3.11.16 \(\int \frac {1}{x (3+3 x+x^2) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx\) [1016]
Optimal. Leaf size=90 \[ -\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}}{\sqrt {3}}\right )}{3^{5/6}}-\frac {\log \left (1-(1+x)^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3} (1+x)-\sqrt [3]{2+(1+x)^3}\right )}{2 \sqrt [3]{3}} \]
[Out]
-1/3*arctan(1/3*(1+2*3^(1/3)*(1+x)/(2+(1+x)^3)^(1/3))*3^(1/2))*3^(1/6)-1/18*ln(1-(1+x)^3)*3^(2/3)+1/6*ln(3^(1/
3)*(1+x)-(2+(1+x)^3)^(1/3))*3^(2/3)
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Rubi [A]
time = 0.04, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {444, 442, 384}
\begin {gather*} -\frac {\text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}+1}{\sqrt {3}}\right )}{3^{5/6}}-\frac {\log \left (1-(x+1)^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3} (x+1)-\sqrt [3]{(x+1)^3+2}\right )}{2 \sqrt [3]{3}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[1/(x*(3 + 3*x + x^2)*(3 + 3*x + 3*x^2 + x^3)^(1/3)),x]
[Out]
-(ArcTan[(1 + (2*3^(1/3)*(1 + x))/(2 + (1 + x)^3)^(1/3))/Sqrt[3]]/3^(5/6)) - Log[1 - (1 + x)^3]/(6*3^(1/3)) +
Log[3^(1/3)*(1 + x) - (2 + (1 + x)^3)^(1/3)]/(2*3^(1/3))
Rule 384
Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Rule 442
Int[((a_.) + (b_.)*(u_)^(n_))^(p_.)*((c_.) + (d_.)*(u_)^(n_))^(q_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1],
Subst[Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x, u], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && LinearQ[u, x] && N
eQ[u, x]
Rule 444
Int[(u_)^(p_.)*(v_)^(q_.)*(x_)^(m_.), x_Symbol] :> Int[NormalizePseudoBinomial[x^(m/p)*u, x]^p*NormalizePseudo
Binomial[v, x]^q, x] /; FreeQ[{p, q}, x] && IntegersQ[p, m/p] && PseudoBinomialPairQ[x^(m/p)*u, v, x]
Rubi steps
\begin {align*} \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx &=\int \frac {1}{\left (-1+(1+x)^3\right ) \sqrt [3]{2+(1+x)^3}} \, dx\\ &=\text {Subst}\left (\int \frac {1}{\left (-1+x^3\right ) \sqrt [3]{2+x^3}} \, dx,x,1+x\right )\\ &=\text {Subst}\left (\int \frac {1}{-1+3 x^3} \, dx,x,\frac {1+x}{\sqrt [3]{2+(1+x)^3}}\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1+\sqrt [3]{3} x} \, dx,x,\frac {1+x}{\sqrt [3]{2+(1+x)^3}}\right )+\frac {1}{3} \text {Subst}\left (\int \frac {-2-\sqrt [3]{3} x}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {1+x}{\sqrt [3]{2+(1+x)^3}}\right )\\ &=\frac {\log \left (1-\frac {\sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{3 \sqrt [3]{3}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {1+x}{\sqrt [3]{2+(1+x)^3}}\right )-\frac {\text {Subst}\left (\int \frac {\sqrt [3]{3}+2\ 3^{2/3} x}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {1+x}{\sqrt [3]{2+(1+x)^3}}\right )}{6 \sqrt [3]{3}}\\ &=\frac {\log \left (1-\frac {\sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (1+\frac {3^{2/3} (1+x)^2}{\left (2+(1+x)^3\right )^{2/3}}+\frac {\sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{6 \sqrt [3]{3}}+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{\sqrt [3]{3}}\\ &=-\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}}{\sqrt {3}}\right )}{3^{5/6}}+\frac {\log \left (1-\frac {\sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{3 \sqrt [3]{3}}-\frac {\log \left (1+\frac {3^{2/3} (1+x)^2}{\left (2+(1+x)^3\right )^{2/3}}+\frac {\sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}\right )}{6 \sqrt [3]{3}}\\ \end {align*}
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Mathematica [A]
time = 0.36, size = 180, normalized size = 2.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{3+3 x+3 x^2+x^3}}{2 \sqrt [3]{3}+2 \sqrt [3]{3} x+\sqrt [3]{3+3 x+3 x^2+x^3}}\right )}{3^{5/6}}+\frac {2 \log \left (\sqrt [3]{3}+\sqrt [3]{3} x-\sqrt [3]{3+3 x+3 x^2+x^3}\right )-\log \left (3^{2/3}+2\ 3^{2/3} x+3^{2/3} x^2+\sqrt [3]{3} (1+x) \sqrt [3]{3+3 x+3 x^2+x^3}+\left (3+3 x+3 x^2+x^3\right )^{2/3}\right )}{6 \sqrt [3]{3}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[1/(x*(3 + 3*x + x^2)*(3 + 3*x + 3*x^2 + x^3)^(1/3)),x]
[Out]
ArcTan[(Sqrt[3]*(3 + 3*x + 3*x^2 + x^3)^(1/3))/(2*3^(1/3) + 2*3^(1/3)*x + (3 + 3*x + 3*x^2 + x^3)^(1/3))]/3^(5
/6) + (2*Log[3^(1/3) + 3^(1/3)*x - (3 + 3*x + 3*x^2 + x^3)^(1/3)] - Log[3^(2/3) + 2*3^(2/3)*x + 3^(2/3)*x^2 +
3^(1/3)*(1 + x)*(3 + 3*x + 3*x^2 + x^3)^(1/3) + (3 + 3*x + 3*x^2 + x^3)^(2/3)])/(6*3^(1/3))
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 9.40, size = 2499, normalized size = 27.77
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result |
size |
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| trager |
\(\text {Expression too large to display}\) |
\(2499\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x,method=_RETURNVERBOSE)
[Out]
1/9*RootOf(_Z^3-9)*ln(-(15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^
2)*RootOf(_Z^3-9)^2*x-12802499577*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2-4041
776571*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3-53038926819*RootOf(RootOf(_Z^3-9)
