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

Optimal. Leaf size=43 \[ -\sqrt {1-x} \sqrt {x+1}+2 \sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt {1-x} \sqrt {x+1}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 43, 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 = {102, 157, 41, 216, 92, 206} \begin {gather*} -\sqrt {1-x} \sqrt {x+1}+2 \sin ^{-1}(x)-\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),x]

[Out]

-(Sqrt[1 - x]*Sqrt[1 + x]) + 2*ArcSin[x] - 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 102

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))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

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

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Mathematica [A]  time = 0.05, size = 64, normalized size = 1.49 \begin {gather*} \frac {x^2+\sqrt {1-x^2} \sin ^{-1}(x)-1}{\sqrt {1-x^2}}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )-2 \sin ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.08, size = 74, normalized size = 1.72 \begin {gather*} -\frac {2 \sqrt {1-x}}{\sqrt {x+1} \left (\frac {1-x}{x+1}+1\right )}-4 \tan ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {x+1}}\right )-2 \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),x]

[Out]

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

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fricas [A]  time = 1.20, size = 57, normalized size = 1.33 \begin {gather*} -\sqrt {x + 1} \sqrt {-x + 1} - 4 \, \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) + \log \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(x + 1)*sqrt(-x + 1) - 4*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) + log((sqrt(x + 1)*sqrt(-x + 1) - 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/(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*(-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))))-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)))+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)))-sqrt(x+1)*sqrt(-x+1)

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maple [A]  time = 0.02, size = 51, normalized size = 1.19 \begin {gather*} \frac {\sqrt {x +1}\, \sqrt {-x +1}\, \left (-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )+2 \arcsin \relax (x )-\sqrt {-x^{2}+1}\right )}{\sqrt {-x^{2}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.95, size = 41, normalized size = 0.95 \begin {gather*} -\sqrt {-x^{2} + 1} + 2 \, \arcsin \relax (x) - \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/(1-x)^(1/2),x, algorithm="maxima")

[Out]

-sqrt(-x^2 + 1) + 2*arcsin(x) - 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\,\sqrt {1-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x + 1)^(3/2)/(x*(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 \sqrt {1 - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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