\(\int \frac {1}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\) [146]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [C] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 19, antiderivative size = 113 \[ \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}}+\frac {\text {arctanh}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}} \]

[Out]

-1/12*arctanh(x)*2^(1/3)+1/4*arctanh(x/(1+2^(1/3)*(-x^2+1)^(1/3)))*2^(1/3)+1/12*arctan(3^(1/2)/x)*2^(1/3)*3^(1
/2)+1/12*arctan((1-2^(1/3)*(-x^2+1)^(1/3))*3^(1/2)/x)*2^(1/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {402} \[ \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [3]{2} \sqrt [3]{1-x^2}+1}\right )}{2\ 2^{2/3}}-\frac {\text {arctanh}(x)}{6\ 2^{2/3}} \]

[In]

Int[1/((1 - x^2)^(1/3)*(3 + x^2)),x]

[Out]

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

Rule 402

Int[1/(((a_) + (b_.)*(x_)^2)^(1/3)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{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[Sqr
t[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^{-1}\left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} \sqrt [3]{1-x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\tanh ^{-1}(x)}{6\ 2^{2/3}}+\frac {\tanh ^{-1}\left (\frac {x}{1+\sqrt [3]{2} \sqrt [3]{1-x^2}}\right )}{2\ 2^{2/3}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=-\frac {9 x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )}{\sqrt [3]{1-x^2} \left (3+x^2\right ) \left (-9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{3},1,\frac {3}{2},x^2,-\frac {x^2}{3}\right )+2 x^2 \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{3},2,\frac {5}{2},x^2,-\frac {x^2}{3}\right )-\operatorname {AppellF1}\left (\frac {3}{2},\frac {4}{3},1,\frac {5}{2},x^2,-\frac {x^2}{3}\right )\right )\right )} \]

[In]

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

[Out]

(-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])))

Maple [C] (verified)

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

Time = 64.53 (sec) , antiderivative size = 704, normalized size of antiderivative = 6.23

method result size
trager \(\text {Expression too large to display}\) \(704\)

[In]

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

[Out]

