3.1.61 \(\int \frac {(1+x)^2}{x^4 \sqrt {1-x^2}} \, dx\) [61]

Optimal. Leaf size=67 \[ -\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}-\frac {5 \sqrt {1-x^2}}{3 x}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1821, 849, 821, 272, 65, 212} \begin {gather*} -\frac {5 \sqrt {1-x^2}}{3 x}-\frac {\sqrt {1-x^2}}{x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {\sqrt {1-x^2}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)^2/(x^4*Sqrt[1 - x^2]),x]

[Out]

-1/3*Sqrt[1 - x^2]/x^3 - Sqrt[1 - x^2]/x^2 - (5*Sqrt[1 - x^2])/(3*x) - ArcTanh[Sqrt[1 - x^2]]

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*} \int \frac {(1+x)^2}{x^4 \sqrt {1-x^2}} \, dx &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {1}{3} \int \frac {-6-5 x}{x^3 \sqrt {1-x^2}} \, dx\\ &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}+\frac {1}{6} \int \frac {10+6 x}{x^2 \sqrt {1-x^2}} \, dx\\ &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}-\frac {5 \sqrt {1-x^2}}{3 x}+\int \frac {1}{x \sqrt {1-x^2}} \, dx\\ &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}-\frac {5 \sqrt {1-x^2}}{3 x}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}-\frac {5 \sqrt {1-x^2}}{3 x}-\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x^2}\right )\\ &=-\frac {\sqrt {1-x^2}}{3 x^3}-\frac {\sqrt {1-x^2}}{x^2}-\frac {5 \sqrt {1-x^2}}{3 x}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 47, normalized size = 0.70 \begin {gather*} \frac {\left (-1-3 x-5 x^2\right ) \sqrt {1-x^2}}{3 x^3}-\log (x)+\log \left (-1+\sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^2/(x^4*Sqrt[1 - x^2]),x]

[Out]

((-1 - 3*x - 5*x^2)*Sqrt[1 - x^2])/(3*x^3) - Log[x] + Log[-1 + Sqrt[1 - x^2]]

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Maple [A]
time = 0.09, size = 56, normalized size = 0.84

method	result	size
trager	\(-\frac {\left (5 x^{2}+3 x +1\right ) \sqrt {-x^{2}+1}}{3 x^{3}}+\ln \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )\)	\(42\)
risch	\(\frac {5 x^{4}+3 x^{3}-4 x^{2}-3 x -1}{3 x^{3} \sqrt {-x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(48\)
default	\(-\frac {\sqrt {-x^{2}+1}}{x^{2}}-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )-\frac {5 \sqrt {-x^{2}+1}}{3 x}-\frac {\sqrt {-x^{2}+1}}{3 x^{3}}\)	\(56\)
meijerg	\(-\frac {\left (2 x^{2}+1\right ) \sqrt {-x^{2}+1}}{3 x^{3}}-\frac {-\frac {\sqrt {\pi }\, \left (-4 x^{2}+8\right )}{8 x^{2}}+\frac {\sqrt {\pi }\, \sqrt {-x^{2}+1}}{x^{2}}+\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )-\frac {\left (1-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2}+\frac {\sqrt {\pi }}{x^{2}}}{\sqrt {\pi }}-\frac {\sqrt {-x^{2}+1}}{x}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)^2/x^4/(-x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 68, normalized size = 1.01 \begin {gather*} -\frac {5 \, \sqrt {-x^{2} + 1}}{3 \, x} - \frac {\sqrt {-x^{2} + 1}}{x^{2}} - \frac {\sqrt {-x^{2} + 1}}{3 \, x^{3}} - \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^4/(-x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-5/3*sqrt(-x^2 + 1)/x - sqrt(-x^2 + 1)/x^2 - 1/3*sqrt(-x^2 + 1)/x^3 - log(2*sqrt(-x^2 + 1)/abs(x) + 2/abs(x))

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Fricas [A]
time = 1.88, size = 48, normalized size = 0.72 \begin {gather*} \frac {3 \, x^{3} \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) - {\left (5 \, x^{2} + 3 \, x + 1\right )} \sqrt {-x^{2} + 1}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)^2/x^4/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/3*(3*x^3*log((sqrt(-x^2 + 1) - 1)/x) - (5*x^2 + 3*x + 1)*sqrt(-x^2 + 1))/x^3

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Sympy [C] Result contains complex when optimal does not.
time = 4.15, size = 128, normalized size = 1.91 \begin {gather*} \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 {i \sqrt {x^{2} - 1}}{x} & \text {for}\: \left |{x^{2}}\right | > 1 \\- \frac {\sqrt {1 - x^{2}}}{x} & \text {otherwise} \end {cases} + 2 \left (\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}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)**2/x**4/(-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((-I*sqrt(x**2 - 1)
/x, Abs(x**2) > 1), (-sqrt(1 - x**2)/x, True)) + 2*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))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (55) = 110\).
time = 1.73, size = 125, normalized size = 1.87 \begin {gather*} -\frac {x^{3} {\left (\frac {6 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {21 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{24 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}} - \frac {7 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{8 \, x} + \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{4 \, x^{2}} - \frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{3}}{24 \, x^{3}} + \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^4/(-x^2+1)^(1/2),x, algorithm="giac")

[Out]

-1/24*x^3*(6*(sqrt(-x^2 + 1) - 1)/x - 21*(sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1)^3 - 7/8*(sqrt(-x
^2 + 1) - 1)/x + 1/4*(sqrt(-x^2 + 1) - 1)^2/x^2 - 1/24*(sqrt(-x^2 + 1) - 1)^3/x^3 + log(-(sqrt(-x^2 + 1) - 1)/
abs(x))

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Mupad [B]
time = 0.03, size = 67, normalized size = 1.00 \begin {gather*} \ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )-\sqrt {1-x^2}\,\left (\frac {2}{3\,x}+\frac {1}{3\,x^3}\right )-\frac {\sqrt {1-x^2}}{x}-\frac {\sqrt {1-x^2}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^2/(x^4*(1 - x^2)^(1/2)),x)

[Out]

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

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