3.11.16 \(\int \frac {1}{x (3+3 x+x^2) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx\) [1016]

3.11.16.1 Optimal result
3.11.16.2 Mathematica [A] (verified)
3.11.16.3 Rubi [A] (verified)
3.11.16.4 Maple [C] (warning: unable to verify)
3.11.16.5 Fricas [B] (verification not implemented)
3.11.16.6 Sympy [F]
3.11.16.7 Maxima [F]
3.11.16.8 Giac [F]
3.11.16.9 Mupad [F(-1)]

3.11.16.1 Optimal result

Integrand size = 31, antiderivative size = 90 \[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=-\frac {\arctan \left (\frac {1+\frac {2 \sqrt [3]{3} (1+x)}{\sqrt [3]{2+(1+x)^3}}}{\sqrt {3}}\right )}{3^{5/6}}-\frac {\log \left (1-(1+x)^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3} (1+x)-\sqrt [3]{2+(1+x)^3}\right )}{2 \sqrt [3]{3}} \]

output
-1/3*arctan(1/3*(1+2*3^(1/3)*(1+x)/(2+(1+x)^3)^(1/3))*3^(1/2))*3^(1/6)-1/1 
8*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)
 
3.11.16.2 Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.00 \[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \sqrt [3]{3+3 x+3 x^2+x^3}}{2 \sqrt [3]{3}+2 \sqrt [3]{3} x+\sqrt [3]{3+3 x+3 x^2+x^3}}\right )}{3^{5/6}}+\frac {2 \log \left (\sqrt [3]{3}+\sqrt [3]{3} x-\sqrt [3]{3+3 x+3 x^2+x^3}\right )-\log \left (3^{2/3}+2\ 3^{2/3} x+3^{2/3} x^2+\sqrt [3]{3} (1+x) \sqrt [3]{3+3 x+3 x^2+x^3}+\left (3+3 x+3 x^2+x^3\right )^{2/3}\right )}{6 \sqrt [3]{3}} \]

input
Integrate[1/(x*(3 + 3*x + x^2)*(3 + 3*x + 3*x^2 + x^3)^(1/3)),x]
 
output
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))
 
3.11.16.3 Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {940, 938, 25, 901}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{x \left (x^2+3 x+3\right ) \sqrt [3]{x^3+3 x^2+3 x+3}} \, dx\)

\(\Big \downarrow \) 940

\(\displaystyle \int \frac {1}{\left ((x+1)^3-1\right ) \sqrt [3]{(x+1)^3+2}}dx\)

\(\Big \downarrow \) 938

\(\displaystyle \int -\frac {1}{\left (1-(x+1)^3\right ) \sqrt [3]{(x+1)^3+2}}d(x+1)\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {1}{\left (1-(x+1)^3\right ) \sqrt [3]{(x+1)^3+2}}d(x+1)\)

\(\Big \downarrow \) 901

\(\displaystyle -\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}+1}{\sqrt {3}}\right )}{3^{5/6}}-\frac {\log \left (1-(x+1)^3\right )}{6 \sqrt [3]{3}}+\frac {\log \left (\sqrt [3]{3} (x+1)-\sqrt [3]{(x+1)^3+2}\right )}{2 \sqrt [3]{3}}\)

input
Int[1/(x*(3 + 3*x + x^2)*(3 + 3*x + 3*x^2 + x^3)^(1/3)),x]
 
output
-(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))
 

3.11.16.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 901
Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit 
h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S 
qrt[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 938
Int[((a_.) + (b_.)*(u_)^(n_))^(p_.)*((c_.) + (d_.)*(u_)^(n_))^(q_.), x_Symb 
ol] :> Simp[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] && NeQ[u 
, x]
 

rule 940
Int[(u_)^(p_.)*(v_)^(q_.)*(x_)^(m_.), x_Symbol] :> Int[NormalizePseudoBinom 
ial[x^(m/p)*u, x]^p*NormalizePseudoBinomial[v, x]^q, x] /; FreeQ[{p, q}, x] 
 && IntegersQ[p, m/p] && PseudoBinomialPairQ[x^(m/p)*u, v, x]
 
3.11.16.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 17.66 (sec) , antiderivative size = 1673, normalized size of antiderivative = 18.59

method result size
trager \(\text {Expression too large to display}\) \(1673\)

input
int(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x,method=_RETURNVERBOSE)
 
