\(\int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx\) [238]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=\frac {5 \sqrt {1+x}}{3 (1-x)^{3/2}}+\frac {14 \sqrt {1+x}}{3 \sqrt {1-x}}-\frac {\sqrt {1+x}}{(1-x)^{3/2} x}-3 \text {arctanh}\left (\sqrt {1-x^2}\right ) \] Output:

5/3*(1+x)^(1/2)/(1-x)^(3/2)+14/3*(1+x)^(1/2)/(1-x)^(1/2)-(1+x)^(1/2)/(1-x) 
^(3/2)/x-3*arctanh((-x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=-\frac {\sqrt {1+x} \left (3-19 x+14 x^2\right )}{3 (1-x)^{3/2} x}-6 \text {arctanh}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \] Input:

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

Output:

-1/3*(Sqrt[1 + x]*(3 - 19*x + 14*x^2))/((1 - x)^(3/2)*x) - 6*ArcTanh[Sqrt[ 
1 + x]/Sqrt[1 - x]]
 

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {110, 27, 168, 25, 169, 27, 103, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}-\frac {2}{3} \int -\frac {4 x+5}{2 (1-x)^{3/2} x^2 \sqrt {x+1}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {4 x+5}{(1-x)^{3/2} x^2 \sqrt {x+1}}dx+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (-\int -\frac {5 x+9}{(1-x)^{3/2} x \sqrt {x+1}}dx-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\int \frac {5 x+9}{(1-x)^{3/2} x \sqrt {x+1}}dx-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {1}{3} \left (-\int -\frac {9}{\sqrt {1-x} x \sqrt {x+1}}dx+\frac {14 \sqrt {x+1}}{\sqrt {1-x}}-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (9 \int \frac {1}{\sqrt {1-x} x \sqrt {x+1}}dx+\frac {14 \sqrt {x+1}}{\sqrt {1-x}}-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 103

\(\displaystyle \frac {1}{3} \left (-9 \int \frac {1}{1-(1-x) (x+1)}d\left (\sqrt {1-x} \sqrt {x+1}\right )+\frac {14 \sqrt {x+1}}{\sqrt {1-x}}-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (-9 \text {arctanh}\left (\sqrt {1-x} \sqrt {x+1}\right )+\frac {14 \sqrt {x+1}}{\sqrt {1-x}}-\frac {5 \sqrt {x+1}}{\sqrt {1-x} x}\right )+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x}\)

Input:

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

Output:

(2*Sqrt[1 + x])/(3*(1 - x)^(3/2)*x) + ((14*Sqrt[1 + x])/Sqrt[1 - x] - (5*S 
qrt[1 + x])/(Sqrt[1 - x]*x) - 9*ArcTanh[Sqrt[1 - x]*Sqrt[1 + x]])/3
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 103

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_ 
))), x_] :> Simp[b*f   Subst[Int[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sq 
rt[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 110

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

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S 
imp[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*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.25

method	result	size
risch	\(\frac {\left (14 x^{3}-5 x^{2}-16 x +3\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{3 \left (-1+x \right ) \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, x \sqrt {1-x}\, \sqrt {1+x}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{\sqrt {1-x}\, \sqrt {1+x}}\)	\(95\)
default	\(-\frac {\left (9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{3}-18 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{2}+14 \sqrt {-x^{2}+1}\, x^{2}+9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x -19 x \sqrt {-x^{2}+1}+3 \sqrt {-x^{2}+1}\right ) \sqrt {1+x}}{3 \left (1-x \right )^{\frac {3}{2}} \sqrt {-x^{2}+1}\, x}\)	\(108\)

Input:

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

Output:

1/3*(14*x^3-5*x^2-16*x+3)/(-1+x)/(-(1+x)*(-1+x))^(1/2)/x*((1+x)*(1-x))^(1/ 
2)/(1-x)^(1/2)/(1+x)^(1/2)-3*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)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=\frac {13 \, x^{3} - 26 \, x^{2} - {\left (14 \, x^{2} - 19 \, x + 3\right )} \sqrt {x + 1} \sqrt {-x + 1} + 9 \, {\left (x^{3} - 2 \, x^{2} + x\right )} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + 13 \, x}{3 \, {\left (x^{3} - 2 \, x^{2} + x\right )}} \] Input:

