\(\int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx\) [833]

Optimal result

Integrand size = 20, antiderivative size = 59 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\frac {2 \sqrt {1+x}}{\sqrt {1-x}}+\frac {(1+x)^{3/2}}{3 (1-x)^{3/2}}-\text {arctanh}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {98, 96, 94, 212} \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=-\text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )+\frac {(x+1)^{3/2}}{3 (1-x)^{3/2}}+\frac {2 \sqrt {x+1}}{\sqrt {1-x}} \]

[In]

Int[Sqrt[1 + x]/((1 - x)^(5/2)*x),x]

[Out]

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

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 96

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

Rule 98

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)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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])

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\frac {(7-5 x) \sqrt {1+x}}{3 (1-x)^{3/2}}-2 \text {arctanh}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \]

[In]

Integrate[Sqrt[1 + x]/((1 - x)^(5/2)*x),x]

[Out]

((7 - 5*x)*Sqrt[1 + x])/(3*(1 - x)^(3/2)) - 2*ArcTanh[Sqrt[1 + x]/Sqrt[1 - x]]

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.47

method	result	size
risch	\(\frac {\left (5 x^{2}-2 x -7\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (-1+x \right ) \sqrt {-\left (-1+x \right ) \left (1+x \right )}\, \sqrt {1-x}\, \sqrt {1+x}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(87\)
default	\(-\frac {\left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{2}-6 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x +5 x \sqrt {-x^{2}+1}+3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )-7 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{3 \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\)	\(93\)

[In]

int((1+x)^(1/2)/(1-x)^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

1/3*(5*x^2-2*x-7)/(-1+x)/(-(-1+x)*(1+x))^(1/2)*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)-arctanh(1/(-x^2+1)^
(1/2))*((1+x)*(1-x))^(1/2)/(1-x)^(1/2)/(1+x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\frac {7 \, x^{2} - {\left (5 \, x - 7\right )} \sqrt {x + 1} \sqrt {-x + 1} + 3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - 14 \, x + 7}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x,x, algorithm="fricas")

[Out]

1/3*(7*x^2 - (5*x - 7)*sqrt(x + 1)*sqrt(-x + 1) + 3*(x^2 - 2*x + 1)*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 14
*x + 7)/(x^2 - 2*x + 1)

Sympy [F]

\[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\int \frac {\sqrt {x + 1}}{x \left (1 - x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((1+x)**(1/2)/(1-x)**(5/2)/x,x)

[Out]

Integral(sqrt(x + 1)/(x*(1 - x)**(5/2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\frac {5 \, x}{3 \, \sqrt {-x^{2} + 1}} + \frac {1}{\sqrt {-x^{2} + 1}} + \frac {4 \, x}{3 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {4}{3 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x,x, algorithm="maxima")

[Out]

5/3*x/sqrt(-x^2 + 1) + 1/sqrt(-x^2 + 1) + 4/3*x/(-x^2 + 1)^(3/2) + 4/3/(-x^2 + 1)^(3/2) - 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. 117 vs. \(2 (45) = 90\).

Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.98 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=-\frac {{\left (5 \, x - 7\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} - \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \]

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x,x, algorithm="giac")

[Out]

-1/3*(5*x - 7)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 - 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. \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x} \, dx=\int \frac {\sqrt {x+1}}{x\,{\left (1-x\right )}^{5/2}} \,d x \]

[In]

int((x + 1)^(1/2)/(x*(1 - x)^(5/2)),x)

[Out]

int((x + 1)^(1/2)/(x*(1 - x)^(5/2)), x)