3.833 \(\int \frac{\sqrt{1+x}}{(1-x)^{5/2} x} \, dx\)

Optimal. Leaf size=59 \[ \frac{(x+1)^{3/2}}{3 (1-x)^{3/2}}+\frac{2 \sqrt{x+1}}{\sqrt{1-x}}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0073131, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {96, 94, 92, 206} \[ \frac{(x+1)^{3/2}}{3 (1-x)^{3/2}}+\frac{2 \sqrt{x+1}}{\sqrt{1-x}}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x]/((1 - x)^(5/2)*x),x]

[Out]

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

Rule 96

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 94

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[p, 1] &&  !SumSimplerQ[m, 1])

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{\sqrt{1+x}}{(1-x)^{5/2} x} \, dx &=\frac{(1+x)^{3/2}}{3 (1-x)^{3/2}}+\int \frac{\sqrt{1+x}}{(1-x)^{3/2} x} \, dx\\ &=\frac{2 \sqrt{1+x}}{\sqrt{1-x}}+\frac{(1+x)^{3/2}}{3 (1-x)^{3/2}}+\int \frac{1}{\sqrt{1-x} x \sqrt{1+x}} \, dx\\ &=\frac{2 \sqrt{1+x}}{\sqrt{1-x}}+\frac{(1+x)^{3/2}}{3 (1-x)^{3/2}}-\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1-x} \sqrt{1+x}\right )\\ &=\frac{2 \sqrt{1+x}}{\sqrt{1-x}}+\frac{(1+x)^{3/2}}{3 (1-x)^{3/2}}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{1+x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0205622, size = 58, normalized size = 0.98 \[ \frac{5 x^2-3 (x-1) \sqrt{1-x^2} \tanh ^{-1}\left (\sqrt{1-x^2}\right )-2 x-7}{3 (x-1) \sqrt{1-x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-7 - 2*x + 5*x^2 - 3*(-1 + x)*Sqrt[1 - x^2]*ArcTanh[Sqrt[1 - x^2]])/(3*(-1 + x)*Sqrt[1 - x^2])

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Maple [B]  time = 0.012, size = 93, normalized size = 1.6 \begin{align*} -{\frac{1}{3\, \left ( -1+x \right ) ^{2}} \left ( 3\,{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ){x}^{2}-6\,{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) x+5\,x\sqrt{-{x}^{2}+1}+3\,{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) -7\,\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)^(1/2)/(1-x)^(5/2)/x,x)

[Out]

-1/3*(3*arctanh(1/(-x^2+1)^(1/2))*x^2-6*arctanh(1/(-x^2+1)^(1/2))*x+5*x*(-x^2+1)^(1/2)+3*arctanh(1/(-x^2+1)^(1
/2))-7*(-x^2+1)^(1/2))*(1-x)^(1/2)*(1+x)^(1/2)/(-1+x)^2/(-x^2+1)^(1/2)

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Maxima [A]  time = 1.02378, size = 95, normalized size = 1.61 \begin{align*} \frac{5 \, x}{3 \, \sqrt{-x^{2} + 1}} + \frac{1}{\sqrt{-x^{2} + 1}} + \frac{4 \, x}{3 \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}}} + \frac{4}{3 \,{\left (-x^{2} + 1\right )}^{\frac{3}{2}}} - \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)^(1/2)/(1-x)^(5/2)/x,x, algorithm="maxima")

[Out]

5/3*x/sqrt(-x^2 + 1) + 1/sqrt(-x^2 + 1) + 4/3*x/(-x^2 + 1)^(3/2) + 4/3/(-x^2 + 1)^(3/2) - log(2*sqrt(-x^2 + 1)
/abs(x) + 2/abs(x))

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Fricas [A]  time = 1.57701, size = 182, normalized size = 3.08 \begin{align*} \frac{7 \, x^{2} -{\left (5 \, x - 7\right )} \sqrt{x + 1} \sqrt{-x + 1} + 3 \,{\left (x^{2} - 2 \, x + 1\right )} \log \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) - 14 \, x + 7}{3 \,{\left (x^{2} - 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(7*x^2 - (5*x - 7)*sqrt(x + 1)*sqrt(-x + 1) + 3*(x^2 - 2*x + 1)*log((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - 14
*x + 7)/(x^2 - 2*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x + 1)/(x*(1 - x)**(5/2)), 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)^(1/2)/(1-x)^(5/2)/x,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError