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

Optimal. Leaf size=134 \[ -\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}+\frac{4 a x+5}{3 c^2 x^2 \sqrt{1-a^2 x^2}}+\frac{a x+1}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{5 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2} \]

[Out]

(1 + a*x)/(3*c^2*x^2*(1 - a^2*x^2)^(3/2)) + (5 + 4*a*x)/(3*c^2*x^2*Sqrt[1 - a^2*x^2]) - (5*Sqrt[1 - a^2*x^2])/
(2*c^2*x^2) - (8*a*Sqrt[1 - a^2*x^2])/(3*c^2*x) - (5*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*c^2)

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Rubi [A]  time = 0.159217, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6148, 823, 835, 807, 266, 63, 208} \[ -\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}+\frac{4 a x+5}{3 c^2 x^2 \sqrt{1-a^2 x^2}}+\frac{a x+1}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{5 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)/(3*c^2*x^2*(1 - a^2*x^2)^(3/2)) + (5 + 4*a*x)/(3*c^2*x^2*Sqrt[1 - a^2*x^2]) - (5*Sqrt[1 - a^2*x^2])/
(2*c^2*x^2) - (8*a*Sqrt[1 - a^2*x^2])/(3*c^2*x) - (5*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*c^2)

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 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[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^3 \left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{1+a x}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{\int \frac{5 a^2+4 a^3 x}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{3 a^2 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{15 a^4+8 a^5 x}{x^3 \sqrt{1-a^2 x^2}} \, dx}{3 a^4 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{\int \frac{-16 a^5-15 a^6 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{6 a^4 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}+\frac{\left (5 a^2\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{2 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}+\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )}{2 c^2}\\ &=\frac{1+a x}{3 c^2 x^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{5+4 a x}{3 c^2 x^2 \sqrt{1-a^2 x^2}}-\frac{5 \sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{8 a \sqrt{1-a^2 x^2}}{3 c^2 x}-\frac{5 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0459144, size = 103, normalized size = 0.77 \[ \frac{16 a^4 x^4-a^3 x^3-23 a^2 x^2-15 a^2 x^2 (a x-1) \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+3 a x+3}{6 c^2 x^2 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 3*a*x - 23*a^2*x^2 - a^3*x^3 + 16*a^4*x^4 - 15*a^2*x^2*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*
x^2]])/(6*c^2*x^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.046, size = 214, normalized size = 1.6 \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{a}{x}\sqrt{-{a}^{2}{x}^{2}+1}}-{\frac{5\,{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{a}{4\,x+4\,{a}^{-1}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{a}{2} \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 ) }-{\frac{7\,a}{4}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-1}}-{\frac{1}{2\,{x}^{2}}\sqrt{-{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^3/(-a^2*c*x^2+c)^2,x)

[Out]

1/c^2*(-a*(-a^2*x^2+1)^(1/2)/x-5/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))+1/4*a/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a)
)^(1/2)+1/2*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))-7/4*a/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-1/2*(-a^2*x^2+1)^(1/2)/x^2)

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

[Out]

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

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Fricas [A]  time = 1.54513, size = 342, normalized size = 2.55 \begin{align*} \frac{14 \, a^{5} x^{5} - 14 \, a^{4} x^{4} - 14 \, a^{3} x^{3} + 14 \, a^{2} x^{2} + 15 \,{\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (16 \, a^{4} x^{4} - a^{3} x^{3} - 23 \, a^{2} x^{2} + 3 \, a x + 3\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{3} c^{2} x^{5} - a^{2} c^{2} x^{4} - a c^{2} x^{3} + c^{2} x^{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^3/(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

1/6*(14*a^5*x^5 - 14*a^4*x^4 - 14*a^3*x^3 + 14*a^2*x^2 + 15*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log((sqrt(
-a^2*x^2 + 1) - 1)/x) - (16*a^4*x^4 - a^3*x^3 - 23*a^2*x^2 + 3*a*x + 3)*sqrt(-a^2*x^2 + 1))/(a^3*c^2*x^5 - a^2
*c^2*x^4 - a*c^2*x^3 + c^2*x^2)

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

[Out]

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

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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 )}^{2} \sqrt{-a^{2} x^{2} + 1} x^{3}}\,{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^3/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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