\(\int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx\) [63]

Optimal result

Integrand size = 20, antiderivative size = 107 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=-\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {6 \sqrt {1-x^2}}{5 x}-\frac {3}{4} \text {arctanh}\left (\sqrt {1-x^2}\right ) \]

[Out]

-3/4*arctanh((-x^2+1)^(1/2))-1/5*(-x^2+1)^(1/2)/x^5-1/2*(-x^2+1)^(1/2)/x^4-3/5*(-x^2+1)^(1/2)/x^3-3/4*(-x^2+1)
^(1/2)/x^2-6/5*(-x^2+1)^(1/2)/x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1821, 849, 821, 272, 65, 212} \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=-\frac {3}{4} \text {arctanh}\left (\sqrt {1-x^2}\right )-\frac {6 \sqrt {1-x^2}}{5 x}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3} \]

[In]

Int[(1 + x)^2/(x^6*Sqrt[1 - x^2]),x]

[Out]

-1/5*Sqrt[1 - x^2]/x^5 - Sqrt[1 - x^2]/(2*x^4) - (3*Sqrt[1 - x^2])/(5*x^3) - (3*Sqrt[1 - x^2])/(4*x^2) - (6*Sq
rt[1 - x^2])/(5*x) - (3*ArcTanh[Sqrt[1 - x^2]])/4

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}}{5 x^5}-\frac {1}{5} \int \frac {-10-9 x}{x^5 \sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}+\frac {1}{20} \int \frac {36+30 x}{x^4 \sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {1}{60} \int \frac {-90-72 x}{x^3 \sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}+\frac {1}{120} \int \frac {144+90 x}{x^2 \sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {6 \sqrt {1-x^2}}{5 x}+\frac {3}{4} \int \frac {1}{x \sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {6 \sqrt {1-x^2}}{5 x}+\frac {3}{8} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {6 \sqrt {1-x^2}}{5 x}-\frac {3}{4} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x^2}\right ) \\ & = -\frac {\sqrt {1-x^2}}{5 x^5}-\frac {\sqrt {1-x^2}}{2 x^4}-\frac {3 \sqrt {1-x^2}}{5 x^3}-\frac {3 \sqrt {1-x^2}}{4 x^2}-\frac {6 \sqrt {1-x^2}}{5 x}-\frac {3}{4} \tanh ^{-1}\left (\sqrt {1-x^2}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.59 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=\frac {\sqrt {1-x^2} \left (-4-10 x-12 x^2-15 x^3-24 x^4\right )}{20 x^5}-\frac {3 \log (x)}{4}+\frac {3}{4} \log \left (-1+\sqrt {1-x^2}\right ) \]

[In]

Integrate[(1 + x)^2/(x^6*Sqrt[1 - x^2]),x]

[Out]

(Sqrt[1 - x^2]*(-4 - 10*x - 12*x^2 - 15*x^3 - 24*x^4))/(20*x^5) - (3*Log[x])/4 + (3*Log[-1 + Sqrt[1 - x^2]])/4

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.50

method	result	size
trager	\(-\frac {\left (24 x^{4}+15 x^{3}+12 x^{2}+10 x +4\right ) \sqrt {-x^{2}+1}}{20 x^{5}}+\frac {3 \ln \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )}{4}\)	\(54\)
risch	\(\frac {24 x^{6}+15 x^{5}-12 x^{4}-5 x^{3}-8 x^{2}-10 x -4}{20 x^{5} \sqrt {-x^{2}+1}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )}{4}\)	\(58\)
default	\(-\frac {\sqrt {-x^{2}+1}}{5 x^{5}}-\frac {3 \sqrt {-x^{2}+1}}{5 x^{3}}-\frac {6 \sqrt {-x^{2}+1}}{5 x}-\frac {\sqrt {-x^{2}+1}}{2 x^{4}}-\frac {3 \sqrt {-x^{2}+1}}{4 x^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )}{4}\)	\(84\)
meijerg	\(-\frac {\left (\frac {8}{3} x^{4}+\frac {4}{3} x^{2}+1\right ) \sqrt {-x^{2}+1}}{5 x^{5}}+\frac {\frac {\sqrt {\pi }\, \left (-7 x^{4}+8 x^{2}+8\right )}{16 x^{4}}-\frac {\sqrt {\pi }\, \left (12 x^{2}+8\right ) \sqrt {-x^{2}+1}}{16 x^{4}}-\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )}{4}+\frac {3 \left (\frac {7}{6}-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{8}-\frac {\sqrt {\pi }}{2 x^{4}}-\frac {\sqrt {\pi }}{2 x^{2}}}{\sqrt {\pi }}-\frac {\left (2 x^{2}+1\right ) \sqrt {-x^{2}+1}}{3 x^{3}}\)	\(152\)

[In]

int((1+x)^2/x^6/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/20*(24*x^4+15*x^3+12*x^2+10*x+4)/x^5*(-x^2+1)^(1/2)+3/4*ln(((-x^2+1)^(1/2)-1)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=\frac {15 \, x^{5} \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) - {\left (24 \, x^{4} + 15 \, x^{3} + 12 \, x^{2} + 10 \, x + 4\right )} \sqrt {-x^{2} + 1}}{20 \, x^{5}} \]

[In]

integrate((1+x)^2/x^6/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/20*(15*x^5*log((sqrt(-x^2 + 1) - 1)/x) - (24*x^4 + 15*x^3 + 12*x^2 + 10*x + 4)*sqrt(-x^2 + 1))/x^5

