3.1264 \(\int e^{-3 \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=149 \[ \frac{a x^3 \sqrt{c-a^2 c x^2}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^2 \sqrt{c-a^2 c x^2}}{2 \sqrt{1-a^2 x^2}}+\frac{4 x \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^2 \sqrt{1-a^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.148544, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 77} \[ \frac{a x^3 \sqrt{c-a^2 c x^2}}{3 \sqrt{1-a^2 x^2}}-\frac{3 x^2 \sqrt{c-a^2 c x^2}}{2 \sqrt{1-a^2 x^2}}+\frac{4 x \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-a^2 c x^2} \log (a x+1)}{a^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

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

Rule 6150

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0481811, size = 63, normalized size = 0.42 \[ \frac{\sqrt{c-a^2 c x^2} \left (-\frac{4 \log (a x+1)}{a^2}+\frac{a x^3}{3}+\frac{4 x}{a}-\frac{3 x^2}{2}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.085, size = 72, normalized size = 0.5 \begin{align*}{\frac{-2\,{x}^{3}{a}^{3}+9\,{a}^{2}{x}^{2}-24\,ax+24\,\ln \left ( ax+1 \right ) }{ \left ( 6\,{a}^{2}{x}^{2}-6 \right ){a}^{2}}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(-c*(a^2*x^2-1))^(1/2)*(-a^2*x^2+1)^(1/2)*(-2*x^3*a^3+9*a^2*x^2-24*a*x+24*ln(a*x+1))/(a^2*x^2-1)/a^2

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.98392, size = 783, normalized size = 5.26 \begin{align*} \left [\frac{12 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \log \left (\frac{a^{6} c x^{6} + 4 \, a^{5} c x^{5} + 5 \, a^{4} c x^{4} - 4 \, a^{2} c x^{2} - 4 \, a c x +{\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 6 \, a^{2} x^{2} + 4 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - 2 \, c}{a^{4} x^{4} + 2 \, a^{3} x^{3} - 2 \, a x - 1}\right ) -{\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{4} x^{2} - a^{2}\right )}}, -\frac{24 \,{\left (a^{2} x^{2} - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 2 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c}}{a^{4} c x^{4} + 2 \, a^{3} c x^{3} - a^{2} c x^{2} - 2 \, a c x}\right ) +{\left (2 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 24 \, a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{4} x^{2} - a^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(12*(a^2*x^2 - 1)*sqrt(c)*log((a^6*c*x^6 + 4*a^5*c*x^5 + 5*a^4*c*x^4 - 4*a^2*c*x^2 - 4*a*c*x + (a^4*x^4 +
4*a^3*x^3 + 6*a^2*x^2 + 4*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - 2*c)/(a^4*x^4 + 2*a^3*x^3 -
2*a*x - 1)) - (2*a^3*x^3 - 9*a^2*x^2 + 24*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^4*x^2 - a^2), -1/6*
(24*(a^2*x^2 - 1)*sqrt(-c)*arctan(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 2*a*x + 2)*sqrt(-a^2*x^2 + 1)*sqrt(-c)/(a^4*
c*x^4 + 2*a^3*c*x^3 - a^2*c*x^2 - 2*a*c*x)) + (2*a^3*x^3 - 9*a^2*x^2 + 24*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*
x^2 + 1))/(a^4*x^2 - a^2)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}} \sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} c x^{2} + c}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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