-1/144*ln((2*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x^2-RootOf(_Z^6+108)^4*x^3-3*RootOf(_Z^6+108)^4*x-36*(-x^2+1)^(
1/3)*RootOf(_Z^6+108)^2*x+216*(-x^2+1)^(2/3)*x+126*RootOf(_Z^6+108)*x^2-54*RootOf(_Z^6+108))/(RootOf(_Z^6+108)
^3*x+18)^2/(RootOf(_Z^6+108)^3*x-18))*RootOf(_Z^6+108)^4-1/24*ln((2*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x^2-Root
Of(_Z^6+108)^4*x^3-3*RootOf(_Z^6+108)^4*x-36*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^2*x+216*(-x^2+1)^(2/3)*x+126*Root
Of(_Z^6+108)*x^2-54*RootOf(_Z^6+108))/(RootOf(_Z^6+108)^3*x+18)^2/(RootOf(_Z^6+108)^3*x-18))*RootOf(_Z^6+108)+
1/216*RootOf(_Z^6+108)^4*ln((72*RootOf(_Z^6+108)^4*x^3-1296*RootOf(_Z^6+108)*x^3-RootOf(_Z^6+108)^4*x^6-225*Ro
otOf(_Z^6+108)^4*x^4+4050*RootOf(_Z^6+108)*x^4-72*x^5*RootOf(_Z^6+108)^4+1296*x^5*RootOf(_Z^6+108)+486*RootOf(
_Z^6+108)-27*RootOf(_Z^6+108)^4+189*RootOf(_Z^6+108)^4*x^2-3402*RootOf(_Z^6+108)*x^2+18*RootOf(_Z^6+108)*x^6-1
08*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x^2+324*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^2*x+6*(-x^2+1)^(1/3)*RootOf(_Z^6+
108)^5*x^5+108*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x^4-3888*(-x^2+1)^(2/3)*x+1296*(-x^2+1)^(2/3)*x^4+9072*(-x^2+
1)^(2/3)*x^3+3888*(-x^2+1)^(2/3)*x^2+144*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x^3-36*(-x^2+1)^(1/3)*RootOf(_Z^6+1
08)^2*x^5-54*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^5*x-648*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^2*x^4-864*(-x^2+1)^(1/3)*
RootOf(_Z^6+108)^2*x^3+648*(-x^2+1)^(1/3)*RootOf(_Z^6+108)^2*x^2)/(x^2+3)^3)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.69 (sec) , antiderivative size = 1232, normalized size of antiderivative = 10.90 \[ \int \frac {1}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/10368*432^(5/6)*(-1)^(1/6)*(sqrt(-3) + 1)*log((432^(5/6)*(-1)^(1/6)*(x^6 - 69*x^4 + 63*x^2 + sqrt(-3)*(x^6 -
 69*x^4 + 63*x^2 - 27) - 27) + 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3 + sqrt(-3)*(5*x^5 - 30*x^3 + 9*x) + 9*x)
 + 1728*(9*x^3 - sqrt(3)*(I*x^4 - 9*I*x^2) - 9*x)*(-x^2 + 1)^(2/3) - 432*(2^(2/3)*(-1)^(1/3)*(x^5 - 18*x^3 - s
qrt(-3)*(x^5 - 18*x^3 + 9*x) + 9*x) + 4*432^(1/6)*(-1)^(5/6)*(x^4 - 3*x^2 - sqrt(-3)*(x^4 - 3*x^2)))*(-x^2 + 1
)^(1/3))/(x^6 + 9*x^4 + 27*x^2 + 27)) - 1/10368*432^(5/6)*(-1)^(1/6)*(sqrt(-3) + 1)*log(-(432^(5/6)*(-1)^(1/6)
*(x^6 - 69*x^4 + 63*x^2 + sqrt(-3)*(x^6 - 69*x^4 + 63*x^2 - 27) - 27) - 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3
 + sqrt(-3)*(5*x^5 - 30*x^3 + 9*x) + 9*x) - 1728*(9*x^3 - sqrt(3)*(-I*x^4 + 9*I*x^2) - 9*x)*(-x^2 + 1)^(2/3) +
 432*(2^(2/3)*(-1)^(1/3)*(x^5 - 18*x^3 - sqrt(-3)*(x^5 - 18*x^3 + 9*x) + 9*x) - 4*432^(1/6)*(-1)^(5/6)*(x^4 -
3*x^2 - sqrt(-3)*(x^4 - 3*x^2)))*(-x^2 + 1)^(1/3))/(x^6 + 9*x^4 + 27*x^2 + 27)) - 1/10368*432^(5/6)*(-1)^(1/6)
*(sqrt(-3) - 1)*log((432^(5/6)*(-1)^(1/6)*(x^6 - 69*x^4 + 63*x^2 - sqrt(-3)*(x^6 - 69*x^4 + 63*x^2 - 27) - 27)
 + 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3 - sqrt(-3)*(5*x^5 - 30*x^3 + 9*x) + 9*x) + 1728*(9*x^3 - sqrt(3)*(I*
x^4 - 9*I*x^2) - 9*x)*(-x^2 + 1)^(2/3) - 432*(2^(2/3)*(-1)^(1/3)*(x^5 - 18*x^3 + sqrt(-3)*(x^5 - 18*x^3 + 9*x)
 + 9*x) + 4*432^(1/6)*(-1)^(5/6)*(x^4 - 3*x^2 + sqrt(-3)*(x^4 - 3*x^2)))*(-x^2 + 1)^(1/3))/(x^6 + 9*x^4 + 27*x
^2 + 27)) + 1/10368*432^(5/6)*(-1)^(1/6)*(sqrt(-3) - 1)*log(-(432^(5/6)*(-1)^(1/6)*(x^6 - 69*x^4 + 63*x^2 - sq
rt(-3)*(x^6 - 69*x^4 + 63*x^2 - 27) - 27) - 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3 - sqrt(-3)*(5*x^5 - 30*x^3
+ 9*x) + 9*x) - 1728*(9*x^3 - sqrt(3)*(-I*x^4 + 9*I*x^2) - 9*x)*(-x^2 + 1)^(2/3) + 432*(2^(2/3)*(-1)^(1/3)*(x^
5 - 18*x^3 + sqrt(-3)*(x^5 - 18*x^3 + 9*x) + 9*x) - 4*432^(1/6)*(-1)^(5/6)*(x^4 - 3*x^2 + sqrt(-3)*(x^4 - 3*x^
2)))*(-x^2 + 1)^(1/3))/(x^6 + 9*x^4 + 27*x^2 + 27)) - 1/5184*432^(5/6)*(-1)^(1/6)*log(-(432^(5/6)*(-1)^(1/6)*(
x^6 - 69*x^4 + 63*x^2 - 27) + 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3 + 9*x) - 864*(9*x^3 - sqrt(3)*(I*x^4 - 9*
I*x^2) - 9*x)*(-x^2 + 1)^(2/3) - 432*(2^(2/3)*(-1)^(1/3)*(x^5 - 18*x^3 + 9*x) + 4*432^(1/6)*(-1)^(5/6)*(x^4 -
3*x^2))*(-x^2 + 1)^(1/3))/(x^6 + 9*x^4 + 27*x^2 + 27)) + 1/5184*432^(5/6)*(-1)^(1/6)*log((432^(5/6)*(-1)^(1/6)
*(x^6 - 69*x^4 + 63*x^2 - 27) - 432*2^(1/3)*(-1)^(2/3)*(5*x^5 - 30*x^3 + 9*x) + 864*(9*x^3 - sqrt(3)*(-I*x^4 +
 9*I*x^2) - 9*x)*(-x^2 + 1)^(2/3) + 432*(2^(2/3)*(-1)^(1/3)*(x^5 - 18*x^3 + 9*x) - 4*432^(1/6)*(-1)^(5/6)*(x^4
 - 3*x^2))*(-x^2 + 1)^(1/3))/(x^6 + 9*x^4 + 27*x^2 + 27))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/((1 - x^2)^(1/3)*(x^2 + 3)),x)

[Out]

int(1/((1 - x^2)^(1/3)*(x^2 + 3)), x)