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.69 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.88 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=\begin {cases} - \frac {\sqrt {1 - x^{2}}}{x} - \frac {\left (1 - x^{2}\right )^{\frac {3}{2}}}{3 x^{3}} & \text {for}\: x > -1 \wedge x < 1 \end {cases} + \begin {cases} - \frac {\sqrt {1 - x^{2}}}{x} - \frac {2 \left (1 - x^{2}\right )^{\frac {3}{2}}}{3 x^{3}} - \frac {\left (1 - x^{2}\right )^{\frac {5}{2}}}{5 x^{5}} & \text {for}\: x > -1 \wedge x < 1 \end {cases} + 2 \left (\begin {cases} - \frac {3 \operatorname {acosh}{\left (\frac {1}{x} \right )}}{8} + \frac {3}{8 x \sqrt {-1 + \frac {1}{x^{2}}}} - \frac {1}{8 x^{3} \sqrt {-1 + \frac {1}{x^{2}}}} - \frac {1}{4 x^{5} \sqrt {-1 + \frac {1}{x^{2}}}} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > 1 \\\frac {3 i \operatorname {asin}{\left (\frac {1}{x} \right )}}{8} - \frac {3 i}{8 x \sqrt {1 - \frac {1}{x^{2}}}} + \frac {i}{8 x^{3} \sqrt {1 - \frac {1}{x^{2}}}} + \frac {i}{4 x^{5} \sqrt {1 - \frac {1}{x^{2}}}} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((1+x)**2/x**6/(-x**2+1)**(1/2),x)

[Out]

Piecewise((-sqrt(1 - x**2)/x - (1 - x**2)**(3/2)/(3*x**3), (x > -1) & (x < 1))) + Piecewise((-sqrt(1 - x**2)/x
- 2*(1 - x**2)**(3/2)/(3*x**3) - (1 - x**2)**(5/2)/(5*x**5), (x > -1) & (x < 1))) + 2*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))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=-\frac {6 \, \sqrt {-x^{2} + 1}}{5 \, x} - \frac {3 \, \sqrt {-x^{2} + 1}}{4 \, x^{2}} - \frac {3 \, \sqrt {-x^{2} + 1}}{5 \, x^{3}} - \frac {\sqrt {-x^{2} + 1}}{2 \, x^{4}} - \frac {\sqrt {-x^{2} + 1}}{5 \, x^{5}} - \frac {3}{4} \, \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

[In]

integrate((1+x)^2/x^6/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-6/5*sqrt(-x^2 + 1)/x - 3/4*sqrt(-x^2 + 1)/x^2 - 3/5*sqrt(-x^2 + 1)/x^3 - 1/2*sqrt(-x^2 + 1)/x^4 - 1/5*sqrt(-x
^2 + 1)/x^5 - 3/4*log(2*sqrt(-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. 199 vs. \(2 (83) = 166\).

Time = 0.28 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.86 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=-\frac {x^{5} {\left (\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {15 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} + \frac {40 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{x^{3}} - \frac {110 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{x^{4}} - 1\right )}}{160 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}} - \frac {11 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{16 \, x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{4 \, x^{2}} - \frac {3 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{32 \, x^{3}} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{4}}{32 \, x^{4}} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{5}}{160 \, x^{5}} + \frac {3}{4} \, \log \left (-\frac {\sqrt {-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \]

[In]

integrate((1+x)^2/x^6/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

-1/160*x^5*(5*(sqrt(-x^2 + 1) - 1)/x - 15*(sqrt(-x^2 + 1) - 1)^2/x^2 + 40*(sqrt(-x^2 + 1) - 1)^3/x^3 - 110*(sq
rt(-x^2 + 1) - 1)^4/x^4 - 1)/(sqrt(-x^2 + 1) - 1)^5 - 11/16*(sqrt(-x^2 + 1) - 1)/x + 1/4*(sqrt(-x^2 + 1) - 1)^
2/x^2 - 3/32*(sqrt(-x^2 + 1) - 1)^3/x^3 + 1/32*(sqrt(-x^2 + 1) - 1)^4/x^4 - 1/160*(sqrt(-x^2 + 1) - 1)^5/x^5 +
3/4*log(-(sqrt(-x^2 + 1) - 1)/abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int \frac {(1+x)^2}{x^6 \sqrt {1-x^2}} \, dx=\frac {3\,\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )}{4}-\sqrt {1-x^2}\,\left (\frac {2}{3\,x}+\frac {1}{3\,x^3}\right )-\sqrt {1-x^2}\,\left (\frac {3}{4\,x^2}+\frac {1}{2\,x^4}\right )-\sqrt {1-x^2}\,\left (\frac {8}{15\,x}+\frac {4}{15\,x^3}+\frac {1}{5\,x^5}\right ) \]

[In]

int((x + 1)^2/(x^6*(1 - x^2)^(1/2)),x)

[Out]

(3*log((1/x^2 - 1)^(1/2) - (1/x^2)^(1/2)))/4 - (1 - x^2)^(1/2)*(2/(3*x) + 1/(3*x^3)) - (1 - x^2)^(1/2)*(3/(4*x
^2) + 1/(2*x^4)) - (1 - x^2)^(1/2)*(8/(15*x) + 4/(15*x^3) + 1/(5*x^5))