\(\int \frac {1}{(1+x) \sqrt [3]{3+2 x+x^2}} \, dx\) [1807]

Optimal result

Integrand size = 18, antiderivative size = 123 \[ \int \frac {1}{(1+x) \sqrt [3]{3+2 x+x^2}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2^{2/3} \sqrt [3]{3+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{3+2 x+x^2}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{3+2 x+x^2}+\sqrt [3]{2} \left (3+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \] Output:

1/4*3^(1/2)*arctan(1/3*3^(1/2)+1/3*2^(2/3)*(x^2+2*x+3)^(1/3)*3^(1/2))*2^(2 
/3)+1/4*ln(-2+2^(2/3)*(x^2+2*x+3)^(1/3))*2^(2/3)-1/8*ln(2+2^(2/3)*(x^2+2*x 
+3)^(1/3)+2^(1/3)*(x^2+2*x+3)^(2/3))*2^(2/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(1+x) \sqrt [3]{3+2 x+x^2}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{3+2 x+x^2}}{\sqrt {3}}\right )+2 \log \left (-2+2^{2/3} \sqrt [3]{3+2 x+x^2}\right )-\log \left (2+2^{2/3} \sqrt [3]{3+2 x+x^2}+\sqrt [3]{2} \left (3+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59, 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, 1082, 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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

((Sqrt[3]*ArcTan[(1 + 2^(2/3)*(3 + x)^(1/3))/Sqrt[3]])/2^(1/3) + (3*Log[-1 
 + 2^(1/3) - x])/(2*2^(1/3)) - Log[(1 + x)^2]/(2*2^(1/3)))/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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[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])] /; Fre 
eQ[{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 = 6.92 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2}+2 x +3\right )^{\frac {1}{3}}+1\right )}{3}\right )+2 \ln \left (\left (x^{2}+2 x +3\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right )-\ln \left (\left (x^{2}+2 x +3\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (x^{2}+2 x +3\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )\right )}{8}\)	\(84\)
trager	\(\text {Expression too large to display}\)	\(1215\)

Input:

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

Output:

1/8*2^(2/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2+2*x+3)^(1/3)+1))+2 
*ln((x^2+2*x+3)^(1/3)-2^(1/3))-ln((x^2+2*x+3)^(2/3)+2^(1/3)*(x^2+2*x+3)^(1 
/3)+2^(2/3)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(1+x) \sqrt [3]{3+2 x+x^2}} \, dx=\frac {1}{2} \cdot 2^{\frac {1}{6}} \sqrt {\frac {3}{2}} \arctan \left (\frac {1}{3} \cdot 2^{\frac {1}{6}} \sqrt {\frac {3}{2}} {\left (2^{\frac {1}{3}} + 2 \, {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 3\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}}\right ) \] Input:

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

Output:

1/2*2^(1/6)*sqrt(3/2)*arctan(1/3*2^(1/6)*sqrt(3/2)*(2^(1/3) + 2*(x^2 + 2*x 
 + 3)^(1/3))) - 1/8*2^(2/3)*log(2^(2/3) + 2^(1/3)*(x^2 + 2*x + 3)^(1/3) + 
(x^2 + 2*x + 3)^(2/3)) + 1/4*2^(2/3)*log(-2^(1/3) + (x^2 + 2*x + 3)^(1/3))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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