\(\int \frac {1}{(b+a x^2) \sqrt [3]{x+x^3}} \, dx\) [2383]

Optimal result

Integrand size = 19, antiderivative size = 191 \[ \int \frac {1}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x}{\sqrt [3]{a-b} x-2 \sqrt [3]{b} \sqrt [3]{x+x^3}}\right )}{2 \sqrt [3]{a-b} b^{2/3}}+\frac {\log \left (\sqrt [3]{a-b} x+\sqrt [3]{b} \sqrt [3]{x+x^3}\right )}{2 \sqrt [3]{a-b} b^{2/3}}-\frac {\log \left ((a-b)^{2/3} x^2-\sqrt [3]{a-b} \sqrt [3]{b} x \sqrt [3]{x+x^3}+b^{2/3} \left (x+x^3\right )^{2/3}\right )}{4 \sqrt [3]{a-b} b^{2/3}} \] Output:

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

Mathematica [A] (verified)

Time = 5.00 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=-\frac {\sqrt [3]{x} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a-b} x^{2/3}}{\sqrt [3]{a-b} x^{2/3}-2 \sqrt [3]{b} \sqrt [3]{1+x^2}}\right )-2 \log \left (\sqrt [3]{a-b} x^{2/3}+\sqrt [3]{b} \sqrt [3]{1+x^2}\right )+\log \left ((a-b)^{2/3} x^{4/3}-\sqrt [3]{a-b} \sqrt [3]{b} x^{2/3} \sqrt [3]{1+x^2}+b^{2/3} \left (1+x^2\right )^{2/3}\right )\right )}{4 \sqrt [3]{a-b} b^{2/3} \sqrt [3]{x+x^3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2467, 368, 965, 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 368

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

\(\Big \downarrow \) 965

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

\(\Big \downarrow \) 901

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 368

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

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 965

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 
 x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 
 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free 
Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x -2 {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}\right )}{3 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {\left (\frac {a -b}{b}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a -b}{b}\right )^{\frac {1}{3}} {\left (x \left (x^{2}+1\right )\right )}^{\frac {1}{3}} x +{\left (x \left (x^{2}+1\right )\right )}^{\frac {2}{3}}}{x^{2}}\right )}{2}}{2 \left (\frac {a -b}{b}\right )^{\frac {1}{3}} b}\)	\(150\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-3/4*(x^3 + x)/((a*x^(7/3) + b*x^(1/3))*(x^2 + 1)^(1/3)) + integrate(3/2*( 
b*x^2 + b)/((a^2*x^(13/3) + 2*a*b*x^(7/3) + b^2*x^(1/3))*(x^2 + 1)^(1/3)), 
 x)
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\frac {3 \, {\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a - b}{b}\right )^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a b^{2} - \sqrt {3} b^{3}\right )}} - \frac {{\left (-a b^{2} + b^{3}\right )}^{\frac {2}{3}} \log \left (\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} + \left (-\frac {a - b}{b}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a b^{2} - b^{3}\right )}} + \frac {\left (-\frac {a - b}{b}\right )^{\frac {2}{3}} \log \left ({\left | -\left (-\frac {a - b}{b}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a - b\right )}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (b+a x^2\right ) \sqrt [3]{x+x^3}} \, dx=\int \frac {1}{x^{\frac {7}{3}} \left (x^{2}+1\right )^{\frac {1}{3}} a +x^{\frac {1}{3}} \left (x^{2}+1\right )^{\frac {1}{3}} b}d x \] Input:

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

Output:

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