\(\int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx\) [278]

Optimal result

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

[Out]

6*x^(1/3)+x-6*arctanh(x^(1/3))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {383, 470, 327, 213} \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=-6 \text {arctanh}\left (\sqrt [3]{x}\right )+x+6 \sqrt [3]{x} \]

[In]

Int[(1 + x^(2/3))/(-1 + x^(2/3)),x]

[Out]

6*x^(1/3) + x - 6*ArcTanh[x^(1/3)]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 383

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, Dis
t[g, Subst[Int[x^(g - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}
, x] && NeQ[b*c - a*d, 0] && FractionQ[n]

Rule 470

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=6 \sqrt [3]{x}+x-6 \text {arctanh}\left (\sqrt [3]{x}\right ) \]

[In]

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

[Out]

6*x^(1/3) + x - 6*ArcTanh[x^(1/3)]

Maple [A] (verified)

Time = 3.97 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(x +6 x^{\frac {1}{3}}+3 \ln \left (x^{\frac {1}{3}}-1\right )-3 \ln \left (1+x^{\frac {1}{3}}\right )\)	\(24\)
default	\(x +6 x^{\frac {1}{3}}+3 \ln \left (x^{\frac {1}{3}}-1\right )-3 \ln \left (1+x^{\frac {1}{3}}\right )\)	\(24\)
trager	\(-2+x +6 x^{\frac {1}{3}}+3 \ln \left (-\frac {2 x^{\frac {2}{3}}-2 x^{\frac {1}{3}}-x +1}{1+x}\right )\)	\(34\)
meijerg	\(-\frac {3 i \left (2 i x^{\frac {1}{3}}-2 i \operatorname {arctanh}\left (x^{\frac {1}{3}}\right )\right )}{2}+\frac {3 i \left (-\frac {2 i x^{\frac {1}{3}} \left (5 x^{\frac {2}{3}}+15\right )}{15}+2 i \operatorname {arctanh}\left (x^{\frac {1}{3}}\right )\right )}{2}\)	\(43\)

[In]

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

[Out]

x+6*x^(1/3)+3*ln(x^(1/3)-1)-3*ln(1+x^(1/3))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=x + 6 \, x^{\frac {1}{3}} - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) + 3 \, \log \left (x^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

x + 6*x^(1/3) - 3*log(x^(1/3) + 1) + 3*log(x^(1/3) - 1)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=6 \sqrt [3]{x} + x + 3 \log {\left (\sqrt [3]{x} - 1 \right )} - 3 \log {\left (\sqrt [3]{x} + 1 \right )} \]

[In]

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

[Out]

6*x**(1/3) + x + 3*log(x**(1/3) - 1) - 3*log(x**(1/3) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=x + 6 \, x^{\frac {1}{3}} - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) + 3 \, \log \left (x^{\frac {1}{3}} - 1\right ) \]

[In]

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

[Out]

x + 6*x^(1/3) - 3*log(x^(1/3) + 1) + 3*log(x^(1/3) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=x + 6 \, x^{\frac {1}{3}} - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) + 3 \, \log \left ({\left | x^{\frac {1}{3}} - 1 \right |}\right ) \]

[In]

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

[Out]

x + 6*x^(1/3) - 3*log(x^(1/3) + 1) + 3*log(abs(x^(1/3) - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {1+x^{2/3}}{-1+x^{2/3}} \, dx=x-6\,\mathrm {atanh}\left (x^{1/3}\right )+6\,x^{1/3} \]

[In]

int((x^(2/3) + 1)/(x^(2/3) - 1),x)

[Out]

x - 6*atanh(x^(1/3)) + 6*x^(1/3)