3.1109 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^2}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=93 \[ \frac{(a x+1)^2}{a^3 \sqrt{c-a^2 c x^2}}+\frac{(a x+6) \sqrt{c-a^2 c x^2}}{2 a^3 c}-\frac{5 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{2 a^3 \sqrt{c}} \]

[Out]

(1 + a*x)^2/(a^3*Sqrt[c - a^2*c*x^2]) + ((6 + a*x)*Sqrt[c - a^2*c*x^2])/(2*a^3*c) - (5*ArcTan[(a*Sqrt[c]*x)/Sq
rt[c - a^2*c*x^2]])/(2*a^3*Sqrt[c])

________________________________________________________________________________________

Rubi [A]  time = 0.248951, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6151, 1635, 780, 217, 203} \[ \frac{(a x+1)^2}{a^3 \sqrt{c-a^2 c x^2}}+\frac{(a x+6) \sqrt{c-a^2 c x^2}}{2 a^3 c}-\frac{5 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{2 a^3 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^3*Sqrt[c - a^2*c*x^2]) + ((6 + a*x)*Sqrt[c - a^2*c*x^2])/(2*a^3*c) - (5*ArcTan[(a*Sqrt[c]*x)/Sq
rt[c - a^2*c*x^2]])/(2*a^3*Sqrt[c])

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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

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

Mathematica [A]  time = 0.102573, size = 94, normalized size = 1.01 \[ \frac{\left (a^2 x^2+3 a x-8\right ) \sqrt{c-a^2 c x^2}+5 \sqrt{c} (a x-1) \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{2 a^3 c (a x-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8 + 3*a*x + a^2*x^2)*Sqrt[c - a^2*c*x^2] + 5*Sqrt[c]*(-1 + a*x)*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(
-1 + a^2*x^2))])/(2*a^3*c*(-1 + a*x))

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 126, normalized size = 1.4 \begin{align*}{\frac{x}{2\,{a}^{2}c}\sqrt{-{a}^{2}c{x}^{2}+c}}-{\frac{5}{2\,{a}^{2}}\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{-{a}^{2}c{x}^{2}+c}}{{a}^{3}c}}-2\,{\frac{1}{{a}^{4}c}\sqrt{-c{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,ac \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x/a^2/c*(-a^2*c*x^2+c)^(1/2)-5/2/a^2/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))+2/a^3/c*(-
a^2*c*x^2+c)^(1/2)-2/a^4/c/(x-1/a)*(-c*a^2*(x-1/a)^2-2*a*c*(x-1/a))^(1/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.82969, size = 410, normalized size = 4.41 \begin{align*} \left [-\frac{5 \,{\left (a x - 1\right )} \sqrt{-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt{-a^{2} c x^{2} + c} a \sqrt{-c} x - c\right ) - 2 \, \sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 3 \, a x - 8\right )}}{4 \,{\left (a^{4} c x - a^{3} c\right )}}, \frac{5 \,{\left (a x - 1\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} c x^{2} + c} a \sqrt{c} x}{a^{2} c x^{2} - c}\right ) + \sqrt{-a^{2} c x^{2} + c}{\left (a^{2} x^{2} + 3 \, a x - 8\right )}}{2 \,{\left (a^{4} c x - a^{3} c\right )}}\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^2/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*(5*(a*x - 1)*sqrt(-c)*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x - c) - 2*sqrt(-a^2*c*x^2 + c
)*(a^2*x^2 + 3*a*x - 8))/(a^4*c*x - a^3*c), 1/2*(5*(a*x - 1)*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(
a^2*c*x^2 - c)) + sqrt(-a^2*c*x^2 + c)*(a^2*x^2 + 3*a*x - 8))/(a^4*c*x - a^3*c)]

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef