3.823 \(\int \frac{1}{\sqrt{-1+x}+\sqrt{(-1+x)^3}} \, dx\)

Optimal. Leaf size=68 \[ \frac{\sqrt{(x-1)^3} \tan ^{-1}\left (\sqrt{x-1}\right )}{(x-1)^{3/2}}+\tan ^{-1}\left (\sqrt{x-1}\right )-\frac{\sqrt{(x-1)^3} \tanh ^{-1}\left (\sqrt{x-1}\right )}{(x-1)^{3/2}}+\tanh ^{-1}\left (\sqrt{x-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.155333, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {6729, 1593, 6725, 329, 212, 206, 203, 15, 298} \[ \frac{\sqrt{(x-1)^3} \tan ^{-1}\left (\sqrt{x-1}\right )}{(x-1)^{3/2}}+\tan ^{-1}\left (\sqrt{x-1}\right )-\frac{\sqrt{(x-1)^3} \tanh ^{-1}\left (\sqrt{x-1}\right )}{(x-1)^{3/2}}+\tanh ^{-1}\left (\sqrt{x-1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6729

Int[(u_.)/((a_.)*(x_)^(m_.) + (b_.)*Sqrt[(c_.)*(x_)^(n_)]), x_Symbol] :> Int[(u*(a*x^m - b*Sqrt[c*x^n]))/(a^2*
x^(2*m) - b^2*c*x^n), x] /; FreeQ[{a, b, c, m, n}, x]

Rule 1593

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 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-1+x}+\sqrt{(-1+x)^3}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\sqrt{x}+\sqrt{x^3}} \, dx,x,-1+x\right )\\ &=\operatorname{Subst}\left (\int \frac{\sqrt{x}-\sqrt{x^3}}{x-x^3} \, dx,x,-1+x\right )\\ &=\operatorname{Subst}\left (\int \frac{\sqrt{x}-\sqrt{x^3}}{x \left (1-x^2\right )} \, dx,x,-1+x\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt{x} \left (-1+x^2\right )}+\frac{\sqrt{x^3}}{x \left (-1+x^2\right )}\right ) \, dx,x,-1+x\right )\\ &=-\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-1+x^2\right )} \, dx,x,-1+x\right )+\operatorname{Subst}\left (\int \frac{\sqrt{x^3}}{x \left (-1+x^2\right )} \, dx,x,-1+x\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\sqrt{-1+x}\right )\right )+\frac{\sqrt{(-1+x)^3} \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+x^2} \, dx,x,-1+x\right )}{(-1+x)^{3/2}}\\ &=\frac{\left (2 \sqrt{(-1+x)^3}\right ) \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\sqrt{-1+x}\right )}{(-1+x)^{3/2}}+\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{-1+x}\right )+\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-1+x}\right )\\ &=\tan ^{-1}\left (\sqrt{-1+x}\right )+\tanh ^{-1}\left (\sqrt{-1+x}\right )-\frac{\sqrt{(-1+x)^3} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{-1+x}\right )}{(-1+x)^{3/2}}+\frac{\sqrt{(-1+x)^3} \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-1+x}\right )}{(-1+x)^{3/2}}\\ &=\tan ^{-1}\left (\sqrt{-1+x}\right )+\frac{\sqrt{(-1+x)^3} \tan ^{-1}\left (\sqrt{-1+x}\right )}{(-1+x)^{3/2}}+\tanh ^{-1}\left (\sqrt{-1+x}\right )-\frac{\sqrt{(-1+x)^3} \tanh ^{-1}\left (\sqrt{-1+x}\right )}{(-1+x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.158107, size = 64, normalized size = 0.94 \[ \left (\frac{\sqrt{(x-1)^3}}{(x-1)^{3/2}}+1\right ) \tan ^{-1}\left (\sqrt{x-1}\right )+\frac{\left ((x-1)^{3/2}-\sqrt{(x-1)^3}\right ) \tanh ^{-1}\left (\sqrt{x-1}\right )}{(x-1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 40, normalized size = 0.6 \begin{align*} 2\,{\arctan \left ( \sqrt{{\frac{\sqrt{ \left ( x-1 \right ) ^{3}}}{ \left ( x-1 \right ) ^{3/2}}}}\sqrt{x-1} \right ){\frac{1}{\sqrt{{\frac{\sqrt{ \left ( x-1 \right ) ^{3}}}{ \left ( x-1 \right ) ^{3/2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(((x-1)^3)^(1/2)/(x-1)^(3/2))^(1/2)*arctan((((x-1)^3)^(1/2)/(x-1)^(3/2))^(1/2)*(x-1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, \sqrt{x - 1} - \int \frac{\sqrt{x - 1}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x - 1) - integrate(sqrt(x - 1)/x, x)

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Fricas [A]  time = 1.9407, size = 31, normalized size = 0.46 \begin{align*} 2 \, \arctan \left (\sqrt{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x - 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x - 1} + \sqrt{\left (x - 1\right )^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12214, size = 11, normalized size = 0.16 \begin{align*} 2 \, \arctan \left (\sqrt{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x - 1))