3.1110 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x}{\sqrt{c-a^2 c x^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.119463, 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, 789, 641, 217, 203} \[ \frac{(a x+1)^2}{a^2 \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{c-a^2 c x^2}}{a^2 c}-\frac{2 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^2/(a^2*Sqrt[c - a^2*c*x^2]) + (2*Sqrt[c - a^2*c*x^2])/(a^2*c) - (2*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2
*c*x^2]])/(a^2*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 789

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g + e*f)*
(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[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}{\sqrt{c-a^2 c x^2}} \, dx &=c \int \frac{x (1+a x)^2}{\left (c-a^2 c x^2\right )^{3/2}} \, dx\\ &=\frac{(1+a x)^2}{a^2 \sqrt{c-a^2 c x^2}}-\frac{2 \int \frac{1+a x}{\sqrt{c-a^2 c x^2}} \, dx}{a}\\ &=\frac{(1+a x)^2}{a^2 \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{c-a^2 c x^2}}{a^2 c}-\frac{2 \int \frac{1}{\sqrt{c-a^2 c x^2}} \, dx}{a}\\ &=\frac{(1+a x)^2}{a^2 \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{c-a^2 c x^2}}{a^2 c}-\frac{2 \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+a x)^2}{a^2 \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{c-a^2 c x^2}}{a^2 c}-\frac{2 \tan ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c-a^2 c x^2}}\right )}{a^2 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.108018, size = 78, normalized size = 0.93 \[ \frac{\frac{(a x-3) \sqrt{c-a^2 c x^2}}{a x-1}+2 \sqrt{c} \tan ^{-1}\left (\frac{a x \sqrt{c-a^2 c x^2}}{\sqrt{c} \left (a^2 x^2-1\right )}\right )}{a^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 103, normalized size = 1.2 \begin{align*}{\frac{1}{{a}^{2}c}\sqrt{-{a}^{2}c{x}^{2}+c}}-2\,{\frac{1}{a\sqrt{{a}^{2}c}}\arctan \left ({\frac{\sqrt{{a}^{2}c}x}{\sqrt{-{a}^{2}c{x}^{2}+c}}} \right ) }-2\,{\frac{1}{{a}^{3}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/(-a^2*c*x^2+c)^(1/2),x)

[Out]

(-a^2*c*x^2+c)^(1/2)/a^2/c-2/a/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))-2/a^3/c/(x-1/a)*(-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 = 2.8286, size = 362, normalized size = 4.31 \begin{align*} \left [-\frac{{\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 ) - \sqrt{-a^{2} c x^{2} + c}{\left (a x - 3\right )}}{a^{3} c x - a^{2} c}, \frac{2 \,{\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 x - 3\right )}}{a^{3} c x - a^{2} c}\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]

[-((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) - sqrt(-a^2*c*x^2 + c)*(a*x -
3))/(a^3*c*x - a^2*c), (2*(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*x - 3))/(a^3*c*x - a^2*c)]

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

[Out]

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

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

[Out]

undef