3.1238 \(\int e^{-2 \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(5-3 a x) \sqrt{c-a^2 c x^2}}{3 a^2}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2} \]

[Out]

-(x^2*Sqrt[c - a^2*c*x^2])/3 - ((5 - 3*a*x)*Sqrt[c - a^2*c*x^2])/(3*a^2) - (Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[
c - a^2*c*x^2]])/a^2

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Rubi [A]  time = 0.183082, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6152, 1809, 780, 217, 203} \[ -\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(5-3 a x) \sqrt{c-a^2 c x^2}}{3 a^2}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*Sqrt[c - a^2*c*x^2])/3 - ((5 - 3*a*x)*Sqrt[c - a^2*c*x^2])/(3*a^2) - (Sqrt[c]*ArcTan[(a*Sqrt[c]*x)/Sqrt[
c - a^2*c*x^2]])/a^2

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]

Rule 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 217

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

Rule 203

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

Rubi steps

\begin{align*} \int e^{-2 \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx &=c \int \frac{x (1-a x)^2}{\sqrt{c-a^2 c x^2}} \, dx\\ &=-\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{\int \frac{x \left (-5 a^2 c+6 a^3 c x\right )}{\sqrt{c-a^2 c x^2}} \, dx}{3 a^2}\\ &=-\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(5-3 a x) \sqrt{c-a^2 c x^2}}{3 a^2}-\frac{c \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{a}\\ &=-\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(5-3 a x) \sqrt{c-a^2 c x^2}}{3 a^2}-\frac{c \operatorname{Subst}\left (\int \frac{1}{1+a^2 c x^2} \, dx,x,\frac{x}{\sqrt{c-a^2 c x^2}}\right )}{a}\\ &=-\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(5-3 a x) \sqrt{c-a^2 c x^2}}{3 a^2}-\frac{\sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2}\\ \end{align*}

Mathematica [A]  time = 0.0813597, size = 80, normalized size = 0.94 \[ \frac{3 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )-\left (a^2 x^2-3 a x+5\right ) \sqrt{c-a^2 c x^2}}{3 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((5 - 3*a*x + a^2*x^2)*Sqrt[c - a^2*c*x^2]) + 3*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*
x^2))])/(3*a^2)

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Maple [B]  time = 0.036, size = 154, normalized size = 1.8 \begin{align*}{\frac{1}{3\,{a}^{2}c} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{x}{a}\sqrt{-{a}^{2}c{x}^{2}+c}}+{\frac{c}{a}\arctan \left ({x\sqrt{{a}^{2}c}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}} \right ){\frac{1}{\sqrt{{a}^{2}c}}}}-2\,{\frac{\sqrt{-c{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,ac \left ( x+{a}^{-1} \right ) }}{{a}^{2}}}-2\,{\frac{c}{a\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-c{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,ac \left ( x+{a}^{-1} \right ) }}} \right ) } \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)^2*(-a^2*x^2+1),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.41889, size = 347, normalized size = 4.08 \begin{align*} \left [-\frac{2 \, \sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} - 3 \, a x + 5\right )} - 3 \, \sqrt{-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right )}{6 \, a^{2}}, -\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} - 3 \, a x + 5\right )} - 3 \, \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right )}{3 \, a^{2}}\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)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

[-1/6*(2*sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 3*a*x + 5) - 3*sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sq
rt(-c)*x - c))/a^2, -1/3*(sqrt(-a^2*c*x^2 + c)*(a^2*x^2 - 3*a*x + 5) - 3*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a
*sqrt(c)*x/(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{- a^{2} c x^{2} + c}}{a x + 1}\, dx - \int \frac{a x^{2} \sqrt{- a^{2} c x^{2} + c}}{a x + 1}\, 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)**2*(-a**2*x**2+1),x)

[Out]

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

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Giac [B]  time = 1.30079, size = 242, normalized size = 2.85 \begin{align*} \frac{{\left (24 \, a^{4} c^{\frac{3}{2}} \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}\left (a\right ) - \frac{{\left (9 \, a^{4}{\left (c - \frac{2 \, c}{a x + 1}\right )}^{2} c^{2} \sqrt{-c + \frac{2 \, c}{a x + 1}} \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right ) + 3 \, a^{4} c^{4} \sqrt{-c + \frac{2 \, c}{a x + 1}} \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right ) + 8 \, a^{4} c^{3}{\left (-c + \frac{2 \, c}{a x + 1}\right )}^{\frac{3}{2}} \mathrm{sgn}\left (\frac{1}{a x + 1}\right ) \mathrm{sgn}\left (a\right )\right )}{\left (a x + 1\right )}^{3}}{c^{3}}\right )}{\left | a \right |}}{12 \, a^{7} c} \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)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

1/12*(24*a^4*c^(3/2)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a*x + 1))*sgn(a) - (9*a^4*(c - 2*c/(a*x +
1))^2*c^2*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) + 3*a^4*c^4*sqrt(-c + 2*c/(a*x + 1))*sgn(1/(a*x +
1))*sgn(a) + 8*a^4*c^3*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x + 1))*sgn(a))*(a*x + 1)^3/c^3)*abs(a)/(a^7*c)