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

Optimal. Leaf size=43 \[ \frac{4 \sqrt{x+1}}{\sqrt{1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0129731, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {98, 21, 105, 41, 216, 92, 206} \[ \frac{4 \sqrt{x+1}}{\sqrt{1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
 /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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 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]

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 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])

Rubi steps

\begin{align*} \int \frac{(1+x)^{3/2}}{(1-x)^{3/2} x} \, dx &=\frac{4 \sqrt{1+x}}{\sqrt{1-x}}-2 \int \frac{-\frac{1}{2}+\frac{x}{2}}{\sqrt{1-x} x \sqrt{1+x}} \, dx\\ &=\frac{4 \sqrt{1+x}}{\sqrt{1-x}}+\int \frac{\sqrt{1-x}}{x \sqrt{1+x}} \, dx\\ &=\frac{4 \sqrt{1+x}}{\sqrt{1-x}}-\int \frac{1}{\sqrt{1-x} \sqrt{1+x}} \, dx+\int \frac{1}{\sqrt{1-x} x \sqrt{1+x}} \, dx\\ &=\frac{4 \sqrt{1+x}}{\sqrt{1-x}}-\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 )\\ &=\frac{4 \sqrt{1+x}}{\sqrt{1-x}}-\sin ^{-1}(x)-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{1+x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0381312, size = 61, normalized size = 1.42 \[ \frac{2 \left (\sqrt{1-x^2} \sin ^{-1}\left (\frac{\sqrt{1-x}}{\sqrt{2}}\right )+2 x+2\right )}{\sqrt{1-x^2}}-\tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 70, normalized size = 1.6 \begin{align*}{\frac{1}{x-1} \left ( -\arcsin \left ( x \right ) x-{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) x+\arcsin \left ( x \right ) +{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) -4\,\sqrt{-{x}^{2}+1} \right ) \sqrt{1-x}\sqrt{1+x}{\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.56226, size = 72, normalized size = 1.67 \begin{align*} \frac{4 \, x}{\sqrt{-x^{2} + 1}} + \frac{4}{\sqrt{-x^{2} + 1}} - \arcsin \left (x\right ) - \log \left (\frac{2 \, \sqrt{-x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.50558, size = 201, normalized size = 4.67 \begin{align*} \frac{2 \,{\left (x - 1\right )} \arctan \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) +{\left (x - 1\right )} \log \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) + 4 \, x - 4 \, \sqrt{x + 1} \sqrt{-x + 1} - 4}{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError