3.1079 \(\int e^{2 \tanh ^{-1}(a x)} x \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(3 a x+5) \sqrt{c-a^2 c x^2}}{3 a^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2} \]

[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 = 0.182244, antiderivative size = 84, 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 = {6151, 1809, 780, 217, 203} \[ -\frac{1}{3} x^2 \sqrt{c-a^2 c x^2}-\frac{(3 a x+5) \sqrt{c-a^2 c x^2}}{3 a^2}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a*x])*x*Sqrt[c - a^2*c*x^2],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 6151

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[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]) && IGtQ[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 = 0.0897938, size = 79, normalized size = 0.94 \[ -\frac{\left (a^2 x^2+3 a x+5\right ) \sqrt{c-a^2 c x^2}+3 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{3 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a*x])*x*Sqrt[c - a^2*c*x^2],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 = 0.038, size = 164, normalized size = 2. \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{1}{{a}^{2}}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) }}+2\,{\frac{c}{a\sqrt{{a}^{2}c}}\arctan \left ({\sqrt{{a}^{2}c}x{\frac{1}{\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((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(1/2),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((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.86243, size = 347, normalized size = 4.13 \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((a*x+1)^2/(-a^2*x^2+1)*x*(-a^2*c*x^2+c)^(1/2),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((a*x+1)**2/(-a**2*x**2+1)*x*(-a**2*c*x**2+c)**(1/2),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 [A]  time = 1.17876, size = 99, normalized size = 1.18 \begin{align*} -\frac{1}{3} \, \sqrt{-a^{2} c x^{2} + c}{\left ({\left (x + \frac{3}{a}\right )} x + \frac{5}{a^{2}}\right )} - \frac{c \log \left ({\left | -\sqrt{-a^{2} c} x + \sqrt{-a^{2} c x^{2} + c} \right |}\right )}{a \sqrt{-c}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(-a^2*c*x^2 + c)*((x + 3/a)*x + 5/a^2) - c*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/(a*sqrt(-
c)*abs(a))