\(\int \frac {x^3 (2+x^3)}{(1+x^3)^{2/3} (1+x^3+x^6)} \, dx\) [7]

Optimal result

Integrand size = 28, antiderivative size = 85 \[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} x^2}{x^2-2 \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (x^2+\sqrt [3]{1+x^3}\right )+\frac {1}{6} \log \left (x^4-x^2 \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]

[Out] -1/3*arctan(3^(1/2)*x^2/(x^2-2*(x^3+1)^(1/3)))*3^(1/2)-1/3*ln(x^2+(x^3 
+1)^(1/3))+1/6*ln(x^4-x^2*(x^3+1)^(1/3)+(x^3+1)^(2/3))
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.40 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.58, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {7281, 778, 936} \[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=\frac {\left (-\sqrt {3}+i\right ) x \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-x^3,-\frac {2 x^3}{1-i \sqrt {3}}\right )}{\sqrt {3}+i}+\frac {\left (\sqrt {3}+i\right ) x \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-x^3,-\frac {2 x^3}{1+i \sqrt {3}}\right )}{-\sqrt {3}+i}+x \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-x^3\right ) \]

[In] Int[(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6)),x]
 

[Out] ((I - Sqrt[3])*x*AppellF1[1/3, 2/3, 1, 4/3, -x^3, (-2*x^3)/(1 - I*Sqrt 
[3])])/(I + Sqrt[3]) + ((I + Sqrt[3])*x*AppellF1[1/3, 2/3, 1, 4/3, -x^ 
3, (-2*x^3)/(1 + I*Sqrt[3])])/(I - Sqrt[3]) + x*Hypergeometric2F1[1/3, 
 2/3, 4/3, -x^3]
 

Rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

Rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

Rule 7281

Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] 
 Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /;  !FalseQ[lst]]
 

Rubi steps \begin{align*} \text {integral}= \int \left (\frac {1}{\left (1+x^3\right )^{2/3}}-\frac {1-x^3}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )}\right ) \, dx \\ = \int \frac {1}{\left (1+x^3\right )^{2/3}} \, dx-\int \frac {1-x^3}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx \\ = x \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-x^3\right )-\int \left (\frac {-1-i \sqrt {3}}{\left (1+x^3\right )^{2/3} \left (1-i \sqrt {3}+2 x^3\right )}+\frac {-1+i \sqrt {3}}{\left (1+x^3\right )^{2/3} \left (1+i \sqrt {3}+2 x^3\right )}\right ) \, dx \\ = x \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-x^3\right )-\left (-1-i \sqrt {3}\right ) \int \frac {1}{\left (1+x^3\right )^{2/3} \left (1-i \sqrt {3}+2 x^3\right )} \, dx-\left (-1+i \sqrt {3}\right ) \int \frac {1}{\left (1+x^3\right )^{2/3} \left (1+i \sqrt {3}+2 x^3\right )} \, dx \\ = -\frac {\left (-1-i \sqrt {3}\right ) x F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {2 x^3}{1-i \sqrt {3}}\right )}{1-i \sqrt {3}}-\frac {\left (-1+i \sqrt {3}\right ) x F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-x^3,-\frac {2 x^3}{1+i \sqrt {3}}\right )}{1+i \sqrt {3}}+x \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00 \[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} x^2}{x^2-2 \sqrt [3]{1+x^3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (x^2+\sqrt [3]{1+x^3}\right )+\frac {1}{6} \log \left (x^4-x^2 \sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \]

[In] Integrate[(x^3*(2 + x^3))/((1 + x^3)^(2/3)*(1 + x^3 + x^6)),x]
 

[Out] -(ArcTan[(Sqrt[3]*x^2)/(x^2 - 2*(1 + x^3)^(1/3))]/Sqrt[3]) - Log[x^2 + 
 (1 + x^3)^(1/3)]/3 + Log[x^4 - x^2*(1 + x^3)^(1/3) + (1 + x^3)^(2/3)] 
/6
 

Maple [C] (verified)

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

Time = 0.53 (sec) , antiderivative size = 292, normalized size of antiderivative = 3.44

method	result	size
trager	\(-\frac {\ln \left (\frac {-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{6}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{4}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x^{2}+2 \left (x^{3}+1\right )^{\frac {1}{3}} x^{4}+\left (x^{3}+1\right )^{\frac {2}{3}} x^{2}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{3}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )}{x^{6}+x^{3}+1}\right )}{3}+\operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (-\frac {-9 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{6}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{6}+6 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}} x^{4}+3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}} x^{2}-\left (x^{3}+1\right )^{\frac {1}{3}} x^{4}-2 \left (x^{3}+1\right )^{\frac {2}{3}} x^{2}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{3}-3 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )}{x^{6}+x^{3}+1}\right )\)	\(292\)

