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

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

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1807, 835, 807, 266, 63, 206} \begin {gather*} -\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 ) \end {gather*}

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 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])

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 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 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 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])

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 = 0.04, size = 73, normalized size = 0.82 \begin {gather*} -\sqrt {1-x^2} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-x^2\right )-\frac {1}{2} \tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {\sqrt {1-x^2} \left (8 x^2+3 x+4\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.17, size = 58, normalized size = 0.65 \begin {gather*} \frac {7}{8} \log \left (\sqrt {1-x^2}-1\right )+\frac {\sqrt {1-x^2} \left (-32 x^3-21 x^2-16 x-6\right )}{24 x^4}-\frac {7 \log (x)}{8} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - x^2]*(-6 - 16*x - 21*x^2 - 32*x^3))/(24*x^4) - (7*Log[x])/8 + (7*Log[-1 + Sqrt[1 - x^2]])/8

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 53, normalized size = 0.60 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

giac [B]  time = 0.18, size = 163, normalized size = 1.83 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 70, normalized size = 0.79 \begin {gather*} -\frac {7 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )}{8}-\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^2/x^5/(-x^2+1)^(1/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.97, size = 82, normalized size = 0.92 \begin {gather*} -\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 {gather*}

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))

________________________________________________________________________________________

mupad [B]  time = 0.03, size = 77, normalized size = 0.87 \begin {gather*} \frac {7\,\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )}{8}-\sqrt {1-x^2}\,\left (\frac {4}{3\,x}+\frac {2}{3\,x^3}\right )-\sqrt {1-x^2}\,\left (\frac {3}{8\,x^2}+\frac {1}{4\,x^4}\right )-\frac {\sqrt {1-x^2}}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^2/(x^5*(1 - x^2)^(1/2)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 11.06, size = 223, normalized size = 2.51 \begin {gather*} 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 {gather*}

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))

________________________________________________________________________________________