\(\int \frac {1}{\sqrt [3]{-1+x^2} (3+x^2)} \, dx\) [2626]

Optimal result

Integrand size = 17, antiderivative size = 231 \[ \int \frac {1}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=\frac {\arctan \left (\frac {2^{2/3} x}{2^{2/3} \sqrt {3}+2 \sqrt {3} \sqrt [3]{-1+x^2}}\right )}{3\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}\left (\frac {3\ 2^{2/3} x \sqrt [3]{-1+x^2}}{-3 \sqrt [3]{2}-\sqrt [3]{2} x^2+3\ 2^{2/3} \sqrt [3]{-1+x^2}-6 \left (-1+x^2\right )^{2/3}}\right )}{6\ 2^{2/3}}-\frac {i \text {arctanh}\left (\frac {2 i \sqrt [3]{2} \sqrt {3} x-i 2^{2/3} \sqrt {3} x \sqrt [3]{-1+x^2}}{-3 \sqrt [3]{2}+\sqrt [3]{2} x^2+3\ 2^{2/3} \sqrt [3]{-1+x^2}-6 \left (-1+x^2\right )^{2/3}}\right )}{6\ 2^{2/3} \sqrt {3}} \] Output:

1/18*arctan(2^(2/3)*x/(2^(2/3)*3^(1/2)+2*(x^2-1)^(1/3)*3^(1/2)))*2^(1/3)*3 
^(1/2)-1/12*arctanh(3*2^(2/3)*x*(x^2-1)^(1/3)/(-3*2^(1/3)-2^(1/3)*x^2+3*2^ 
(2/3)*(x^2-1)^(1/3)-6*(x^2-1)^(2/3)))*2^(1/3)-1/36*I*arctanh((2*I*2^(1/3)* 
3^(1/2)*x-I*2^(2/3)*3^(1/2)*x*(x^2-1)^(1/3))/(-3*2^(1/3)+2^(1/3)*x^2+3*2^( 
2/3)*(x^2-1)^(1/3)-6*(x^2-1)^(2/3)))*2^(1/3)*3^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 4.44 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=-\frac {9 x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{-1+x^2} \left (3+x^2\right ) \left (-9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )+2 x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},x^2,-\frac {x^2}{3}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},x^2,-\frac {x^2}{3}\right )\right )\right )} \] Input:

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

Output:

(-9*x*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2])/((-1 + x^2)^(1/3)*(3 + x^ 
2)*(-9*AppellF1[1/2, 1/3, 1, 3/2, x^2, -1/3*x^2] + 2*x^2*(AppellF1[3/2, 1/ 
3, 2, 5/2, x^2, -1/3*x^2] - AppellF1[3/2, 4/3, 1, 5/2, x^2, -1/3*x^2])))
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.59, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {305}

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.

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

\(\Big \downarrow \) 305

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 305

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit 
h[{q = Rt[-b/a, 2]}, Simp[q*(ArcTan[Sqrt[3]/(q*x)]/(2*2^(2/3)*Sqrt[3]*a^(1/ 
3)*d)), x] + (Simp[q*(ArcTanh[(a^(1/3)*q*x)/(a^(1/3) + 2^(1/3)*(a + b*x^2)^ 
(1/3))]/(2*2^(2/3)*a^(1/3)*d)), x] - Simp[q*(ArcTanh[q*x]/(6*2^(2/3)*a^(1/3 
)*d)), x] + Simp[q*(ArcTan[Sqrt[3]*((a^(1/3) - 2^(1/3)*(a + b*x^2)^(1/3))/( 
a^(1/3)*q*x))]/(2*2^(2/3)*Sqrt[3]*a^(1/3)*d)), x])] /; FreeQ[{a, b, c, d}, 
x] && NeQ[b*c - a*d, 0] && EqQ[b*c + 3*a*d, 0] && NegQ[b/a]
 

Maple [C] (verified)

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

Time = 11.42 (sec) , antiderivative size = 874, normalized size of antiderivative = 3.78

method	result	size
trager	\(\text {Expression too large to display}\)	\(874\)

Input:

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

Output:

-1/432*ln((RootOf(_Z^6+108)^4*x^6-72*x^5*RootOf(_Z^6+108)^4+225*RootOf(_Z^ 
6+108)^4*x^4+36*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^5+72*RootOf(_Z^6+108)^4 
*x^3-648*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^4-189*RootOf(_Z^6+108)^4*x^2+8 
64*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^3+648*(x^2-1)^(2/3)*x^4+648*RootOf(_ 
Z^6+108)^2*(x^2-1)^(1/3)*x^2-4536*x^3*(x^2-1)^(2/3)+27*RootOf(_Z^6+108)^4- 
324*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x+1944*(x^2-1)^(2/3)*x^2+1944*x*(x^2- 
1)^(2/3))/(x^2+3)^3)*RootOf(_Z^6+108)^4-1/72*ln((RootOf(_Z^6+108)^4*x^6-72 
*x^5*RootOf(_Z^6+108)^4+225*RootOf(_Z^6+108)^4*x^4+36*RootOf(_Z^6+108)^2*( 
x^2-1)^(1/3)*x^5+72*RootOf(_Z^6+108)^4*x^3-648*RootOf(_Z^6+108)^2*(x^2-1)^ 
(1/3)*x^4-189*RootOf(_Z^6+108)^4*x^2+864*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)* 
x^3+648*(x^2-1)^(2/3)*x^4+648*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^2-4536*x^ 
3*(x^2-1)^(2/3)+27*RootOf(_Z^6+108)^4-324*RootOf(_Z^6+108)^2*(x^2-1)^(1/3) 
*x+1944*(x^2-1)^(2/3)*x^2+1944*x*(x^2-1)^(2/3))/(x^2+3)^3)*RootOf(_Z^6+108 
)+1/36*RootOf(_Z^6+108)*ln(-(-486*RootOf(_Z^6+108)-4050*RootOf(_Z^6+108)*x 
^4+72*x^5*RootOf(_Z^6+108)^4-1296*x^5*RootOf(_Z^6+108)+RootOf(_Z^6+108)^4* 
x^6+225*RootOf(_Z^6+108)^4*x^4-189*RootOf(_Z^6+108)^4*x^2+3402*RootOf(_Z^6 
+108)*x^2-72*RootOf(_Z^6+108)^4*x^3+1296*RootOf(_Z^6+108)*x^3-9072*x^3*(x^ 
2-1)^(2/3)+3888*x*(x^2-1)^(2/3)+27*RootOf(_Z^6+108)^4-36*RootOf(_Z^6+108)^ 
2*(x^2-1)^(1/3)*x^5-648*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^4-864*RootOf(_Z 
^6+108)^2*(x^2-1)^(1/3)*x^3+648*RootOf(_Z^6+108)^2*(x^2-1)^(1/3)*x^2+32...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1108 vs. \(2 (164) = 328\).

Time = 0.55 (sec) , antiderivative size = 1108, normalized size of antiderivative = 4.80 \[ \int \frac {1}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/24*(-1/432)^(1/6)*(sqrt(-3) + 1)*log((2*(-1/2)^(2/3)*(5*x^5 - 30*x^3 + s 
qrt(-3)*(5*x^5 - 30*x^3 + 9*x) + 9*x) + 12*(3*x^3 + sqrt(-1/3)*(x^4 - 9*x^ 
2) - 3*x)*(x^2 - 1)^(2/3) - 2*(x^2 - 1)^(1/3)*(864*(-1/432)^(5/6)*(x^4 - 3 
*x^2 - sqrt(-3)*(x^4 - 3*x^2)) - (-1/2)^(1/3)*(x^5 - 18*x^3 - sqrt(-3)*(x^ 
5 - 18*x^3 + 9*x) + 9*x)) - (-1/432)^(1/6)*(x^6 - 69*x^4 + 63*x^2 + sqrt(- 
3)*(x^6 - 69*x^4 + 63*x^2 - 27) - 27))/(x^6 + 9*x^4 + 27*x^2 + 27)) - 1/24 
*(-1/432)^(1/6)*(sqrt(-3) + 1)*log((2*(-1/2)^(2/3)*(5*x^5 - 30*x^3 + sqrt( 
-3)*(5*x^5 - 30*x^3 + 9*x) + 9*x) + 12*(3*x^3 - sqrt(-1/3)*(x^4 - 9*x^2) - 
 3*x)*(x^2 - 1)^(2/3) + 2*(x^2 - 1)^(1/3)*(864*(-1/432)^(5/6)*(x^4 - 3*x^2 
 - sqrt(-3)*(x^4 - 3*x^2)) + (-1/2)^(1/3)*(x^5 - 18*x^3 - sqrt(-3)*(x^5 - 
18*x^3 + 9*x) + 9*x)) + (-1/432)^(1/6)*(x^6 - 69*x^4 + 63*x^2 + sqrt(-3)*( 
x^6 - 69*x^4 + 63*x^2 - 27) - 27))/(x^6 + 9*x^4 + 27*x^2 + 27)) - 1/24*(-1 
/432)^(1/6)*(sqrt(-3) - 1)*log((2*(-1/2)^(2/3)*(5*x^5 - 30*x^3 - sqrt(-3)* 
(5*x^5 - 30*x^3 + 9*x) + 9*x) + 12*(3*x^3 + sqrt(-1/3)*(x^4 - 9*x^2) - 3*x 
)*(x^2 - 1)^(2/3) - 2*(x^2 - 1)^(1/3)*(864*(-1/432)^(5/6)*(x^4 - 3*x^2 + s 
qrt(-3)*(x^4 - 3*x^2)) - (-1/2)^(1/3)*(x^5 - 18*x^3 + sqrt(-3)*(x^5 - 18*x 
^3 + 9*x) + 9*x)) - (-1/432)^(1/6)*(x^6 - 69*x^4 + 63*x^2 - sqrt(-3)*(x^6 
- 69*x^4 + 63*x^2 - 27) - 27))/(x^6 + 9*x^4 + 27*x^2 + 27)) + 1/24*(-1/432 
)^(1/6)*(sqrt(-3) - 1)*log((2*(-1/2)^(2/3)*(5*x^5 - 30*x^3 - sqrt(-3)*(5*x 
^5 - 30*x^3 + 9*x) + 9*x) + 12*(3*x^3 - sqrt(-1/3)*(x^4 - 9*x^2) - 3*x)...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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