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

Optimal. Leaf size=35 \[ \frac{4 (x+1)}{\sqrt{1-x^2}}-\tanh ^{-1}\left (\sqrt{1-x^2}\right )-\sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0606364, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1805, 844, 216, 266, 63, 206} \[ \frac{4 (x+1)}{\sqrt{1-x^2}}-\tanh ^{-1}\left (\sqrt{1-x^2}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^3/(x*(1 - x^2)^(3/2)),x]

[Out]

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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])

Rubi steps

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

Mathematica [C]  time = 0.0211454, size = 47, normalized size = 1.34 \[ \frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};1-x^2\right )-\sqrt{1-x^2} \sin ^{-1}(x)+4 x+3}{\sqrt{1-x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 4*x - Sqrt[1 - x^2]*ArcSin[x] + Hypergeometric2F1[-1/2, 1, 1/2, 1 - x^2])/Sqrt[1 - x^2]

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Maple [A]  time = 0.008, size = 41, normalized size = 1.2 \begin{align*} 4\,{\frac{x}{\sqrt{-{x}^{2}+1}}}-\arcsin \left ( x \right ) +4\,{\frac{1}{\sqrt{-{x}^{2}+1}}}-{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5344, size = 72, normalized size = 2.06 \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/x/(-x^2+1)^(3/2),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 = 1.48969, size = 161, normalized size = 4.6 \begin{align*} \frac{2 \,{\left (x - 1\right )} \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) +{\left (x - 1\right )} \log \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) + 4 \, x - 4 \, \sqrt{-x^{2} + 1} - 4}{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(x - 1)*arctan((sqrt(-x^2 + 1) - 1)/x) + (x - 1)*log((sqrt(-x^2 + 1) - 1)/x) + 4*x - 4*sqrt(-x^2 + 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 )^{3}}{x \left (- \left (x - 1\right ) \left (x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12818, size = 59, normalized size = 1.69 \begin{align*} \frac{8}{\frac{\sqrt{-x^{2} + 1} - 1}{x} + 1} - \arcsin \left (x\right ) + \log \left (-\frac{\sqrt{-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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