^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)-10200674203*RootOf(_Z^3-9)*x^3-30602022609*RootOf(_Z^3-9)*x^2-30602022609*Root
Of(_Z^3-9)*x-32311070361*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x^3-96933211083*RootOf(RootOf(_Z
^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x^2-96933211083*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x-16
744502937*RootOf(_Z^3-9)+6809512275*(x^3+3*x^2+3*x+3)^(2/3)*x+2837496903*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9
)^2+6809512275*(x^3+3*x^2+3*x+3)^(2/3)+5486785533*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootO
f(_Z^3-9)^2*x^2+1732189959*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x^2+548678553
3*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x+1732189959*RootOf(RootOf(_Z^3-9)^2
+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x+15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9
*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^2+2837496903*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2*x^2+567499380
6*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2*x-20428536825*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*
RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)-20428536825*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)*RootOf(RootOf(_Z^3-9
)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x^2-40857073650*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_
Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)*x+1828928511*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3
-9)^2*x^3+577396653*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x^3)/x/(x^2+3*x+3))-
1/9*ln((15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9
)^2*x+23573489562*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2+4041776571*RootOf(Ro
otOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3-74088110052*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_
Z^3-9)+81*_Z^2)-10778070856*RootOf(_Z^3-9)*x^3-32334212568*RootOf(_Z^3-9)*x^2-32334212568*RootOf(_Z^3-9)*x-628
62638832*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x^3-188587916496*RootOf(RootOf(_Z^3-9)^2+9*_Z*Ro
otOf(_Z^3-9)+81*_Z^2)*x^2-188587916496*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x-12702726366*Root
Of(_Z^3-9)+8512490709*(x^3+3*x^2+3*x+3)^(2/3)*x+2269837425*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2+8512490709
*(x^3+3*x^2+3*x+3)^(2/3)-10102924098*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x
^2-1732189959*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x^2-10102924098*RootOf(Roo
tOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x-1732189959*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf
(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x+15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_
Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^2+2269837425*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2*x^2+4539674850*(x^3+3*x^2
+3*x+3)^(1/3)*RootOf(_Z^3-9)^2*x-25537472127*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-
9)+81*_Z^2)*RootOf(_Z^3-9)-25537472127*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)*RootOf(RootOf(_Z^3-9)^2+9*_Z*Roo
tOf(_Z^3-9)+81*_Z^2)*x^2-51074944254*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z
^2)*RootOf(_Z^3-9)*x-3367641366*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x^3-57
7396653*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x^3)/x/(x^2+3*x+3))*RootOf(_Z^3-
9)-ln((15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)
^2*x+23573489562*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2+4041776571*RootOf(Roo
tOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3-74088110052*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z
^3-9)+81*_Z^2)-10778070856*RootOf(_Z^3-9)*x^3-32334212568*RootOf(_Z^3-9)*x^2-32334212568*RootOf(_Z^3-9)*x-6286
2638832*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x^3-188587916496*RootOf(RootOf(_Z^3-9)^2+9*_Z*Roo
tOf(_Z^3-9)+81*_Z^2)*x^2-188587916496*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x-12702726366*RootO
f(_Z^3-9)+8512490709*(x^3+3*x^2+3*x+3)^(2/3)*x+2269837425*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2+8512490709*
(x^3+3*x^2+3*x+3)^(2/3)-10102924098*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x^
2-1732189959*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x^2-10102924098*RootOf(Root
Of(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x-1732189959*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(
_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)^3*x+15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z
^3-9)+81*_Z^2)*RootOf(_Z^3-9)^2+2269837425*(x^3...
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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(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="maxima")
[Out]
integrate(1/((x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 3*x + 3)*x), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 458 vs.
\(2 (71) = 142\).
time = 5.25, size = 458, normalized size = 5.09 \begin {gather*} -\frac {1}{54} \cdot 3^{\frac {2}{3}} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (7 \, x^{4} + 28 \, x^{3} + 42 \, x^{2} + 30 \, x + 9\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (31 \, x^{6} + 186 \, x^{5} + 465 \, x^{4} + 666 \, x^{3} + 603 \, x^{2} + 324 \, x + 81\right )} + 9 \, {\left (5 \, x^{5} + 25 \, x^{4} + 50 \, x^{3} + 54 \, x^{2} + 33 \, x + 9\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}}}{x^{6} + 6 \, x^{5} + 15 \, x^{4} + 18 \, x^{3} + 9 \, x^{2}}\right ) + \frac {1}{27} \cdot 3^{\frac {2}{3}} \log \left (\frac {2 \cdot 3^{\frac {2}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x\right )} - 9 \cdot 3^{\frac {1}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 1\right )} + 9 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} {\left (x + 1\right )}}{x^{3} + 3 \, x^{2} + 3 \, x}\right ) - \frac {1}{9} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {3^{\frac {1}{6}} {\left (12 \cdot 3^{\frac {2}{3}} {\left (7 \, x^{7} + 49 \, x^{6} + 147 \, x^{5} + 240 \, x^{4} + 225 \, x^{3} + 117 \, x^{2} + 27 \, x\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} - 3^{\frac {1}{3}} {\left (127 \, x^{9} + 1143 \, x^{8} + 4572 \, x^{7} + 11070 \, x^{6} + 18414 \, x^{5} + 22032 \, x^{4} + 18900 \, x^{3} + 11178 \, x^{2} + 4131 \, x + 729\right )} - 18 \, {\left (31 \, x^{8} + 248 \, x^{7} + 868 \, x^{6} + 1782 \, x^{5} + 2400 \, x^{4} + 2196 \, x^{3} + 1332 \, x^{2} + 486 \, x + 81\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (251 \, x^{9} + 2259 \, x^{8} + 9036 \, x^{7} + 21546 \, x^{6} + 34398 \, x^{5} + 38556 \, x^{4} + 30348 \, x^{3} + 16038 \, x^{2} + 5103 \, x + 729\right )}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="fricas")
[Out]
-1/54*3^(2/3)*log((3*3^(2/3)*(7*x^4 + 28*x^3 + 42*x^2 + 30*x + 9)*(x^3 + 3*x^2 + 3*x + 3)^(2/3) + 3^(1/3)*(31*
x^6 + 186*x^5 + 465*x^4 + 666*x^3 + 603*x^2 + 324*x + 81) + 9*(5*x^5 + 25*x^4 + 50*x^3 + 54*x^2 + 33*x + 9)*(x
^3 + 3*x^2 + 3*x + 3)^(1/3))/(x^6 + 6*x^5 + 15*x^4 + 18*x^3 + 9*x^2)) + 1/27*3^(2/3)*log((2*3^(2/3)*(x^3 + 3*x
^2 + 3*x) - 9*3^(1/3)*(x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 2*x + 1) + 9*(x^3 + 3*x^2 + 3*x + 3)^(2/3)*(x + 1))
/(x^3 + 3*x^2 + 3*x)) - 1/9*3^(1/6)*arctan(1/3*3^(1/6)*(12*3^(2/3)*(7*x^7 + 49*x^6 + 147*x^5 + 240*x^4 + 225*x
^3 + 117*x^2 + 27*x)*(x^3 + 3*x^2 + 3*x + 3)^(2/3) - 3^(1/3)*(127*x^9 + 1143*x^8 + 4572*x^7 + 11070*x^6 + 1841
4*x^5 + 22032*x^4 + 18900*x^3 + 11178*x^2 + 4131*x + 729) - 18*(31*x^8 + 248*x^7 + 868*x^6 + 1782*x^5 + 2400*x
^4 + 2196*x^3 + 1332*x^2 + 486*x + 81)*(x^3 + 3*x^2 + 3*x + 3)^(1/3))/(251*x^9 + 2259*x^8 + 9036*x^7 + 21546*x
^6 + 34398*x^5 + 38556*x^4 + 30348*x^3 + 16038*x^2 + 5103*x + 729))
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \left (x^{2} + 3 x + 3\right ) \sqrt [3]{x^{3} + 3 x^{2} + 3 x + 3}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x/(x**2+3*x+3)/(x**3+3*x**2+3*x+3)**(1/3),x)
[Out]
Integral(1/(x*(x**2 + 3*x + 3)*(x**3 + 3*x**2 + 3*x + 3)**(1/3)), x)
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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(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="giac")
[Out]
integrate(1/((x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 3*x + 3)*x), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,\left (x^2+3\,x+3\right )\,{\left (x^3+3\,x^2+3\,x+3\right )}^{1/3}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(x*(3*x + x^2 + 3)*(3*x + 3*x^2 + x^3 + 3)^(1/3)),x)
[Out]
int(1/(x*(3*x + x^2 + 3)*(3*x + 3*x^2 + x^3 + 3)^(1/3)), x)
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