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

Output:

1/3*(13*x^3 - 26*x^2 - (14*x^2 - 19*x + 3)*sqrt(x + 1)*sqrt(-x + 1) + 9*(x 
^3 - 2*x^2 + x)*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + 13*x)/(x^3 - 2*x^2 
 + x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=\frac {14 \, x}{3 \, \sqrt {-x^{2} + 1}} + \frac {3}{\sqrt {-x^{2} + 1}} + \frac {7 \, x}{3 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {4}{3 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} - \frac {1}{{\left (-x^{2} + 1\right )}^{\frac {3}{2}} x} - 3 \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \] Input:

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

Output:

14/3*x/sqrt(-x^2 + 1) + 3/sqrt(-x^2 + 1) + 7/3*x/(-x^2 + 1)^(3/2) + 4/3/(- 
x^2 + 1)^(3/2) - 1/((-x^2 + 1)^(3/2)*x) - 3*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. 211 vs. \(2 (58) = 116\).

Time = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.78 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=-\frac {{\left (11 \, x - 13\right )} \sqrt {x + 1} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}^{2}} - \frac {4 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}}{{\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{2} - 4} - 3 \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} + 2 \right |}\right ) + 3 \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} + \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}} - 2 \right |}\right ) \] Input:

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

Output:

-1/3*(11*x - 13)*sqrt(x + 1)*sqrt(-x + 1)/(x - 1)^2 - 4*((sqrt(2) - sqrt(- 
x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))/(((sqrt(2) - s 
qrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^2 - 4) - 
3*log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqrt(x + 1)/(sqrt(2) - s 
qrt(-x + 1)) + 2)) + 3*log(abs(-(sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) + sqr 
t(x + 1)/(sqrt(2) - sqrt(-x + 1)) - 2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=\int \frac {\sqrt {x+1}}{x^2\,{\left (1-x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.80 \[ \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^2} \, dx=\frac {-9 \sqrt {1-x}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right ) x^{2}+9 \sqrt {1-x}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right ) x +9 \sqrt {1-x}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right ) x^{2}-9 \sqrt {1-x}\, \mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right ) x -9 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right ) x^{2}+9 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )-1\right ) x +9 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right ) x^{2}-9 \sqrt {1-x}\, \mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )}{2}\right )+1\right ) x +14 \sqrt {x +1}\, x^{2}-19 \sqrt {x +1}\, x +3 \sqrt {x +1}}{3 \sqrt {1-x}\, x \left (x -1\right )} \] Input:

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

Output:

( - 9*sqrt( - x + 1)*log( - sqrt(2) + tan(asin(sqrt( - x + 1)/sqrt(2))/2) 
- 1)*x**2 + 9*sqrt( - x + 1)*log( - sqrt(2) + tan(asin(sqrt( - x + 1)/sqrt 
(2))/2) - 1)*x + 9*sqrt( - x + 1)*log( - sqrt(2) + tan(asin(sqrt( - x + 1) 
/sqrt(2))/2) + 1)*x**2 - 9*sqrt( - x + 1)*log( - sqrt(2) + tan(asin(sqrt( 
- x + 1)/sqrt(2))/2) + 1)*x - 9*sqrt( - x + 1)*log(sqrt(2) + tan(asin(sqrt 
( - x + 1)/sqrt(2))/2) - 1)*x**2 + 9*sqrt( - x + 1)*log(sqrt(2) + tan(asin 
(sqrt( - x + 1)/sqrt(2))/2) - 1)*x + 9*sqrt( - x + 1)*log(sqrt(2) + tan(as 
in(sqrt( - x + 1)/sqrt(2))/2) + 1)*x**2 - 9*sqrt( - x + 1)*log(sqrt(2) + t 
an(asin(sqrt( - x + 1)/sqrt(2))/2) + 1)*x + 14*sqrt(x + 1)*x**2 - 19*sqrt( 
x + 1)*x + 3*sqrt(x + 1))/(3*sqrt( - x + 1)*x*(x - 1))