3.970 \(\int \frac{e^{\tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=91 \[ \frac{\sqrt{1-a^2 x^2}}{2 a^2 c (1-a x) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2 c \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.151133, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6153, 6150, 77, 207} \[ \frac{\sqrt{1-a^2 x^2}}{2 a^2 c (1-a x) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{2 a^2 c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 207

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

Rubi steps

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

Mathematica [A]  time = 0.0348936, size = 60, normalized size = 0.66 \[ \frac{\sqrt{1-a^2 x^2} \left (\frac{1}{2 a^2 (1-a x)}-\frac{\tanh ^{-1}(a x)}{2 a^2}\right )}{c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 88, normalized size = 1. \begin{align*}{\frac{ax\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) xa-\ln \left ( ax+1 \right ) +\ln \left ( ax-1 \right ) +2}{ \left ( 4\,{a}^{2}{x}^{2}-4 \right ){c}^{2}{a}^{2} \left ( ax-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(a*x*ln(a*x+1)-ln(a*x-1)*x*a-ln(a*x+1)+ln(a*x-1)+2)/(a^2*x^2-1)/
c^2/a^2/(a*x-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -a \int -\frac{x^{2}}{{\left (a^{2} c^{\frac{3}{2}} x^{2} - c^{\frac{3}{2}}\right )}{\left (a x + 1\right )}{\left (a x - 1\right )}}\,{d x} - \frac{1}{2 \,{\left (a^{4} c^{\frac{3}{2}} x^{2} - a^{2} c^{\frac{3}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*integrate(-x^2/((a^2*c^(3/2)*x^2 - c^(3/2))*(a*x + 1)*(a*x - 1)), x) - 1/2/(a^4*c^(3/2)*x^2 - a^2*c^(3/2))

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Fricas [A]  time = 2.10398, size = 710, normalized size = 7.8 \begin{align*} \left [\frac{4 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} a x +{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt{c} \log \left (-\frac{a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} + 4 \,{\left (a^{3} x^{3} + a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} \sqrt{c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right )}{8 \,{\left (a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - a^{3} c^{2} x + a^{2} c^{2}\right )}}, \frac{2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} a x -{\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \sqrt{-c} \arctan \left (\frac{2 \, \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1} a \sqrt{-c} x}{a^{4} c x^{4} - c}\right )}{4 \,{\left (a^{5} c^{2} x^{3} - a^{4} c^{2} x^{2} - a^{3} c^{2} x + a^{2} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(4*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*x + (a^3*x^3 - a^2*x^2 - a*x + 1)*sqrt(c)*log(-(a^6*c*x^6 +
5*a^4*c*x^4 - 5*a^2*c*x^2 + 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 -
3*a^4*x^4 + 3*a^2*x^2 - 1)))/(a^5*c^2*x^3 - a^4*c^2*x^2 - a^3*c^2*x + a^2*c^2), 1/4*(2*sqrt(-a^2*c*x^2 + c)*sq
rt(-a^2*x^2 + 1)*a*x - (a^3*x^3 - a^2*x^2 - a*x + 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)
*a*sqrt(-c)*x/(a^4*c*x^4 - c)))/(a^5*c^2*x^3 - a^4*c^2*x^2 - a^3*c^2*x + a^2*c^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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