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

Optimal. Leaf size=135 \[ \frac{a x+1}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}+\frac{5 a x+8}{5 c^3 x \sqrt{1-a^2 x^2}}+\frac{5 a x+6}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}-\frac{a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3} \]

[Out]

(1 + a*x)/(5*c^3*x*(1 - a^2*x^2)^(5/2)) + (6 + 5*a*x)/(15*c^3*x*(1 - a^2*x^2)^(3/2)) + (8 + 5*a*x)/(5*c^3*x*Sq
rt[1 - a^2*x^2]) - (16*Sqrt[1 - a^2*x^2])/(5*c^3*x) - (a*ArcTanh[Sqrt[1 - a^2*x^2]])/c^3

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Rubi [A]  time = 0.170999, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6148, 823, 807, 266, 63, 208} \[ \frac{a x+1}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}+\frac{5 a x+8}{5 c^3 x \sqrt{1-a^2 x^2}}+\frac{5 a x+6}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}-\frac{a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)/(5*c^3*x*(1 - a^2*x^2)^(5/2)) + (6 + 5*a*x)/(15*c^3*x*(1 - a^2*x^2)^(3/2)) + (8 + 5*a*x)/(5*c^3*x*Sq
rt[1 - a^2*x^2]) - (16*Sqrt[1 - a^2*x^2])/(5*c^3*x) - (a*ArcTanh[Sqrt[1 - a^2*x^2]])/c^3

Rule 6148

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a^2*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] || Gt
Q[c, 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^3} \, dx &=\frac{\int \frac{1+a x}{x^2 \left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{\int \frac{6 a^2+5 a^3 x}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx}{5 a^2 c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{\int \frac{24 a^4+15 a^5 x}{x^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{15 a^4 c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{8+5 a x}{5 c^3 x \sqrt{1-a^2 x^2}}+\frac{\int \frac{48 a^6+15 a^7 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{15 a^6 c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{8+5 a x}{5 c^3 x \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}+\frac{a \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{8+5 a x}{5 c^3 x \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}+\frac{a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{2 c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{8+5 a x}{5 c^3 x \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}-\frac{\operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{a c^3}\\ &=\frac{1+a x}{5 c^3 x \left (1-a^2 x^2\right )^{5/2}}+\frac{6+5 a x}{15 c^3 x \left (1-a^2 x^2\right )^{3/2}}+\frac{8+5 a x}{5 c^3 x \sqrt{1-a^2 x^2}}-\frac{16 \sqrt{1-a^2 x^2}}{5 c^3 x}-\frac{a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.0589028, size = 121, normalized size = 0.9 \[ \frac{48 a^5 x^5-33 a^4 x^4-87 a^3 x^3+52 a^2 x^2-15 a x (a x-1)^2 (a x+1) \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+38 a x-15}{15 c^3 x (a x-1)^2 (a x+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15 + 38*a*x + 52*a^2*x^2 - 87*a^3*x^3 - 33*a^4*x^4 + 48*a^5*x^5 - 15*a*x*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2
*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(15*c^3*x*(-1 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2])

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Maple [B]  time = 0.052, size = 314, normalized size = 2.3 \begin{align*} -{\frac{1}{{c}^{3}} \left ({\frac{1}{x}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{1}{4\,a} \left ({\frac{1}{5\,a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-3}}-{\frac{2\,a}{5} \left ({\frac{1}{3\,a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}-{\frac{1}{3}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}} \right ) } \right ) }+{\frac{1}{24\,a \left ( x+{a}^{-1} \right ) ^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{23}{48\,x+48\,{a}^{-1}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+a{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) -{\frac{1}{4\,a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{27}{16}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-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^2/(-a^2*c*x^2+c)^3,x)

[Out]

-1/c^3*((-a^2*x^2+1)^(1/2)/x+1/4/a*(1/5/a/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-2/5*a*(1/3/a/(x-1/a)^2*
(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/3/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)))+1/24/a/(x+1/a)^2*(-a^2*(x+
1/a)^2+2*a*(x+1/a))^(1/2)+23/48/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+a*arctanh(1/(-a^2*x^2+1)^(1/2))-1/4
/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+27/16/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))

