\(\int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx\) [213]

Optimal result

Integrand size = 19, antiderivative size = 33 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=6 \sqrt [6]{1+x}+3 \sqrt [3]{1+x}+6 \log \left (1-\sqrt [6]{1+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2037, 1607, 272, 45} \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=3 \sqrt [3]{x+1}+6 \sqrt [6]{x+1}+6 \log \left (1-\sqrt [6]{x+1}\right ) \]

[In]

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

[Out]

6*(1 + x)^(1/6) + 3*(1 + x)^(1/3) + 6*Log[1 - (1 + x)^(1/6)]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2037

Int[((a_.)*(u_)^(j_.) + (b_.)*(u_)^(n_.))^(p_), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(a*x^j + b
*x^n)^p, x], x, u], x] /; FreeQ[{a, b, j, n, p}, x] && LinearQ[u, x] && NeQ[u, x]

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=3 \left (2 \sqrt [6]{1+x}+\sqrt [3]{1+x}+2 \log \left (-1+\sqrt [6]{1+x}\right )\right ) \]

[In]

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

[Out]

3*(2*(1 + x)^(1/6) + (1 + x)^(1/3) + 2*Log[-1 + (1 + x)^(1/6)])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(3 \left (1+x \right )^{\frac {1}{3}}+6 \left (1+x \right )^{\frac {1}{6}}+6 \ln \left (\left (1+x \right )^{\frac {1}{6}}-1\right )\)	\(26\)
default	\(6 \left (1+x \right )^{\frac {1}{6}}+3 \left (1+x \right )^{\frac {1}{3}}+\ln \left (x \right )-\ln \left (\left (1+x \right )^{\frac {1}{3}}+\left (1+x \right )^{\frac {1}{6}}+1\right )+2 \ln \left (\left (1+x \right )^{\frac {1}{6}}-1\right )-2 \ln \left (\left (1+x \right )^{\frac {1}{6}}+1\right )+\ln \left (\left (1+x \right )^{\frac {1}{3}}-\left (1+x \right )^{\frac {1}{6}}+1\right )-\ln \left (1+\sqrt {1+x}\right )+\ln \left (-1+\sqrt {1+x}\right )-\ln \left (\left (1+x \right )^{\frac {2}{3}}+\left (1+x \right )^{\frac {1}{3}}+1\right )+2 \ln \left (\left (1+x \right )^{\frac {1}{3}}-1\right )\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=3 \, {\left (x + 1\right )}^{\frac {1}{3}} + 6 \, {\left (x + 1\right )}^{\frac {1}{6}} + 6 \, \log \left ({\left (x + 1\right )}^{\frac {1}{6}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=6 \sqrt [6]{x + 1} + 3 \sqrt [3]{x + 1} + 6 \log {\left (\sqrt [6]{x + 1} - 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=3 \, {\left (x + 1\right )}^{\frac {1}{3}} + 6 \, {\left (x + 1\right )}^{\frac {1}{6}} + 6 \, \log \left ({\left (x + 1\right )}^{\frac {1}{6}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=3 \, {\left (x + 1\right )}^{\frac {1}{3}} + 6 \, {\left (x + 1\right )}^{\frac {1}{6}} + 6 \, \log \left ({\left | {\left (x + 1\right )}^{\frac {1}{6}} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \[ \int \frac {1}{-\sqrt {1+x}+(1+x)^{2/3}} \, dx=6\,\ln \left ({\left (x+1\right )}^{1/6}-1\right )+3\,{\left (x+1\right )}^{1/3}+6\,{\left (x+1\right )}^{1/6} \]

[In]

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

[Out]

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