\(\int \frac {1+2 x}{\sqrt [3]{-1+x^2} (3+x^2)} \, dx\) [3072]

Optimal result

Integrand size = 22, antiderivative size = 499 \[ \int \frac {1+2 x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=-\frac {\sqrt [6]{-253+1260 i \sqrt {3}} \arctan \left (\frac {3 \sqrt [3]{-1+x^2}}{-2^{2/3} \sqrt {3}-i 2^{2/3} x+\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{6\ 2^{2/3}}-\frac {\sqrt [6]{-253-1260 i \sqrt {3}} \arctan \left (\frac {3 \sqrt [3]{-1+x^2}}{-2^{2/3} \sqrt {3}+i 2^{2/3} x+\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{6\ 2^{2/3}}+\frac {(-1)^{2/3} \sqrt [3]{-54+35 i \sqrt {3}} \log \left (3\ 2^{2/3}-i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )}{6\ 6^{2/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{-54-35 i \sqrt {3}} \log \left (3\ 2^{2/3}+i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )}{6\ 6^{2/3}}+\frac {\sqrt [3]{54+35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}+2 i \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}-i 2^{2/3} \sqrt {3} x \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}\right )}{12\ 6^{2/3}}+\frac {\sqrt [3]{54-35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}-2 i \sqrt [3]{2} \sqrt {3} x-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}+i 2^{2/3} \sqrt {3} x \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}\right )}{12\ 6^{2/3}} \]

[Out]

-1/12*(-253+1260*I*3^(1/2))^(1/6)*arctan(3*(x^2-1)^(1/3)/(-2^(2/3)*3^(1/2)-I*2^(2/3)*x+(x^2-1)^(1/3)*3^(1/2)))
*2^(1/3)-1/12*(-253-1260*I*3^(1/2))^(1/6)*arctan(3*(x^2-1)^(1/3)/(-2^(2/3)*3^(1/2)+I*2^(2/3)*x+(x^2-1)^(1/3)*3
^(1/2)))*2^(1/3)+1/36*(-1)^(2/3)*(-54+35*I*3^(1/2))^(1/3)*ln(3*2^(2/3)-I*2^(2/3)*3^(1/2)*x+6*(x^2-1)^(1/3))*6^
(1/3)-1/36*(-1)^(1/3)*(-54-35*I*3^(1/2))^(1/3)*ln(3*2^(2/3)+I*2^(2/3)*3^(1/2)*x+6*(x^2-1)^(1/3))*6^(1/3)+1/72*
(54+35*I*3^(1/2))^(1/3)*ln(3*2^(1/3)+2*I*2^(1/3)*3^(1/2)*x-2^(1/3)*x^2-3*2^(2/3)*(x^2-1)^(1/3)-I*2^(2/3)*3^(1/
2)*x*(x^2-1)^(1/3)+6*(x^2-1)^(2/3))*6^(1/3)+1/72*(54-35*I*3^(1/2))^(1/3)*ln(3*2^(1/3)-2*I*2^(1/3)*3^(1/2)*x-2^
(1/3)*x^2-3*2^(2/3)*(x^2-1)^(1/3)+I*2^(2/3)*3^(1/2)*x*(x^2-1)^(1/3)+6*(x^2-1)^(2/3))*6^(1/3)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.43, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1024, 402, 455, 58, 631, 210, 31} \[ \int \frac {1+2 x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\sqrt [3]{2} \sqrt [3]{x^2-1}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3} \left ((-1)^{2/3} \sqrt [3]{2} \sqrt [3]{x^2-1}+1\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{2} \sqrt [3]{x^2-1}+\sqrt [3]{-1}}\right )+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}(x)+\frac {\log \left (x^2+3\right )}{2\ 2^{2/3}}-\frac {3 \log \left (\sqrt [3]{x^2-1}+2^{2/3}\right )}{2\ 2^{2/3}} \]

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 58

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-(b*c - a*d)/b, 3]}, Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && NegQ
[(b*c - a*d)/b]

Rule 210

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 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]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1024

Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Dist[g, Int[(a + c
*x^2)^p*(d + f*x^2)^q, x], x] + Dist[h, Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h,
p, q}, x]

