3.8.30 \(\int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^3} \, dx\) [730]

Optimal. Leaf size=69 \[ -\frac {3 \sqrt {1-x} \sqrt {1+x}}{2 x}-\frac {\sqrt {1-x} (1+x)^{3/2}}{2 x^2}-\frac {3}{2} \tanh ^{-1}\left (\sqrt {1-x} \sqrt {1+x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {96, 94, 212} \begin {gather*} -\frac {\sqrt {1-x} (x+1)^{3/2}}{2 x^2}-\frac {3 \sqrt {1-x} \sqrt {x+1}}{2 x}-\frac {3}{2} \tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.12, size = 54, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {1-x} \left (1+5 x+4 x^2\right )}{2 x^2 \sqrt {1+x}}-3 \tanh ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[1 - x]*(1 + 5*x + 4*x^2))/(x^2*Sqrt[1 + x]) - 3*ArcTanh[Sqrt[1 - x]/Sqrt[1 + x]]

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Maple [A]
time = 0.08, size = 64, normalized size = 0.93

method	result	size
default	\(-\frac {\sqrt {1+x}\, \sqrt {1-x}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) x^{2}+4 x \sqrt {-x^{2}+1}+\sqrt {-x^{2}+1}\right )}{2 x^{2} \sqrt {-x^{2}+1}}\)	\(64\)
risch	\(\frac {\sqrt {1+x}\, \left (-1+x \right ) \left (1+4 x \right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 x^{2} \sqrt {-\left (1+x \right ) \left (-1+x \right )}\, \sqrt {1-x}}-\frac {3 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right ) \sqrt {\left (1+x \right ) \left (1-x \right )}}{2 \sqrt {1-x}\, \sqrt {1+x}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(1+x)^(1/2)*(1-x)^(1/2)*(3*arctanh(1/(-x^2+1)^(1/2))*x^2+4*x*(-x^2+1)^(1/2)+(-x^2+1)^(1/2))/x^2/(-x^2+1)^
(1/2)

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Maxima [A]
time = 0.51, size = 54, normalized size = 0.78 \begin {gather*} -\frac {2 \, \sqrt {-x^{2} + 1}}{x} - \frac {\sqrt {-x^{2} + 1}}{2 \, x^{2}} - \frac {3}{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)^(3/2)/x^3/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-2*sqrt(-x^2 + 1)/x - 1/2*sqrt(-x^2 + 1)/x^2 - 3/2*log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

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Fricas [A]
time = 1.60, size = 50, normalized size = 0.72 \begin {gather*} \frac {3 \, x^{2} \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - {\left (4 \, x + 1\right )} \sqrt {x + 1} \sqrt {-x + 1}}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(3*x^2*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - (4*x + 1)*sqrt(x + 1)*sqrt(-x + 1))/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2)/x^3/(1-x)^(1/2),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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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