3.1242 \(\int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx\)

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

[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 = 0.24, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6152, 1807, 807, 266, 63, 208} \[ \frac {2 a \sqrt {c-a^2 c x^2}}{x}-\frac {\sqrt {c-a^2 c x^2}}{2 x^2}-\frac {3}{2} a^2 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[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 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 208

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

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 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])

Rule 6152

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]

Rubi steps

\begin {align*} \int \frac {e^{-2 \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x^3} \, dx &=c \int \frac {(1-a x)^2}{x^3 \sqrt {c-a^2 c x^2}} \, dx\\ &=-\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 ) \operatorname {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} \operatorname {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 = 0.15, size = 76, normalized size = 0.97 \[ \frac {1}{2} \left (\frac {(4 a x-1) \sqrt {c-a^2 c x^2}}{x^2}-3 a^2 \sqrt {c} \log \left (\sqrt {c} \sqrt {c-a^2 c x^2}+c\right )+3 a^2 \sqrt {c} \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[c - a^2*c*x^2]/(E^(2*ArcTanh[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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fricas [A]  time = 0.64, size = 149, normalized size = 1.91 \[ \left [\frac {3 \, a^{2} \sqrt {c} x^{2} \log \left (-\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + 2 \, \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{4 \, x^{2}}, -\frac {3 \, a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) - \sqrt {-a^{2} c x^{2} + c} {\left (4 \, a x - 1\right )}}{2 \, x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a^2*c*x^2+c)^(1/2)/(a*x+1)^2*(-a^2*x^2+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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giac [B]  time = 3.09, size = 152, normalized size = 1.95 \[ \frac {1}{4} \, {\left (\frac {12 \, a c \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)}{\sqrt {-c}} - \frac {{\left (3 \, \pi a c - 8 \, a c\right )} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)}{\sqrt {-c}} + \frac {3 \, a c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a) - 5 \, a c {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\relax (a)}{{\left (c - \frac {c}{a x + 1}\right )}^{2}}\right )} {\left | a \right |} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(12*a*c*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(-c))*sgn(1/(a*x + 1))*sgn(a)/sqrt(-c) - (3*pi*a*c - 8*a*c)*sg
n(1/(a*x + 1))*sgn(a)/sqrt(-c) + (3*a*c^2*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) - 5*a*c*(-c + 2*c/(
a*x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn(a))/(c - c/(a*x + 1))^2)*abs(a)

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maple [B]  time = 0.05, size = 231, normalized size = 2.96 \[ -\frac {3 \sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right ) a^{2}}{2}+\frac {3 \sqrt {-a^{2} c \,x^{2}+c}\, a^{2}}{2}+\frac {2 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{c x}+2 a^{3} x \sqrt {-a^{2} c \,x^{2}+c}+\frac {2 a^{3} c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}-\frac {\left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 c \,x^{2}}-2 a^{2} \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}-\frac {2 a^{3} c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 a c \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2} c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {\sqrt {-a^{2} c x^{2} + c} {\left (a^{2} x^{2} - 1\right )}}{{\left (a x + 1\right )}^{2} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {\sqrt {- a^{2} c x^{2} + c}}{a x^{4} + x^{3}}\right )\, dx - \int \frac {a x \sqrt {- a^{2} c x^{2} + c}}{a x^{4} + x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-sqrt(-a**2*c*x**2 + c)/(a*x**4 + x**3), x) - Integral(a*x*sqrt(-a**2*c*x**2 + c)/(a*x**4 + x**3), x
)

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