3.731 \(\int \frac{(1+x)^{3/2}}{\sqrt{1-x} x^4} \, dx\)

Optimal. Leaf size=88 \[ -\frac{\sqrt{1-x} (x+1)^{5/2}}{3 x^3}-\frac{\sqrt{1-x} (x+1)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} \sqrt{x+1}}{x}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

[Out]

-((Sqrt[1 - x]*Sqrt[1 + x])/x) - (Sqrt[1 - x]*(1 + x)^(3/2))/(3*x^2) - (Sqrt[1 - x]*(1 + x)^(5/2))/(3*x^3) - A
rcTanh[Sqrt[1 - x]*Sqrt[1 + x]]

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Rubi [A]  time = 0.0164854, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {96, 94, 92, 206} \[ -\frac{\sqrt{1-x} (x+1)^{5/2}}{3 x^3}-\frac{\sqrt{1-x} (x+1)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} \sqrt{x+1}}{x}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{x+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^(3/2)/(Sqrt[1 - x]*x^4),x]

[Out]

-((Sqrt[1 - x]*Sqrt[1 + x])/x) - (Sqrt[1 - x]*(1 + x)^(3/2))/(3*x^2) - (Sqrt[1 - x]*(1 + x)^(5/2))/(3*x^3) - A
rcTanh[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{(1+x)^{3/2}}{\sqrt{1-x} x^4} \, dx &=-\frac{\sqrt{1-x} (1+x)^{5/2}}{3 x^3}+\frac{2}{3} \int \frac{(1+x)^{3/2}}{\sqrt{1-x} x^3} \, dx\\ &=-\frac{\sqrt{1-x} (1+x)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} (1+x)^{5/2}}{3 x^3}+\int \frac{\sqrt{1+x}}{\sqrt{1-x} x^2} \, dx\\ &=-\frac{\sqrt{1-x} \sqrt{1+x}}{x}-\frac{\sqrt{1-x} (1+x)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} (1+x)^{5/2}}{3 x^3}+\int \frac{1}{\sqrt{1-x} x \sqrt{1+x}} \, dx\\ &=-\frac{\sqrt{1-x} \sqrt{1+x}}{x}-\frac{\sqrt{1-x} (1+x)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} (1+x)^{5/2}}{3 x^3}-\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1-x} \sqrt{1+x}\right )\\ &=-\frac{\sqrt{1-x} \sqrt{1+x}}{x}-\frac{\sqrt{1-x} (1+x)^{3/2}}{3 x^2}-\frac{\sqrt{1-x} (1+x)^{5/2}}{3 x^3}-\tanh ^{-1}\left (\sqrt{1-x} \sqrt{1+x}\right )\\ \end{align*}

Mathematica [A]  time = 0.02106, size = 66, normalized size = 0.75 \[ -\frac{-5 x^4-3 x^3+4 x^2+3 \sqrt{1-x^2} x^3 \tanh ^{-1}\left (\sqrt{1-x^2}\right )+3 x+1}{3 x^3 \sqrt{1-x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^(3/2)/(Sqrt[1 - x]*x^4),x]

[Out]

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

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Maple [A]  time = 0.013, size = 78, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{x}^{3}}\sqrt{1-x}\sqrt{1+x} \left ( 3\,{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ){x}^{3}+5\,{x}^{2}\sqrt{-{x}^{2}+1}+3\,x\sqrt{-{x}^{2}+1}+\sqrt{-{x}^{2}+1} \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^(3/2)/x^4/(1-x)^(1/2),x)

[Out]

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

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Maxima [A]  time = 3.11486, size = 92, normalized size = 1.05 \begin{align*} -\frac{5 \, \sqrt{-x^{2} + 1}}{3 \, x} - \frac{\sqrt{-x^{2} + 1}}{x^{2}} - \frac{\sqrt{-x^{2} + 1}}{3 \, x^{3}} - \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)/x^4/(1-x)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.70746, size = 135, normalized size = 1.53 \begin{align*} \frac{3 \, x^{3} \log \left (\frac{\sqrt{x + 1} \sqrt{-x + 1} - 1}{x}\right ) -{\left (5 \, x^{2} + 3 \, x + 1\right )} \sqrt{x + 1} \sqrt{-x + 1}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^(3/2)/x^4/(1-x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**(3/2)/x**4/(1-x)**(1/2),x)

[Out]

Exception raised: ValueError

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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)/x^4/(1-x)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError