3.9.23 \(\int \frac {1}{\sqrt {-1+x}+\sqrt {(-1+x)^3}} \, dx\) [823]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 = {6861, 1607, 6857, 335, 218, 212, 209, 15, 304} \begin {gather*} \frac {\sqrt {(x-1)^3} \text {ArcTan}\left (\sqrt {x-1}\right )}{(x-1)^{3/2}}+\text {ArcTan}\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 ) \end {gather*}

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 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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 304

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]

Rule 335

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 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 6857

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 6861

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]

Rubi steps

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

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Mathematica [A]
time = 1.76, size = 67, normalized size = 0.99 \begin {gather*} \tan ^{-1}\left (\sqrt {-1+x}\right )+\tan ^{-1}\left (\frac {\sqrt {-1+3 x-3 x^2+x^3}}{-1+x}\right )+\tanh ^{-1}\left (\sqrt {-1+x}\right )-\tanh ^{-1}\left (\frac {\sqrt {-1+3 x-3 x^2+x^3}}{-1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.03, size = 40, normalized size = 0.59

method	result	size
default	\(\frac {2 \arctan \left (\sqrt {\frac {\sqrt {\left (-1+x \right )^{3}}}{\left (-1+x \right )^{\frac {3}{2}}}}\, \sqrt {-1+x}\right )}{\sqrt {\frac {\sqrt {\left (-1+x \right )^{3}}}{\left (-1+x \right )^{\frac {3}{2}}}}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.33, size = 8, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x - 1}\right ) \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x - 1} + \sqrt {\left (x - 1\right )^{3}}}\, dx \end {gather*}

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 = 3.01, size = 8, normalized size = 0.12 \begin {gather*} 2 \, \arctan \left (\sqrt {x - 1}\right ) \end {gather*}

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))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {\sqrt {x-1}-\sqrt {{\left (x-1\right )}^3}}{{\left (x-1\right )}^3-x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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