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx+\int \frac {1}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx \\ & = -\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\text {Subst}\left (\int \frac {1}{\sqrt [3]{-1+x} (3+x)} \, dx,x,x^2\right ) \\ & = -\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}+\frac {3}{2} \text {Subst}\left (\int \frac {1}{2 \sqrt [3]{2}-2^{2/3} x+x^2} \, dx,x,\sqrt [3]{-1+x^2}\right )-\frac {3 \text {Subst}\left (\int \frac {1}{2^{2/3}+x} \, dx,x,\sqrt [3]{-1+x^2}\right )}{2\ 2^{2/3}} \\ & = -\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}+\sqrt [3]{-1+x^2}\right )}{2\ 2^{2/3}}+\frac {3 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{2^{2/3}} \\ & = -\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3}}{x}\right )}{2\ 2^{2/3} \sqrt {3}}-\frac {\sqrt {3} \arctan \left (\frac {1-\sqrt [3]{2} \sqrt [3]{-1+x^2}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {(-1)^{2/3} \arctan \left (\frac {\sqrt {3} \left (1+(-1)^{2/3} \sqrt [3]{2} \sqrt [3]{-1+x^2}\right )}{x}\right )}{2\ 2^{2/3} \sqrt {3}}+\frac {1}{6} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}(x)-\frac {1}{2} \left (-\frac {1}{2}\right )^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{-1} x}{\sqrt [3]{-1}+\sqrt [3]{2} \sqrt [3]{-1+x^2}}\right )+\frac {\log \left (3+x^2\right )}{2\ 2^{2/3}}-\frac {3 \log \left (2^{2/3}+\sqrt [3]{-1+x^2}\right )}{2\ 2^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.87 (sec) , antiderivative size = 444, normalized size of antiderivative = 0.89 \[ \int \frac {1+2 x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=\frac {-6 \sqrt [6]{-253+1260 i \sqrt {3}} \arctan \left (\frac {3 \sqrt [3]{-1+x^2}}{-i 2^{2/3} x+\sqrt {3} \left (-2^{2/3}+\sqrt [3]{-1+x^2}\right )}\right )-6 \sqrt [6]{-253-1260 i \sqrt {3}} \arctan \left (\frac {3 \sqrt [3]{-1+x^2}}{i 2^{2/3} x+\sqrt {3} \left (-2^{2/3}+\sqrt [3]{-1+x^2}\right )}\right )+\sqrt [3]{3} \left (2 (-1)^{2/3} \sqrt [3]{-54+35 i \sqrt {3}} \log \left (3\ 2^{2/3}-i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )-2 \sqrt [3]{54+35 i \sqrt {3}} \log \left (3\ 2^{2/3}+i 2^{2/3} \sqrt {3} x+6 \sqrt [3]{-1+x^2}\right )+\sqrt [3]{54+35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}-i \sqrt [3]{2} \sqrt {3} x \left (-2+\sqrt [3]{2} \sqrt [3]{-1+x^2}\right )\right )+\sqrt [3]{54-35 i \sqrt {3}} \log \left (3 \sqrt [3]{2}-\sqrt [3]{2} x^2-3\ 2^{2/3} \sqrt [3]{-1+x^2}+6 \left (-1+x^2\right )^{2/3}+i \sqrt [3]{2} \sqrt {3} x \left (-2+\sqrt [3]{2} \sqrt [3]{-1+x^2}\right )\right )\right )}{36\ 2^{2/3}} \]

[In]

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

[Out]

(-6*(-253 + (1260*I)*Sqrt[3])^(1/6)*ArcTan[(3*(-1 + x^2)^(1/3))/((-I)*2^(2/3)*x + Sqrt[3]*(-2^(2/3) + (-1 + x^
2)^(1/3)))] - 6*(-253 - (1260*I)*Sqrt[3])^(1/6)*ArcTan[(3*(-1 + x^2)^(1/3))/(I*2^(2/3)*x + Sqrt[3]*(-2^(2/3) +
(-1 + x^2)^(1/3)))] + 3^(1/3)*(2*(-1)^(2/3)*(-54 + (35*I)*Sqrt[3])^(1/3)*Log[3*2^(2/3) - I*2^(2/3)*Sqrt[3]*x
+ 6*(-1 + x^2)^(1/3)] - 2*(54 + (35*I)*Sqrt[3])^(1/3)*Log[3*2^(2/3) + I*2^(2/3)*Sqrt[3]*x + 6*(-1 + x^2)^(1/3)
] + (54 + (35*I)*Sqrt[3])^(1/3)*Log[3*2^(1/3) - 2^(1/3)*x^2 - 3*2^(2/3)*(-1 + x^2)^(1/3) + 6*(-1 + x^2)^(2/3)
- I*2^(1/3)*Sqrt[3]*x*(-2 + 2^(1/3)*(-1 + x^2)^(1/3))] + (54 - (35*I)*Sqrt[3])^(1/3)*Log[3*2^(1/3) - 2^(1/3)*x
^2 - 3*2^(2/3)*(-1 + x^2)^(1/3) + 6*(-1 + x^2)^(2/3) + I*2^(1/3)*Sqrt[3]*x*(-2 + 2^(1/3)*(-1 + x^2)^(1/3))]))/
(36*2^(2/3))

Maple [F(-1)]

Timed out.

hanged

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {1+2 x}{\sqrt [3]{-1+x^2} \left (3+x^2\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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