\(\int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx\) [62]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=-\frac {\sqrt {1-x^2}}{4 x^4}-\frac {2 \sqrt {1-x^2}}{3 x^3}-\frac {7 \sqrt {1-x^2}}{8 x^2}-\frac {4 \sqrt {1-x^2}}{3 x}-\frac {7}{8} \text {arctanh}\left (\sqrt {1-x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1821, 849, 821, 272, 65, 212} \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=-\frac {7}{8} \text {arctanh}\left (\sqrt {1-x^2}\right )-\frac {4 \sqrt {1-x^2}}{3 x}-\frac {7 \sqrt {1-x^2}}{8 x^2}-\frac {\sqrt {1-x^2}}{4 x^4}-\frac {2 \sqrt {1-x^2}}{3 x^3} \]

[In]

Int[(1 + x)^2/(x^5*Sqrt[1 - x^2]),x]

[Out]

-1/4*Sqrt[1 - x^2]/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 65

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 272

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 821

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 849

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 1821

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*} \text {integral}& = -\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} \text {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} \text {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 [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.65 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=\frac {\sqrt {1-x^2} \left (-6-16 x-21 x^2-32 x^3\right )}{24 x^4}-\frac {7 \log (x)}{8}+\frac {7}{8} \log \left (-1+\sqrt {1-x^2}\right ) \]

[In]

Integrate[(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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.55

method	result	size
trager	\(-\frac {\left (32 x^{3}+21 x^{2}+16 x +6\right ) \sqrt {-x^{2}+1}}{24 x^{4}}-\frac {7 \ln \left (\frac {\sqrt {-x^{2}+1}+1}{x}\right )}{8}\)	\(49\)
risch	\(\frac {32 x^{5}+21 x^{4}-16 x^{3}-15 x^{2}-16 x -6}{24 x^{4} \sqrt {-x^{2}+1}}-\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )}{8}\)	\(53\)
default	\(-\frac {\sqrt {-x^{2}+1}}{4 x^{4}}-\frac {7 \sqrt {-x^{2}+1}}{8 x^{2}}-\frac {7 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )}{8}-\frac {2 \sqrt {-x^{2}+1}}{3 x^{3}}-\frac {4 \sqrt {-x^{2}+1}}{3 x}\)	\(70\)
meijerg	\(\frac {\frac {\sqrt {\pi }\, \left (-7 x^{4}+8 x^{2}+8\right )}{16 x^{4}}-\frac {\sqrt {\pi }\, \left (12 x^{2}+8\right ) \sqrt {-x^{2}+1}}{16 x^{4}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )}{4}+\frac {3 \left (\frac {7}{6}-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{8}-\frac {\sqrt {\pi }}{2 x^{4}}-\frac {\sqrt {\pi }}{2 x^{2}}}{2 \sqrt {\pi }}-\frac {2 \left (2 x^{2}+1\right ) \sqrt {-x^{2}+1}}{3 x^{3}}-\frac {-\frac {\sqrt {\pi }\, \left (-4 x^{2}+8\right )}{8 x^{2}}+\frac {\sqrt {\pi }\, \sqrt {-x^{2}+1}}{x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )-\frac {\left (1-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}+\frac {\sqrt {\pi }}{x^{2}}}{2 \sqrt {\pi }}\)	\(208\)

[In]

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

[Out]

-1/24*(32*x^3+21*x^2+16*x+6)/x^4*(-x^2+1)^(1/2)-7/8*ln(((-x^2+1)^(1/2)+1)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=\frac {21 \, x^{4} \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) - {\left (32 \, x^{3} + 21 \, x^{2} + 16 \, x + 6\right )} \sqrt {-x^{2} + 1}}{24 \, x^{4}} \]

[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

Sympy [A] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.51 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=2 \left (\begin {cases} - \frac {\sqrt {1 - x^{2}}}{x} - \frac {\left (1 - x^{2}\right )^{\frac {3}{2}}}{3 x^{3}} & \text {for}\: x > -1 \wedge x < 1 \end {cases}\right ) + \begin {cases} - \frac {\operatorname {acosh}{\left (\frac {1}{x} \right )}}{2} + \frac {1}{2 x \sqrt {-1 + \frac {1}{x^{2}}}} - \frac {1}{2 x^{3} \sqrt {-1 + \frac {1}{x^{2}}}} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > 1 \\\frac {i \operatorname {asin}{\left (\frac {1}{x} \right )}}{2} - \frac {i \sqrt {1 - \frac {1}{x^{2}}}}{2 x} & \text {otherwise} \end {cases} + \begin {cases} - \frac {3 \operatorname {acosh}{\left (\frac {1}{x} \right )}}{8} + \frac {3}{8 x \sqrt {-1 + \frac {1}{x^{2}}}} - \frac {1}{8 x^{3} \sqrt {-1 + \frac {1}{x^{2}}}} - \frac {1}{4 x^{5} \sqrt {-1 + \frac {1}{x^{2}}}} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > 1 \\\frac {3 i \operatorname {asin}{\left (\frac {1}{x} \right )}}{8} - \frac {3 i}{8 x \sqrt {1 - \frac {1}{x^{2}}}} + \frac {i}{8 x^{3} \sqrt {1 - \frac {1}{x^{2}}}} + \frac {i}{4 x^{5} \sqrt {1 - \frac {1}{x^{2}}}} & \text {otherwise} \end {cases} \]

[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 +
1/(2*x*sqrt(-1 + x**(-2))) - 1/(2*x**3*sqrt(-1 + x**(-2))), 1/Abs(x**2) > 1), (I*asin(1/x)/2 - I*sqrt(1 - 1/x
**2)/(2*x), 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*sq
rt(1 - 1/x**2)) + I/(4*x**5*sqrt(1 - 1/x**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=-\frac {4 \, \sqrt {-x^{2} + 1}}{3 \, x} - \frac {7 \, \sqrt {-x^{2} + 1}}{8 \, x^{2}} - \frac {2 \, \sqrt {-x^{2} + 1}}{3 \, x^{3}} - \frac {\sqrt {-x^{2} + 1}}{4 \, x^{4}} - \frac {7}{8} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[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))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (69) = 138\).

Time = 0.29 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.83 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=\frac {x^{4} {\left (\frac {16 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {48 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + \frac {144 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{x^{3}} - 3\right )}}{192 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}} - \frac {3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{4 \, x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{4 \, x^{2}} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{12 \, x^{3}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{64 \, x^{4}} + \frac {7}{8} \, \log \left (-\frac {\sqrt {-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.87 \[ \int \frac {(1+x)^2}{x^5 \sqrt {1-x^2}} \, dx=\frac {7\,\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )}{8}-\sqrt {1-x^2}\,\left (\frac {4}{3\,x}+\frac {2}{3\,x^3}\right )-\sqrt {1-x^2}\,\left (\frac {3}{8\,x^2}+\frac {1}{4\,x^4}\right )-\frac {\sqrt {1-x^2}}{2\,x^2} \]

[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)