output
1/9*RootOf(_Z^3-9)*ln(-(34139998872*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z 
^3-9)+81*_Z^2)*x^3+102419996616*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9 
)+81*_Z^2)*x^2-6984737648*RootOf(_Z^3-9)*x^3-20954212944*RootOf(_Z^3-9)*x^ 
2-20954212944*RootOf(_Z^3-9)*x+102419996616*RootOf(RootOf(_Z^3-9)^2+9*_Z*R 
ootOf(_Z^3-9)+81*_Z^2)*x-15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(_Z^3-9 
)^2*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*x+1828928511*Root 
Of(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)^2*RootOf(_Z^3-9)^2*x^3-37 
4182374*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9 
)^3*x^3-8232012228*RootOf(_Z^3-9)+2619276618*RootOf(_Z^3-9)^3*RootOf(RootO 
f(_Z^3-9)^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)+40236427242*RootOf(RootOf(_Z^3-9) 
^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)+20428536825*RootOf(RootOf(_Z^3-9)^2+9*_Z*R 
ootOf(_Z^3-9)+81*_Z^2)*RootOf(_Z^3-9)*(x^3+3*x^2+3*x+3)^(1/3)*x^2+40857073 
650*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)*RootOf(RootOf(_Z^3-9)^2+9*_Z*Ro 
otOf(_Z^3-9)+81*_Z^2)*x+5107334328*(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)*x-8512490709*(x^3+3*x^2+3*x+3)^(2/3)- 
15322002984*(x^3+3*x^2+3*x+3)^(2/3)*RootOf(_Z^3-9)^2*RootOf(RootOf(_Z^3-9) 
^2+9*_Z*RootOf(_Z^3-9)+81*_Z^2)+5107334328*RootOf(_Z^3-9)^2*(x^3+3*x^2+3*x 
+3)^(1/3)*x^2+10214668656*(x^3+3*x^2+3*x+3)^(1/3)*RootOf(_Z^3-9)^2*x+20428 
536825*(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)-12802499577*RootOf(RootOf(_Z^3-9)^2+9*_Z*RootO...
 
3.11.16.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (71) = 142\).

Time = 5.03 (sec) , antiderivative size = 458, normalized size of antiderivative = 5.09 \[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=-\frac {1}{54} \cdot 3^{\frac {2}{3}} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (7 \, x^{4} + 28 \, x^{3} + 42 \, x^{2} + 30 \, x + 9\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (31 \, x^{6} + 186 \, x^{5} + 465 \, x^{4} + 666 \, x^{3} + 603 \, x^{2} + 324 \, x + 81\right )} + 9 \, {\left (5 \, x^{5} + 25 \, x^{4} + 50 \, x^{3} + 54 \, x^{2} + 33 \, x + 9\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}}}{x^{6} + 6 \, x^{5} + 15 \, x^{4} + 18 \, x^{3} + 9 \, x^{2}}\right ) + \frac {1}{27} \cdot 3^{\frac {2}{3}} \log \left (\frac {2 \cdot 3^{\frac {2}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x\right )} - 9 \cdot 3^{\frac {1}{3}} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 1\right )} + 9 \, {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} {\left (x + 1\right )}}{x^{3} + 3 \, x^{2} + 3 \, x}\right ) - \frac {1}{9} \cdot 3^{\frac {1}{6}} \arctan \left (\frac {3^{\frac {1}{6}} {\left (12 \cdot 3^{\frac {2}{3}} {\left (7 \, x^{7} + 49 \, x^{6} + 147 \, x^{5} + 240 \, x^{4} + 225 \, x^{3} + 117 \, x^{2} + 27 \, x\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {2}{3}} - 3^{\frac {1}{3}} {\left (127 \, x^{9} + 1143 \, x^{8} + 4572 \, x^{7} + 11070 \, x^{6} + 18414 \, x^{5} + 22032 \, x^{4} + 18900 \, x^{3} + 11178 \, x^{2} + 4131 \, x + 729\right )} - 18 \, {\left (31 \, x^{8} + 248 \, x^{7} + 868 \, x^{6} + 1782 \, x^{5} + 2400 \, x^{4} + 2196 \, x^{3} + 1332 \, x^{2} + 486 \, x + 81\right )} {\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (251 \, x^{9} + 2259 \, x^{8} + 9036 \, x^{7} + 21546 \, x^{6} + 34398 \, x^{5} + 38556 \, x^{4} + 30348 \, x^{3} + 16038 \, x^{2} + 5103 \, x + 729\right )}}\right ) \]

input
integrate(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="fricas")
 
output
-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 + 60 
3*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 + 18414*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 + 343 
98*x^5 + 38556*x^4 + 30348*x^3 + 16038*x^2 + 5103*x + 729))
 
3.11.16.6 Sympy [F]

\[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=\int \frac {1}{x \left (x^{2} + 3 x + 3\right ) \sqrt [3]{x^{3} + 3 x^{2} + 3 x + 3}}\, dx \]

input
integrate(1/x/(x**2+3*x+3)/(x**3+3*x**2+3*x+3)**(1/3),x)
 
output
Integral(1/(x*(x**2 + 3*x + 3)*(x**3 + 3*x**2 + 3*x + 3)**(1/3)), x)
 
3.11.16.7 Maxima [F]

\[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}} {\left (x^{2} + 3 \, x + 3\right )} x} \,d x } \]

input
integrate(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="maxima")
 
output
integrate(1/((x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 3*x + 3)*x), x)
 
3.11.16.8 Giac [F]

\[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=\int { \frac {1}{{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac {1}{3}} {\left (x^{2} + 3 \, x + 3\right )} x} \,d x } \]

input
integrate(1/x/(x^2+3*x+3)/(x^3+3*x^2+3*x+3)^(1/3),x, algorithm="giac")
 
output
integrate(1/((x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 3*x + 3)*x), x)
 
3.11.16.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \left (3+3 x+x^2\right ) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx=\int \frac {1}{x\,\left (x^2+3\,x+3\right )\,{\left (x^3+3\,x^2+3\,x+3\right )}^{1/3}} \,d x \]

input
int(1/(x*(3*x + x^2 + 3)*(3*x + 3*x^2 + x^3 + 3)^(1/3)),x)
 
output
int(1/(x*(3*x + x^2 + 3)*(3*x + 3*x^2 + x^3 + 3)^(1/3)), x)