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Maxima [A]  time = 1.05485, size = 240, normalized size = 1.78 \begin{align*} -\frac{\frac{15 \, a^{2} \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right )}{c^{3}} - \frac{15 \, a^{2} \log \left (\sqrt{-a^{2} x^{2} + 1} - 1\right )}{c^{3}} - \frac{2 \,{\left (15 \,{\left (a^{2} x^{2} - 1\right )}^{2} a^{2} - 5 \,{\left (a^{2} x^{2} - 1\right )} a^{2} + 3 \, a^{2}\right )}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{5}{2}} c^{3}}}{30 \, a} + \frac{16 \, a^{6} x^{6} - 40 \, a^{4} x^{4} + 30 \, a^{2} x^{2} - 5}{5 \,{\left (a^{4} c^{3} x^{5} - 2 \, a^{2} c^{3} x^{3} + c^{3} x\right )} \sqrt{a x + 1} \sqrt{-a x + 1}} \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^2/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-1/30*(15*a^2*log(sqrt(-a^2*x^2 + 1) + 1)/c^3 - 15*a^2*log(sqrt(-a^2*x^2 + 1) - 1)/c^3 - 2*(15*(a^2*x^2 - 1)^2
*a^2 - 5*(a^2*x^2 - 1)*a^2 + 3*a^2)/((-a^2*x^2 + 1)^(5/2)*c^3))/a + 1/5*(16*a^6*x^6 - 40*a^4*x^4 + 30*a^2*x^2
- 5)/((a^4*c^3*x^5 - 2*a^2*c^3*x^3 + c^3*x)*sqrt(a*x + 1)*sqrt(-a*x + 1))

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Fricas [A]  time = 1.70131, size = 464, normalized size = 3.44 \begin{align*} \frac{23 \, a^{6} x^{6} - 23 \, a^{5} x^{5} - 46 \, a^{4} x^{4} + 46 \, a^{3} x^{3} + 23 \, a^{2} x^{2} - 23 \, a x + 15 \,{\left (a^{6} x^{6} - a^{5} x^{5} - 2 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + a^{2} x^{2} - a x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (48 \, a^{5} x^{5} - 33 \, a^{4} x^{4} - 87 \, a^{3} x^{3} + 52 \, a^{2} x^{2} + 38 \, a x - 15\right )} \sqrt{-a^{2} x^{2} + 1}}{15 \,{\left (a^{5} c^{3} x^{6} - a^{4} c^{3} x^{5} - 2 \, a^{3} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{3} + a c^{3} x^{2} - c^{3} x\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^2/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

1/15*(23*a^6*x^6 - 23*a^5*x^5 - 46*a^4*x^4 + 46*a^3*x^3 + 23*a^2*x^2 - 23*a*x + 15*(a^6*x^6 - a^5*x^5 - 2*a^4*
x^4 + 2*a^3*x^3 + a^2*x^2 - a*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (48*a^5*x^5 - 33*a^4*x^4 - 87*a^3*x^3 + 52*
a^2*x^2 + 38*a*x - 15)*sqrt(-a^2*x^2 + 1))/(a^5*c^3*x^6 - a^4*c^3*x^5 - 2*a^3*c^3*x^4 + 2*a^2*c^3*x^3 + a*c^3*
x^2 - c^3*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{- a^{6} x^{7} \sqrt{- a^{2} x^{2} + 1} + 3 a^{4} x^{5} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{3} \sqrt{- a^{2} x^{2} + 1} + x \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{1}{- a^{6} x^{8} \sqrt{- a^{2} x^{2} + 1} + 3 a^{4} x^{6} \sqrt{- a^{2} x^{2} + 1} - 3 a^{2} x^{4} \sqrt{- a^{2} x^{2} + 1} + x^{2} \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{3}} \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**2/(-a**2*c*x**2+c)**3,x)

[Out]

(Integral(a/(-a**6*x**7*sqrt(-a**2*x**2 + 1) + 3*a**4*x**5*sqrt(-a**2*x**2 + 1) - 3*a**2*x**3*sqrt(-a**2*x**2
+ 1) + x*sqrt(-a**2*x**2 + 1)), x) + Integral(1/(-a**6*x**8*sqrt(-a**2*x**2 + 1) + 3*a**4*x**6*sqrt(-a**2*x**2
+ 1) - 3*a**2*x**4*sqrt(-a**2*x**2 + 1) + x**2*sqrt(-a**2*x**2 + 1)), x))/c**3

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

[Out]

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