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

Optimal. Leaf size=99 \[ \frac{x (1-a x) \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{2 a \sqrt{1-a x} \sqrt{a x+1}} \]

[Out]

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

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Rubi [A]  time = 0.213416, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6159, 6129, 50, 41, 216} \[ \frac{x (1-a x) \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}}}{2 a}+\frac{3 x \sqrt{c-\frac{c}{a^2 x^2}} \sin ^{-1}(a x)}{2 a \sqrt{1-a x} \sqrt{a x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6159

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

Rule 6129

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

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

Rubi steps

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

Mathematica [A]  time = 0.0564665, size = 100, normalized size = 1.01 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (\sqrt{a x+1} \left (a^2 x^2-5 a x+4\right )-6 \sqrt{1-a x} \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )\right )}{2 a \sqrt{1-a x} \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(Sqrt[1 + a*x]*(4 - 5*a*x + a^2*x^2) - 6*Sqrt[1 - a*x]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])
)/(2*a*Sqrt[1 - a*x]*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.109, size = 147, normalized size = 1.5 \begin{align*} -{\frac{x}{2\,{a}^{2}}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}} \left ( x\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}{a}^{2}-\sqrt{c}\ln \left ( x\sqrt{c}+\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}} \right ) +4\,\sqrt{c}\ln \left ({\frac{1}{\sqrt{c}} \left ( \sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}\sqrt{c}+cx \right ) } \right ) -4\,\sqrt{{\frac{ \left ( ax-1 \right ) \left ( ax+1 \right ) c}{{a}^{2}}}}a \right ){\frac{1}{\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*(a^2*x^2-1)/a^2/x^2)^(1/2)*x*(x*(c*(a^2*x^2-1)/a^2)^(1/2)*a^2-c^(1/2)*ln(x*c^(1/2)+(c*(a^2*x^2-1)/a^2)
^(1/2))+4*c^(1/2)*ln((((a*x-1)*(a*x+1)*c/a^2)^(1/2)*c^(1/2)+c*x)/c^(1/2))-4*((a*x-1)*(a*x+1)*c/a^2)^(1/2)*a)/(
c*(a^2*x^2-1)/a^2)^(1/2)/a^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (a^{2} x^{2} - 1\right )} \sqrt{c - \frac{c}{a^{2} x^{2}}} x}{{\left (a x + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.28018, size = 406, normalized size = 4.1 \begin{align*} \left [-\frac{2 \,{\left (a^{2} x^{2} - 4 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \sqrt{c} \log \left (2 \, a^{2} c x^{2} - 2 \, a^{2} \sqrt{c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c\right )}{4 \, a^{2}}, -\frac{{\left (a^{2} x^{2} - 4 \, a x\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - 3 \, \sqrt{-c} \arctan \left (\frac{a^{2} \sqrt{-c} x^{2} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}\right )}{2 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*(a^2*x^2 - 4*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 3*sqrt(c)*log(2*a^2*c*x^2 - 2*a^2*sqrt(c)*x^2*sqr
t((a^2*c*x^2 - c)/(a^2*x^2)) - c))/a^2, -1/2*((a^2*x^2 - 4*a*x)*sqrt((a^2*c*x^2 - c)/(a^2*x^2)) - 3*sqrt(-c)*a
rctan(a^2*sqrt(-c)*x^2*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^2 - c)))/a^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{x \sqrt{c - \frac{c}{a^{2} x^{2}}}}{a x + 1}\, dx - \int \frac{a x^{2} \sqrt{c - \frac{c}{a^{2} x^{2}}}}{a x + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-x*sqrt(c - c/(a**2*x**2))/(a*x + 1), x) - Integral(a*x**2*sqrt(c - c/(a**2*x**2))/(a*x + 1), x)

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Giac [A]  time = 1.22271, size = 143, normalized size = 1.44 \begin{align*} -\frac{1}{4} \,{\left (2 \, \sqrt{a^{2} c x^{2} - c}{\left (\frac{x \mathrm{sgn}\left (x\right )}{a^{2}} - \frac{4 \, \mathrm{sgn}\left (x\right )}{a^{3}}\right )} - \frac{6 \, \sqrt{c} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} - c} \right |}\right ) \mathrm{sgn}\left (x\right )}{a^{2}{\left | a \right |}} + \frac{{\left (3 \, a \sqrt{c} \log \left ({\left | c \right |}\right ) + 8 \, \sqrt{-c}{\left | a \right |}\right )} \mathrm{sgn}\left (x\right )}{a^{3}{\left | a \right |}}\right )}{\left | a \right |} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(a^2*c*x^2 - c)*(x*sgn(x)/a^2 - 4*sgn(x)/a^3) - 6*sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2
- c)))*sgn(x)/(a^2*abs(a)) + (3*a*sqrt(c)*log(abs(c)) + 8*sqrt(-c)*abs(a))*sgn(x)/(a^3*abs(a)))*abs(a)