3.8.18 \(\int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\) [718]

Optimal. Leaf size=78 \[ \frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{x}+\frac {3}{2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6302, 6287, 1821, 821, 272, 65, 214} \begin {gather*} -\frac {2 a \sqrt {c-a^2 c x^2}}{x}+\frac {\sqrt {c-a^2 c x^2}}{2 x^2}+\frac {3}{2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a^2*c*x^2]/(E^(2*ArcCoth[a*x])*x^3),x]

[Out]

Sqrt[c - a^2*c*x^2]/(2*x^2) - (2*a*Sqrt[c - a^2*c*x^2])/x + (3*a^2*Sqrt[c]*ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]
])/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 214

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

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 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])

Rule 6287

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/c^(n/2), Int[x^m*
((c + d*x^2)^(p + n/2)/(1 - a*x)^n), x], x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p
] || GtQ[c, 0]) && ILtQ[n/2, 0]

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx &=-\int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\\ &=-\left (c \int \frac {(1-a x)^2}{x^3 \sqrt {c-a^2 c x^2}} \, dx\right )\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 x^2}+\frac {1}{2} \int \frac {4 a c-3 a^2 c x}{x^2 \sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{x}-\frac {1}{2} \left (3 a^2 c\right ) \int \frac {1}{x \sqrt {c-a^2 c x^2}} \, dx\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{x}-\frac {1}{4} \left (3 a^2 c\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c-a^2 c x}} \, dx,x,x^2\right )\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{x}+\frac {3}{2} \text {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c-a^2 c x^2}\right )\\ &=\frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {2 a \sqrt {c-a^2 c x^2}}{x}+\frac {3}{2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 76, normalized size = 0.97 \begin {gather*} \frac {1}{2} \left (\frac {(1-4 a x) \sqrt {c-a^2 c x^2}}{x^2}-3 a^2 \sqrt {c} \log (x)+3 a^2 \sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c - a^2*c*x^2]/(E^(2*ArcCoth[a*x])*x^3),x]

[Out]

(((1 - 4*a*x)*Sqrt[c - a^2*c*x^2])/x^2 - 3*a^2*Sqrt[c]*Log[x] + 3*a^2*Sqrt[c]*Log[c + Sqrt[c]*Sqrt[c - a^2*c*x
^2]])/2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(229\) vs. \(2(64)=128\).
time = 0.22, size = 230, normalized size = 2.95

method	result	size
risch	\(\frac {\left (4 a^{3} x^{3}-a^{2} x^{2}-4 a x +1\right ) c}{2 x^{2} \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {3 a^{2} \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )}{2}\)	\(79\)
default	\(2 a^{2} \left (\sqrt {-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c}+\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \left (x +\frac {1}{a}\right )^{2}+2 \left (x +\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )+2 a \left (-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )\right )+\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}-\frac {3 a^{2} \left (\sqrt {-a^{2} c \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )\right )}{2}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-a^2*c*x^2+c)^(1/2)*(a*x-1)/(a*x+1)/x^3,x,method=_RETURNVERBOSE)

[Out]

2*a^2*((-a^2*c*(x+1/a)^2+2*(x+1/a)*a*c)^(1/2)+a*c/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*(x+1/a)^2+2*(x+
1/a)*a*c)^(1/2)))+2*a*(-1/c/x*(-a^2*c*x^2+c)^(3/2)-2*a^2*(1/2*x*(-a^2*c*x^2+c)^(1/2)+1/2*c/(a^2*c)^(1/2)*arcta
n((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))))+1/2/c/x^2*(-a^2*c*x^2+c)^(3/2)-3/2*a^2*((-a^2*c*x^2+c)^(1/2)-c^(1/2)
*ln((2*c+2*c^(1/2)*(-a^2*c*x^2+c)^(1/2))/x))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*(a*x-1)/(a*x+1)/x^3,x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*(a*x - 1)/((a*x + 1)*x^3), x)

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Fricas [A]
time = 0.40, size = 149, normalized size = 1.91 \begin {gather*} \left [\frac {3 \, a^{2} \sqrt {c} x^{2} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) - 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{4 \, x^{2}}, \frac {3 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{2 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*(a*x-1)/(a*x+1)/x^3,x, algorithm="fricas")

[Out]

[1/4*(3*a^2*sqrt(c)*x^2*log(-(a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*sqrt(c) - 2*c)/x^2) - 2*sqrt(-a^2*c*x^2 + c)*
(4*a*x - 1))/x^2, 1/2*(3*a^2*sqrt(-c)*x^2*arctan(sqrt(-a^2*c*x^2 + c)*sqrt(-c)/(a^2*c*x^2 - c)) - sqrt(-a^2*c*
x^2 + c)*(4*a*x - 1))/x^2]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x - 1\right )}{x^{3} \left (a x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a**2*c*x**2+c)**(1/2)*(a*x-1)/(a*x+1)/x**3,x)

[Out]

Integral(sqrt(-c*(a*x - 1)*(a*x + 1))*(a*x - 1)/(x**3*(a*x + 1)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (64) = 128\).
time = 0.42, size = 200, normalized size = 2.56 \begin {gather*} -\frac {3 \, a^{2} c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {{\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{3} a^{2} c + 4 \, {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} a \sqrt {-c} c {\left | a \right |} + {\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )} a^{2} c^{2} - 4 \, a \sqrt {-c} c^{2} {\left | a \right |}}{{\left ({\left (\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}\right )}^{2} - c\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)*(a*x-1)/(a*x+1)/x^3,x, algorithm="giac")

[Out]

-3*a^2*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + ((sqrt(-a^2*c)*x - sqrt(-a^2*c*x
^2 + c))^3*a^2*c + 4*(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2*a*sqrt(-c)*c*abs(a) + (sqrt(-a^2*c)*x - sqrt(-a
^2*c*x^2 + c))*a^2*c^2 - 4*a*sqrt(-c)*c^2*abs(a))/((sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))^2 - c)^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a\,x-1\right )}{x^3\,\left (a\,x+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a^2*c*x^2)^(1/2)*(a*x - 1))/(x^3*(a*x + 1)),x)

[Out]

int(((c - a^2*c*x^2)^(1/2)*(a*x - 1))/(x^3*(a*x + 1)), x)

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