3.981 \(\int \frac{e^{\tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=137 \[ \frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2 c^2 \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.16511, antiderivative size = 137, 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} \[ \frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (a x+1) \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^(5/2),x]

[Out]

Sqrt[1 - a^2*x^2]/(8*a^2*c^2*(1 - a*x)^2*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(8*a^2*c^2*(1 + a*x)*Sqrt[c
- a^2*c*x^2]) - (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*a^2*c^2*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 )^{5/2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)} x}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x}{(1-a x)^3 (1+a x)^2} \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \left (-\frac{1}{4 a (-1+a x)^3}-\frac{1}{8 a (1+a x)^2}+\frac{1}{8 a \left (-1+a^2 x^2\right )}\right ) \, dx}{c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (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}{8 a c^2 \sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (1-a x)^2 \sqrt{c-a^2 c x^2}}+\frac{\sqrt{1-a^2 x^2}}{8 a^2 c^2 (1+a x) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \tanh ^{-1}(a x)}{8 a^2 c^2 \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0647348, size = 60, normalized size = 0.44 \[ \frac{\sqrt{1-a^2 x^2} \left (\frac{1}{a x+1}+\frac{1}{(a x-1)^2}-\tanh ^{-1}(a x)\right )}{8 a^2 c^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*x)/(c - a^2*c*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.095, size = 161, normalized size = 1.2 \begin{align*}{\frac{{a}^{3}{x}^{3}\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ){x}^{3}{a}^{3}-\ln \left ( ax+1 \right ){a}^{2}{x}^{2}+\ln \left ( ax-1 \right ){a}^{2}{x}^{2}-2\,{a}^{2}{x}^{2}-ax\ln \left ( ax+1 \right ) +\ln \left ( ax-1 \right ) xa+2\,ax+\ln \left ( ax+1 \right ) -\ln \left ( ax-1 \right ) -4}{ \left ( 16\,{a}^{2}{x}^{2}-16 \right ){c}^{3}{a}^{2} \left ( ax+1 \right ) \left ( ax-1 \right ) ^{2}}\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)^(5/2),x)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(a^3*x^3*ln(a*x+1)-ln(a*x-1)*x^3*a^3-ln(a*x+1)*a^2*x^2+ln(a*x-1
)*a^2*x^2-2*a^2*x^2-a*x*ln(a*x+1)+ln(a*x-1)*x*a+2*a*x+ln(a*x+1)-ln(a*x-1)-4)/(a^2*x^2-1)/c^3/a^2/(a*x+1)/(a*x-
1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \int -\frac{x^{2}}{{\left (a^{4} c^{\frac{5}{2}} x^{4} - 2 \, a^{2} c^{\frac{5}{2}} x^{2} + c^{\frac{5}{2}}\right )}{\left (a x + 1\right )}{\left (a x - 1\right )}}\,{d x} + \frac{1}{4 \,{\left (a^{6} c^{\frac{5}{2}} x^{4} - 2 \, a^{4} c^{\frac{5}{2}} x^{2} + a^{2} c^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

a*integrate(-x^2/((a^4*c^(5/2)*x^4 - 2*a^2*c^(5/2)*x^2 + c^(5/2))*(a*x + 1)*(a*x - 1)), x) + 1/4/(a^6*c^(5/2)*
x^4 - 2*a^4*c^(5/2)*x^2 + a^2*c^(5/2))

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Fricas [A]  time = 2.08522, size = 936, normalized size = 6.83 \begin{align*} \left [\frac{{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, 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 ) + 4 \,{\left (2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{32 \,{\left (a^{7} c^{3} x^{5} - a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{4} c^{3} x^{2} + a^{3} c^{3} x - a^{2} c^{3}\right )}}, -\frac{{\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, 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 ) - 2 \,{\left (2 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - a x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{16 \,{\left (a^{7} c^{3} x^{5} - a^{6} c^{3} x^{4} - 2 \, a^{5} c^{3} x^{3} + 2 \, a^{4} c^{3} x^{2} + a^{3} c^{3} x - a^{2} c^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/32*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*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)) + 4*(2*a^3*x^3 - 3*a^2*x^2 - a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^7*c^3*x^5 - a^6*c^3*x^4 - 2
*a^5*c^3*x^3 + 2*a^4*c^3*x^2 + a^3*c^3*x - a^2*c^3), -1/16*((a^5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*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)) - 2*(2*a^3*x^3 - 3
*a^2*x^2 - a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^7*c^3*x^5 - a^6*c^3*x^4 - 2*a^5*c^3*x^3 + 2*a^4*c^
3*x^2 + a^3*c^3*x - a^2*c^3)]

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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{5}{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)**(5/2),x)

[Out]

Integral(x*(a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(5/2)), 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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((a*x + 1)*x/((-a^2*c*x^2 + c)^(5/2)*sqrt(-a^2*x^2 + 1)), x)