\(\int \frac {1}{(1+x) \sqrt [3]{2+2 x+x^2}} \, dx\) [1156]

Optimal result

Integrand size = 18, antiderivative size = 86 \[ \int \frac {1}{(1+x) \sqrt [3]{2+2 x+x^2}} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{2+2 x+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (-1+\sqrt [3]{2+2 x+x^2}\right )-\frac {1}{4} \log \left (1+\sqrt [3]{2+2 x+x^2}+\left (2+2 x+x^2\right )^{2/3}\right ) \] Output:

1/2*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(x^2+2*x+2)^(1/3)*3^(1/2))+1/2*ln(-1+(x 
^2+2*x+2)^(1/3))-1/4*ln(1+(x^2+2*x+2)^(1/3)+(x^2+2*x+2)^(2/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(1+x) \sqrt [3]{2+2 x+x^2}} \, dx=\frac {1}{4} \left (2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{2+2 x+x^2}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{2+2 x+x^2}\right )-\log \left (1+\sqrt [3]{2+2 x+x^2}+\left (2+2 x+x^2\right )^{2/3}\right )\right ) \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1118, 243, 67, 16, 1083, 217}

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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{2} \left (-\frac {3}{2} \int -\frac {1}{x}d\sqrt [3]{x+2}+\frac {3}{2} \int \frac {1}{(x+1)^4+x+2}d\sqrt [3]{x+2}-\frac {1}{2} \log \left ((x+1)^2\right )\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x+1)^4+x+2}d\sqrt [3]{x+2}+\frac {3 \log (-x)}{2}-\frac {1}{2} \log \left ((x+1)^2\right )\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{-(x+1)^4-3}d\left (2 \sqrt [3]{x+2}+1\right )+\frac {3 \log (-x)}{2}-\frac {1}{2} \log \left ((x+1)^2\right )\right )\)

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(-\frac {\ln \left (1+\left (x^{2}+2 x +2\right )^{\frac {1}{3}}+\left (x^{2}+2 x +2\right )^{\frac {2}{3}}\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{2}+\frac {\ln \left (-1+\left (x^{2}+2 x +2\right )^{\frac {1}{3}}\right )}{2}\)	\(67\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {2}{3}}-8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}+2 x +2\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+18 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -2 x^{2}-9 \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+28 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-4 x -7}{\left (1+x \right )^{2}}\right )}{2}-\frac {\ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {2}{3}}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+2 x +2\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+34 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +15 x^{2}+24 \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+28 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+30 x +35}{\left (1+x \right )^{2}}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{2}-\frac {\ln \left (-\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {2}{3}}+8 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +17 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}+2 x +2\right )^{\frac {2}{3}}+15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+34 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +15 x^{2}+24 \left (x^{2}+2 x +2\right )^{\frac {1}{3}}+28 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+30 x +35}{\left (1+x \right )^{2}}\right )}{2}\)	\(432\)

Input:

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

Output:

-1/4*ln(1+(x^2+2*x+2)^(1/3)+(x^2+2*x+2)^(2/3))+1/2*3^(1/2)*arctan(1/3*(2*( 
x^2+2*x+2)^(1/3)+1)*3^(1/2))+1/2*ln(-1+(x^2+2*x+2)^(1/3))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1+x) \sqrt [3]{2+2 x+x^2}} \, dx=\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} + 2 \, x + 2\right )}^{\frac {2}{3}} + {\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} + 2 \, x + 2\right )}^{\frac {1}{3}} - 1\right ) \] Input:

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

Output:

1/2*sqrt(3)*arctan(2/3*sqrt(3)*(x^2 + 2*x + 2)^(1/3) + 1/3*sqrt(3)) - 1/4* 
log((x^2 + 2*x + 2)^(2/3) + (x^2 + 2*x + 2)^(1/3) + 1) + 1/2*log((x^2 + 2* 
x + 2)^(1/3) - 1)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(1/((x**2 + 2*x + 2)**(1/3)*x + (x**2 + 2*x + 2)**(1/3)),x)