[In] int(x^3*(x^3+2)/(x^3+1)^(2/3)/(x^6+x^3+1),x,method=_RETURNVERBOSE)
 

[Out] -1/3*ln((-9*RootOf(9*_Z^2-3*_Z+1)^2*x^6-3*RootOf(9*_Z^2-3*_Z+1)*(x^3+1 
)^(1/3)*x^4+3*RootOf(9*_Z^2-3*_Z+1)*(x^3+1)^(2/3)*x^2+2*(x^3+1)^(1/3)* 
x^4+(x^3+1)^(2/3)*x^2+3*RootOf(9*_Z^2-3*_Z+1)*x^3+3*RootOf(9*_Z^2-3*_Z 
+1))/(x^6+x^3+1))+RootOf(9*_Z^2-3*_Z+1)*ln(-(-9*RootOf(9*_Z^2-3*_Z+1)^ 
2*x^6+3*RootOf(9*_Z^2-3*_Z+1)*x^6+6*RootOf(9*_Z^2-3*_Z+1)*(x^3+1)^(1/3 
)*x^4+3*RootOf(9*_Z^2-3*_Z+1)*(x^3+1)^(2/3)*x^2-(x^3+1)^(1/3)*x^4-2*(x 
^3+1)^(2/3)*x^2-3*RootOf(9*_Z^2-3*_Z+1)*x^3-3*RootOf(9*_Z^2-3*_Z+1))/( 
x^6+x^3+1))
 

Fricas [A] (verification not implemented)

none

Time = 1.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x^{2} - 2 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x^{2}}\right ) - \frac {1}{6} \, \log \left (\frac {x^{6} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{4} + x^{3} + 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} x^{2} + 1}{x^{6} + x^{3} + 1}\right ) \]

[In] integrate(x^3*(x^3+2)/(x^3+1)^(2/3)/(x^6+x^3+1),x, algorithm="fricas")
 

[Out] -1/3*sqrt(3)*arctan(-1/3*(sqrt(3)*x^2 - 2*sqrt(3)*(x^3 + 1)^(1/3))/x^2 
) - 1/6*log((x^6 + 3*(x^3 + 1)^(1/3)*x^4 + x^3 + 3*(x^3 + 1)^(2/3)*x^2 
 + 1)/(x^6 + x^3 + 1))
 

Sympy [F]

\[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=\int \frac {x^{3} \left (x^{3} + 2\right )}{\left (\left (x + 1\right ) \left (x^{2} - x + 1\right )\right )^{\frac {2}{3}} \left (x^{6} + x^{3} + 1\right )}\, dx \]

[In] integrate(x**3*(x**3+2)/(x**3+1)**(2/3)/(x**6+x**3+1),x)
 

[Out] Integral(x**3*(x**3 + 2)/(((x + 1)*(x**2 - x + 1))**(2/3)*(x**6 + x**3 
 + 1)), x)
 

Maxima [F]

\[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 2\right )} x^{3}}{{\left (x^{6} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}} \,d x } \]

[In] integrate(x^3*(x^3+2)/(x^3+1)^(2/3)/(x^6+x^3+1),x, algorithm="maxima")
 

[Out] integrate((x^3 + 2)*x^3/((x^6 + x^3 + 1)*(x^3 + 1)^(2/3)), x)
 

Giac [F]

\[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 2\right )} x^{3}}{{\left (x^{6} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}} \,d x } \]

[In] integrate(x^3*(x^3+2)/(x^3+1)^(2/3)/(x^6+x^3+1),x, algorithm="giac")
 

[Out] integrate((x^3 + 2)*x^3/((x^6 + x^3 + 1)*(x^3 + 1)^(2/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (2+x^3\right )}{\left (1+x^3\right )^{2/3} \left (1+x^3+x^6\right )} \, dx=\int \frac {x^3\,\left (x^3+2\right )}{{\left (x^3+1\right )}^{2/3}\,\left (x^6+x^3+1\right )} \,d x \]

[In] int((x^3*(x^3 + 2))/((x^3 + 1)^(2/3)*(x^3 + x^6 + 1)),x)
 

[Out] int((x^3*(x^3 + 2))/((x^3 + 1)^(2/3)*(x^3 + x^6 + 1)), x)