\(\int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx\) [1745]

Optimal result

Integrand size = 17, antiderivative size = 118 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {1}{2} a \left (x+x^3\right )^{2/3}+\frac {1}{6} \left (-\sqrt {3} a+3 \sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )+\frac {1}{6} (a-3 b) \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{12} (-a+3 b) \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \] Output:

1/2*a*(x^3+x)^(2/3)+1/6*(-3^(1/2)*a+3*3^(1/2)*b)*arctan(3^(1/2)*x/(x+2*(x^ 
3+x)^(1/3)))+1/6*(a-3*b)*ln(-x+(x^3+x)^(1/3))+1/12*(-a+3*b)*ln(x^2+x*(x^3+ 
x)^(1/3)+(x^3+x)^(2/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.73 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {a x \left (1+x^2\right )}{2 \sqrt [3]{x+x^3}}-\frac {(a-3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{1+x^2}}\right )}{2 \sqrt {3} \sqrt [3]{x+x^3}}+\frac {(a-3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (-x^{2/3}+\sqrt [3]{1+x^2}\right )}{6 \sqrt [3]{x+x^3}}+\frac {(-a+3 b) \sqrt [3]{x} \sqrt [3]{1+x^2} \log \left (x^{4/3}+x^{2/3} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )}{12 \sqrt [3]{x+x^3}} \] Input:

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

Output:

(a*x*(1 + x^2))/(2*(x + x^3)^(1/3)) - ((a - 3*b)*x^(1/3)*(1 + x^2)^(1/3)*A 
rcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*(1 + x^2)^(1/3))])/(2*Sqrt[3]*(x + x^ 
3)^(1/3)) + ((a - 3*b)*x^(1/3)*(1 + x^2)^(1/3)*Log[-x^(2/3) + (1 + x^2)^(1 
/3)])/(6*(x + x^3)^(1/3)) + ((-a + 3*b)*x^(1/3)*(1 + x^2)^(1/3)*Log[x^(4/3 
) + x^(2/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)])/(12*(x + x^3)^(1/3))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.89, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2450, 2009}

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.

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

\(\Big \downarrow \) 2450

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2450

Int[(Pq_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[Expan 
dIntegrand[Pq*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j, n, p}, x] && (Po 
lyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IntegerQ[p] && NeQ[n, j]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 1.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.31

method	result	size
meijerg	\(\frac {3 a \,x^{\frac {8}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -x^{2}\right )}{8}+\frac {3 b \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(36\)
risch	\(\frac {a \left (x^{2}+1\right ) x}{2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}}+\frac {3 b \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}-\frac {a \,x^{\frac {2}{3}} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{2}\right )}{2}\)	\(54\)
pseudoelliptic	\(-\frac {x \left (\frac {\left (-a +3 b \right ) \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x^{2}}{x^{2}}\right )}{2}+\sqrt {3}\, \left (a -3 b \right ) \arctan \left (\frac {\left (2 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )+\left (a -3 b \right ) \ln \left (\frac {{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}-x}{x}\right )+3 {\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}} a \right )}{6 \left (-{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}+x \right ) \left ({\left (\left (x^{2}+1\right ) x \right )}^{\frac {2}{3}}+x \left (x +{\left (\left (x^{2}+1\right ) x \right )}^{\frac {1}{3}}\right )\right )}\)	\(150\)
trager	\(\frac {a \left (x^{3}+x \right )^{\frac {2}{3}}}{2}-\frac {\left (a -3 b \right ) \left (6 \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}-96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-3\right ) \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-6 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +174 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}+36 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )+8\right )+\ln \left (180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2} x^{2}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-144 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -114 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right ) x^{2}-180 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-9 x \left (x^{3}+x \right )^{\frac {1}{3}}-4 x^{2}-96 \operatorname {RootOf}\left (36 \textit {\_Z}^{2}+6 \textit {\_Z} +1\right )-3\right )\right )}{6}\)	\(436\)

Input:

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

Output:

3/8*a*x^(8/3)*hypergeom([1/3,4/3],[7/3],-x^2)+3/2*b*x^(2/3)*hypergeom([1/3 
,1/3],[4/3],-x^2)
 

Fricas [A] (verification not implemented)

Time = 43.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=-\frac {1}{6} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{12} \, {\left (a - 3 \, b\right )} \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) + \frac {1}{2} \, {\left (x^{3} + x\right )}^{\frac {2}{3}} a \] Input:

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

Output:

-1/6*sqrt(3)*(a - 3*b)*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(5 
39*x^2 + 507) - 1274*sqrt(3)*(x^3 + x)^(2/3))/(2205*x^2 + 2197)) + 1/12*(a 
 - 3*b)*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1) + 1/2*(x^3 + x)^( 
2/3)*a
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\frac {1}{2} \, a x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + \frac {1}{6} \, \sqrt {3} {\left (a - 3 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{12} \, {\left (a - 3 \, b\right )} \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{6} \, {\left (a - 3 \, b\right )} \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \] Input:

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

Output:

1/2*a*x^2*(1/x^2 + 1)^(2/3) + 1/6*sqrt(3)*(a - 3*b)*arctan(1/3*sqrt(3)*(2* 
(1/x^2 + 1)^(1/3) + 1)) - 1/12*(a - 3*b)*log((1/x^2 + 1)^(2/3) + (1/x^2 + 
1)^(1/3) + 1) + 1/6*(a - 3*b)*log(abs((1/x^2 + 1)^(1/3) - 1))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {b+a x^2}{\sqrt [3]{x+x^3}} \, dx=\left (\int \frac {x^{\frac {5}{3}}}{\left (x^{2}+1\right )^{\frac {1}{3}}}d x \right ) a +\left (\int \frac {1}{x^{\frac {1}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x \right ) b \] Input:

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

Output:

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