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

Optimal. Leaf size=44 \[ -\frac {\sqrt {1-x} \sqrt {x+1}}{x}+\sin ^{-1}(x)-2 \tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {98, 157, 41, 216, 92, 206} \begin {gather*} -\frac {\sqrt {1-x} \sqrt {x+1}}{x}+\sin ^{-1}(x)-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^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 92

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 98

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 157

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

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*} \int \frac {(1+x)^{3/2}}{\sqrt {1-x} x^2} \, dx &=-\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 \operatorname {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*}

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Mathematica [A]  time = 0.05, size = 66, normalized size = 1.50 \begin {gather*} \frac {x}{\sqrt {1-x^2}}-\frac {1}{\sqrt {1-x^2} x}-2 \tanh ^{-1}\left (\sqrt {1-x^2}\right )+2 \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )+2 \sin ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 74, normalized size = 1.68 \begin {gather*} \frac {2 \sqrt {1-x}}{\sqrt {x+1} \left (\frac {1-x}{x+1}-1\right )}-2 \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right )-4 \tanh ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 65, normalized size = 1.48 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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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^2/(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 [-93.616423693]Warning, choosing root of [1,0,-4,0,%%%{4,[2]%%%}] at parameter
s values [-17.8804557086]-4*(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))/(-
(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^2+4)-2*(-1/2*pi-atan(sqrt(x+1)
*((-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1))^2-1)/(-2*sqrt(-x+1)+2*sqrt(2))))-2*ln(abs(2*sqrt(x+1)/(-2*sqrt(-x
+1)+2*sqrt(2))+2-1/2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))+2*ln(abs(2*sqrt(x+1)/(-2*sqrt(-x+1)+2*sqrt(2))-2-1/
2*(-2*sqrt(-x+1)+2*sqrt(2))/sqrt(x+1)))

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maple [A]  time = 0.01, size = 55, normalized size = 1.25 \begin {gather*} \frac {\sqrt {x +1}\, \sqrt {-x +1}\, \left (-2 x \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )+x \arcsin \relax (x )-\sqrt {-x^{2}+1}\right )}{\sqrt {-x^{2}+1}\, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.93, size = 42, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {-x^{2} + 1}}{x} + \arcsin \relax (x) - 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^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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (x+1\right )}^{3/2}}{x^2\,\sqrt {1-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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^{2} \sqrt {1 - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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