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

Optimal. Leaf size=89 \[ -\frac{4 \sqrt{1-x^2}}{3 x}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{7}{8} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

[Out]

-Sqrt[1 - x^2]/(4*x^4) - (2*Sqrt[1 - x^2])/(3*x^3) - (7*Sqrt[1 - x^2])/(8*x^2) - (4*Sqrt[1 - x^2])/(3*x) - (7*
ArcTanh[Sqrt[1 - x^2]])/8

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Rubi [A]  time = 0.0895911, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1807, 835, 807, 266, 63, 206} \[ -\frac{4 \sqrt{1-x^2}}{3 x}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{7}{8} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^2/(x^5*Sqrt[1 - x^2]),x]

[Out]

-Sqrt[1 - x^2]/(4*x^4) - (2*Sqrt[1 - x^2])/(3*x^3) - (7*Sqrt[1 - x^2])/(8*x^2) - (4*Sqrt[1 - x^2])/(3*x) - (7*
ArcTanh[Sqrt[1 - x^2]])/8

Rule 1807

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

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

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)^2}{x^5 \sqrt{1-x^2}} \, dx &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{1}{4} \int \frac{-8-7 x}{x^4 \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}+\frac{1}{12} \int \frac{21+16 x}{x^3 \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{1}{24} \int \frac{-32-21 x}{x^2 \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{4 \sqrt{1-x^2}}{3 x}+\frac{7}{8} \int \frac{1}{x \sqrt{1-x^2}} \, dx\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{4 \sqrt{1-x^2}}{3 x}+\frac{7}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{4 \sqrt{1-x^2}}{3 x}-\frac{7}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1-x^2}\right )\\ &=-\frac{\sqrt{1-x^2}}{4 x^4}-\frac{2 \sqrt{1-x^2}}{3 x^3}-\frac{7 \sqrt{1-x^2}}{8 x^2}-\frac{4 \sqrt{1-x^2}}{3 x}-\frac{7}{8} \tanh ^{-1}\left (\sqrt{1-x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.0400475, size = 73, normalized size = 0.82 \[ -\sqrt{1-x^2} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};1-x^2\right )-\frac{\sqrt{1-x^2} \left (8 x^2+3 x+4\right )}{6 x^3}-\frac{1}{2} \tanh ^{-1}\left (\sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - x^2]*(4 + 3*x + 8*x^2))/(6*x^3) - ArcTanh[Sqrt[1 - x^2]]/2 - Sqrt[1 - x^2]*Hypergeometric2F1[1/2, 3
, 3/2, 1 - x^2]

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Maple [A]  time = 0.053, size = 70, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,{x}^{3}}\sqrt{-{x}^{2}+1}}-{\frac{4}{3\,x}\sqrt{-{x}^{2}+1}}-{\frac{1}{4\,{x}^{4}}\sqrt{-{x}^{2}+1}}-{\frac{7}{8\,{x}^{2}}\sqrt{-{x}^{2}+1}}-{\frac{7}{8}{\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)^2/x^5/(-x^2+1)^(1/2),x)

[Out]

-2/3*(-x^2+1)^(1/2)/x^3-4/3*(-x^2+1)^(1/2)/x-1/4/x^4*(-x^2+1)^(1/2)-7/8/x^2*(-x^2+1)^(1/2)-7/8*arctanh(1/(-x^2
+1)^(1/2))

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Maxima [A]  time = 1.49207, size = 111, normalized size = 1.25 \begin{align*} -\frac{4 \, \sqrt{-x^{2} + 1}}{3 \, x} - \frac{7 \, \sqrt{-x^{2} + 1}}{8 \, x^{2}} - \frac{2 \, \sqrt{-x^{2} + 1}}{3 \, x^{3}} - \frac{\sqrt{-x^{2} + 1}}{4 \, x^{4}} - \frac{7}{8} \, \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)^2/x^5/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-4/3*sqrt(-x^2 + 1)/x - 7/8*sqrt(-x^2 + 1)/x^2 - 2/3*sqrt(-x^2 + 1)/x^3 - 1/4*sqrt(-x^2 + 1)/x^4 - 7/8*log(2*s
qrt(-x^2 + 1)/abs(x) + 2/abs(x))

