\(\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx\) [728]

Optimal result

Integrand size = 20, antiderivative size = 43 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {1-x} \sqrt {1+x}+2 \arcsin (x)-\text {arctanh}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

2*arcsin(x)-arctanh((1-x)^(1/2)*(1+x)^(1/2))-(1-x)^(1/2)*(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {104, 163, 41, 222, 94, 212} \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=2 \arcsin (x)-\text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\sqrt {1-x} \sqrt {x+1} \]

[In]

Int[(1 + x)^(3/2)/(Sqrt[1 - x]*x),x]

[Out]

-(Sqrt[1 - x]*Sqrt[1 + x]) + 2*ArcSin[x] - ArcTanh[Sqrt[1 - x]*Sqrt[1 + x]]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 104

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(m + n + p + 1))), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

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 222

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.26 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {1-x^2}-4 \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )-2 \text {arctanh}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \]

[In]

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

[Out]

-Sqrt[1 - x^2] - 4*ArcTan[Sqrt[1 - x]/Sqrt[1 + x]] - 2*ArcTanh[Sqrt[1 - x]/Sqrt[1 + x]]

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (-\sqrt {-x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )+2 \arcsin \left (x \right )\right )}{\sqrt {-x^{2}+1}}\)	\(51\)

[In]

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

[Out]

(1+x)^(1/2)*(1-x)^(1/2)/(-x^2+1)^(1/2)*(-(-x^2+1)^(1/2)-arctanh(1/(-x^2+1)^(1/2))+2*arcsin(x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.33 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {x + 1} \sqrt {-x + 1} - 4 \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-sqrt(x + 1)*sqrt(-x + 1) - 4*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.95 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=-\sqrt {-x^{2} + 1} + 2 \, \arcsin \left (x\right ) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[In]

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

[Out]

-sqrt(-x^2 + 1) + 2*arcsin(x) - log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (35) = 70\).

Time = 0.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 3.67 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x} \, dx=2 \, \pi - \sqrt {x + 1} \sqrt {-x + 1} + 4 \, \arctan \left (\frac {\sqrt {x + 1} {\left (\frac {{\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{2}}{x + 1} - 1\right )}}{2 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}\right ) - \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \]

[In]

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

[Out]

2*pi - sqrt(x + 1)*sqrt(-x + 1) + 4*arctan(1/2*sqrt(x + 1)*((sqrt(2) - sqrt(-x + 1))^2/(x + 1) - 1)/(sqrt(2) -
sqrt(-x + 1))) - log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)) + 2)) +
log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)) - 2))

Mupad [F(-1)]

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

[In]

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

[Out]

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