*x^3+48*RootOf(_Z^3-2)^2*x^3-72*RootOf(RootO
f(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*(2*x^3+1)^(1/3)*x^2+69*(2*x^3+1)^(1/3)*RootOf(_Z^3-2)*x^2-69*(2*x^3+1
)^(2/3)*x+51*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)+8*RootOf(_Z^3-2)^2)/(4*x^3+1)
)

________________________________________________________________________________________

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

________________________________________________________________________________________

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)

________________________________________________________________________________________

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

________________________________________________________________________________________