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Fricas [A]  time = 1.81457, size = 126, normalized size = 1.42 \begin{align*} \frac{21 \, x^{4} \log \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) -{\left (32 \, x^{3} + 21 \, x^{2} + 16 \, x + 6\right )} \sqrt{-x^{2} + 1}}{24 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(21*x^4*log((sqrt(-x^2 + 1) - 1)/x) - (32*x^3 + 21*x^2 + 16*x + 6)*sqrt(-x^2 + 1))/x^4

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Sympy [A]  time = 12.4501, size = 223, normalized size = 2.51 \begin{align*} 2 \left (\begin{cases} - \frac{\sqrt{1 - x^{2}}}{x} - \frac{\left (1 - x^{2}\right )^{\frac{3}{2}}}{3 x^{3}} & \text{for}\: x > -1 \wedge x < 1 \end{cases}\right ) + \begin{cases} - \frac{\operatorname{acosh}{\left (\frac{1}{x} \right )}}{2} - \frac{\sqrt{-1 + \frac{1}{x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{x^{2}}\right |} > 1 \\\frac{i \operatorname{asin}{\left (\frac{1}{x} \right )}}{2} - \frac{i}{2 x \sqrt{1 - \frac{1}{x^{2}}}} + \frac{i}{2 x^{3} \sqrt{1 - \frac{1}{x^{2}}}} & \text{otherwise} \end{cases} + \begin{cases} - \frac{3 \operatorname{acosh}{\left (\frac{1}{x} \right )}}{8} + \frac{3}{8 x \sqrt{-1 + \frac{1}{x^{2}}}} - \frac{1}{8 x^{3} \sqrt{-1 + \frac{1}{x^{2}}}} - \frac{1}{4 x^{5} \sqrt{-1 + \frac{1}{x^{2}}}} & \text{for}\: \frac{1}{\left |{x^{2}}\right |} > 1 \\\frac{3 i \operatorname{asin}{\left (\frac{1}{x} \right )}}{8} - \frac{3 i}{8 x \sqrt{1 - \frac{1}{x^{2}}}} + \frac{i}{8 x^{3} \sqrt{1 - \frac{1}{x^{2}}}} + \frac{i}{4 x^{5} \sqrt{1 - \frac{1}{x^{2}}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*Piecewise((-sqrt(1 - x**2)/x - (1 - x**2)**(3/2)/(3*x**3), (x > -1) & (x < 1))) + Piecewise((-acosh(1/x)/2 -
sqrt(-1 + x**(-2))/(2*x), 1/Abs(x**2) > 1), (I*asin(1/x)/2 - I/(2*x*sqrt(1 - 1/x**2)) + I/(2*x**3*sqrt(1 - 1/
x**2)), True)) + Piecewise((-3*acosh(1/x)/8 + 3/(8*x*sqrt(-1 + x**(-2))) - 1/(8*x**3*sqrt(-1 + x**(-2))) - 1/(
4*x**5*sqrt(-1 + x**(-2))), 1/Abs(x**2) > 1), (3*I*asin(1/x)/8 - 3*I/(8*x*sqrt(1 - 1/x**2)) + I/(8*x**3*sqrt(1
- 1/x**2)) + I/(4*x**5*sqrt(1 - 1/x**2)), True))

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Giac [B]  time = 1.1037, size = 220, normalized size = 2.47 \begin{align*} \frac{x^{4}{\left (\frac{16 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{x} - \frac{48 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + \frac{144 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{3}}{x^{3}} - 3\right )}}{192 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{4}} - \frac{3 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}}{4 \, x} + \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{4 \, x^{2}} - \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{3}}{12 \, x^{3}} + \frac{{\left (\sqrt{-x^{2} + 1} - 1\right )}^{4}}{64 \, x^{4}} + \frac{7}{8} \, \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)^2/x^5/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

1/192*x^4*(16*(sqrt(-x^2 + 1) - 1)/x - 48*(sqrt(-x^2 + 1) - 1)^2/x^2 + 144*(sqrt(-x^2 + 1) - 1)^3/x^3 - 3)/(sq
rt(-x^2 + 1) - 1)^4 - 3/4*(sqrt(-x^2 + 1) - 1)/x + 1/4*(sqrt(-x^2 + 1) - 1)^2/x^2 - 1/12*(sqrt(-x^2 + 1) - 1)^
3/x^3 + 1/64*(sqrt(-x^2 + 1) - 1)^4/x^4 + 7/8*log(-(sqrt(-x^2 + 1) - 1)/abs(x))