\(\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx\) [729]

Optimal result

Integrand size = 20, antiderivative size = 44 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=-\frac {\sqrt {1-x} \sqrt {1+x}}{x}+\arcsin (x)-2 \text {arctanh}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, 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 = {100, 163, 41, 222, 94, 212} \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=\arcsin (x)-2 \text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )-\frac {\sqrt {1-x} \sqrt {x+1}}{x} \]

[In]

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

[Out]

-((Sqrt[1 - x]*Sqrt[1 + x])/x) + ArcSin[x] - 2*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 100

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.34 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=-\frac {\sqrt {1-x^2}+2 x \arctan \left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )+4 x \text {arctanh}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right )}{x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.65 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\left (-2 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x +\arcsin \left (x \right ) x -\sqrt {-x^{2}+1}\right ) \sqrt {1+x}\, \sqrt {1-x}}{x \sqrt {-x^{2}+1}}\)	\(55\)
risch	\(\frac {\left (-1+x \right ) \sqrt {1+x}\, \sqrt {\left (1+x \right ) \left (1-x \right )}}{x \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}}+\frac {\left (\arcsin \left (x \right )-2 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(81\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=-\frac {2 \, x \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 2 \, x \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + \sqrt {x + 1} \sqrt {-x + 1}}{x} \]

[In]

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

[Out]

-(2*x*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 2*x*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + sqrt(x + 1)*sqrt(
-x + 1))/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=-\frac {\sqrt {-x^{2} + 1}}{x} + \arcsin \left (x\right ) - 2 \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[In]

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

[Out]

-sqrt(-x^2 + 1)/x + arcsin(x) - 2*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. 236 vs. \(2 (36) = 72\).

Time = 0.34 (sec) , antiderivative size = 236, normalized size of antiderivative = 5.36 \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=\pi - \frac {4 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}}{{\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{2} - 4} + 2 \, \arctan \left (\frac {\sqrt {x + 1} {\left (\frac {{\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{2}}{x + 1} - 1\right )}}{2 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}\right ) - 2 \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + 2 \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \]

[In]

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

[Out]

pi - 4*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))/(((sqrt(2) - sqrt(-x + 1)
)/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^2 - 4) + 2*arctan(1/2*sqrt(x + 1)*((sqrt(2) - sqrt(-x +
1))^2/(x + 1) - 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)) + 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. \[ \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx=\int \frac {{\left (x+1\right )}^{3/2}}{x^2\,\sqrt {1-x}} \,d x \]

[In]

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

[Out]

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