3.9.35 \(\int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^3} \, dx\) [835]

Optimal. Leaf size=112 \[ \frac {26 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}-\frac {11}{2} \tanh ^{-1}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

-11/2*arctanh((1-x)^(1/2)*(1+x)^(1/2))+2/3*(1+x)^(1/2)/(1-x)^(3/2)/x^2+26/3*(1+x)^(1/2)/(1-x)^(1/2)-7/6*(1+x)^
(1/2)/x^2/(1-x)^(1/2)-19/6*(1+x)^(1/2)/x/(1-x)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {101, 156, 157, 12, 94, 212} \begin {gather*} -\frac {7 \sqrt {x+1}}{6 \sqrt {1-x} x^2}+\frac {2 \sqrt {x+1}}{3 (1-x)^{3/2} x^2}+\frac {26 \sqrt {x+1}}{3 \sqrt {1-x}}-\frac {19 \sqrt {x+1}}{6 \sqrt {1-x} x}-\frac {11}{2} \tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x]/((1 - x)^(5/2)*x^3),x]

[Out]

(26*Sqrt[1 + x])/(3*Sqrt[1 - x]) + (2*Sqrt[1 + x])/(3*(1 - x)^(3/2)*x^2) - (7*Sqrt[1 + x])/(6*Sqrt[1 - x]*x^2)
- (19*Sqrt[1 + x])/(6*Sqrt[1 - x]*x) - (11*ArcTanh[Sqrt[1 - x]*Sqrt[1 + x]])/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 101

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[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] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] + Dist[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 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> 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] + Dist[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 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*} \int \frac {\sqrt {1+x}}{(1-x)^{5/2} x^3} \, dx &=\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {2}{3} \int \frac {-\frac {7}{2}-3 x}{(1-x)^{3/2} x^3 \sqrt {1+x}} \, dx\\ &=\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}+\frac {1}{3} \int \frac {\frac {19}{2}+7 x}{(1-x)^{3/2} x^2 \sqrt {1+x}} \, dx\\ &=\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}-\frac {1}{3} \int \frac {-\frac {33}{2}-\frac {19 x}{2}}{(1-x)^{3/2} x \sqrt {1+x}} \, dx\\ &=\frac {26 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}+\frac {1}{3} \int \frac {33}{2 \sqrt {1-x} x \sqrt {1+x}} \, dx\\ &=\frac {26 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}+\frac {11}{2} \int \frac {1}{\sqrt {1-x} x \sqrt {1+x}} \, dx\\ &=\frac {26 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}-\frac {11}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x} \sqrt {1+x}\right )\\ &=\frac {26 \sqrt {1+x}}{3 \sqrt {1-x}}+\frac {2 \sqrt {1+x}}{3 (1-x)^{3/2} x^2}-\frac {7 \sqrt {1+x}}{6 \sqrt {1-x} x^2}-\frac {19 \sqrt {1+x}}{6 \sqrt {1-x} x}-\frac {11}{2} \tanh ^{-1}\left (\sqrt {1-x} \sqrt {1+x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 59, normalized size = 0.53 \begin {gather*} -\frac {\sqrt {1+x} \left (3+12 x-71 x^2+52 x^3\right )}{6 (1-x)^{3/2} x^2}-11 \tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {1-x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x]/((1 - x)^(5/2)*x^3),x]

[Out]

-1/6*(Sqrt[1 + x]*(3 + 12*x - 71*x^2 + 52*x^3))/((1 - x)^(3/2)*x^2) - 11*ArcTanh[Sqrt[1 + x]/Sqrt[1 - x]]

________________________________________________________________________________________

Maple [A]
time = 0.08, size = 129, normalized size = 1.15

method	result	size
risch	\(\frac {\left (52 x^{4}-19 x^{3}-59 x^{2}+15 x +3\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{6 x^{2} \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \left (-1+x \right ) \sqrt {1-x}\, \sqrt {1+x}}-\frac {11 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {1-x}\, \sqrt {1+x}}\)	\(100\)
default	\(-\frac {\left (33 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{4}-66 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{3}+52 x^{3} \sqrt {-x^{2}+1}+33 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{2}-71 x^{2} \sqrt {-x^{2}+1}+12 x \sqrt {-x^{2}+1}+3 \sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {1+x}}{6 x^{2} \left (-1+x \right )^{2} \sqrt {-x^{2}+1}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^(1/2)/(1-x)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/6*(33*arctanh(1/(-x^2+1)^(1/2))*x^4-66*arctanh(1/(-x^2+1)^(1/2))*x^3+52*x^3*(-x^2+1)^(1/2)+33*arctanh(1/(-x
^2+1)^(1/2))*x^2-71*x^2*(-x^2+1)^(1/2)+12*x*(-x^2+1)^(1/2)+3*(-x^2+1)^(1/2))*(1-x)^(1/2)*(1+x)^(1/2)/x^2/(-1+x
)^2/(-x^2+1)^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 100, normalized size = 0.89 \begin {gather*} \frac {26 \, x}{3 \, \sqrt {-x^{2} + 1}} + \frac {11}{2 \, \sqrt {-x^{2} + 1}} + \frac {13 \, x}{3 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} + \frac {11}{6 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}}} - \frac {3}{{\left (-x^{2} + 1\right )}^{\frac {3}{2}} x} - \frac {1}{2 \, {\left (-x^{2} + 1\right )}^{\frac {3}{2}} x^{2}} - \frac {11}{2} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x^3,x, algorithm="maxima")

[Out]

26/3*x/sqrt(-x^2 + 1) + 11/2/sqrt(-x^2 + 1) + 13/3*x/(-x^2 + 1)^(3/2) + 11/6/(-x^2 + 1)^(3/2) - 3/((-x^2 + 1)^
(3/2)*x) - 1/2/((-x^2 + 1)^(3/2)*x^2) - 11/2*log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

________________________________________________________________________________________

Fricas [A]
time = 0.52, size = 95, normalized size = 0.85 \begin {gather*} \frac {38 \, x^{4} - 76 \, x^{3} + 38 \, x^{2} - {\left (52 \, x^{3} - 71 \, x^{2} + 12 \, x + 3\right )} \sqrt {x + 1} \sqrt {-x + 1} + 33 \, {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right )}{6 \, {\left (x^{4} - 2 \, x^{3} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x^3,x, algorithm="fricas")

[Out]

1/6*(38*x^4 - 76*x^3 + 38*x^2 - (52*x^3 - 71*x^2 + 12*x + 3)*sqrt(x + 1)*sqrt(-x + 1) + 33*(x^4 - 2*x^3 + x^2)
*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x))/(x^4 - 2*x^3 + x^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + 1}}{x^{3} \left (1 - x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(1/2)/(1-x)**(5/2)/x**3,x)

[Out]

Integral(sqrt(x + 1)/(x**3*(1 - x)**(5/2)), x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(1/2)/(1-x)^(5/2)/x^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,-4,0,%%%{
4,[2]%%%}] at parameters values [5.38357630698]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [81.11954429

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x+1}}{x^3\,{\left (1-x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^(1/2)/(x^3*(1 - x)^(5/2)),x)

[Out]

int((x + 1)^(1/2)/(x^3*(1 - x)^(5/2)), x)

________________________________________________________________________________________