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

Optimal. Leaf size=99 \[ \frac{4 a (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}+\frac{a (11 a x+9)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{3 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^2} \]

[Out]

(4*a*(1 + a*x))/(3*c^2*(1 - a^2*x^2)^(3/2)) + (a*(9 + 11*a*x))/(3*c^2*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(
c^2*x) - (3*a*ArcTanh[Sqrt[1 - a^2*x^2]])/c^2

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Rubi [A]  time = 0.266828, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {6128, 852, 1805, 807, 266, 63, 208} \[ \frac{4 a (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}+\frac{a (11 a x+9)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{3 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(1 + a*x))/(3*c^2*(1 - a^2*x^2)^(3/2)) + (a*(9 + 11*a*x))/(3*c^2*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x^2]/(
c^2*x) - (3*a*ArcTanh[Sqrt[1 - a^2*x^2]])/c^2

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[2*p]

Rule 852

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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 (c-a c x)^2} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{x^2 (c-a c x)^3} \, dx\\ &=\frac{\int \frac{(c+a c x)^3}{x^2 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^5}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{\int \frac{-3 c^3-9 a c^3 x-8 a^2 c^3 x^2}{x^2 \left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^5}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (9+11 a x)}{3 c^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{3 c^3+9 a c^3 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{3 c^5}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (9+11 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}+\frac{(3 a) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{c^2}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (9+11 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{2 c^2}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (9+11 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}-\frac{3 \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^2}\\ &=\frac{4 a (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a (9+11 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{c^2 x}-\frac{3 a \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{c^2}\\ \end{align*}

Mathematica [A]  time = 0.0415937, size = 91, normalized size = 0.92 \[ \frac{14 a^3 x^3-5 a^2 x^2-9 a x (a x-1) \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-16 a x+3}{3 c^2 x (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 - 16*a*x - 5*a^2*x^2 + 14*a^3*x^3 - 9*a*x*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])/(3*c^2*x
*(-1 + a*x)*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.045, size = 118, normalized size = 1.2 \begin{align*}{\frac{1}{{c}^{2}} \left ( -{\frac{1}{x}\sqrt{-{a}^{2}{x}^{2}+1}}-3\,a{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) +{\frac{2}{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{11}{3}\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*c*x+c)^2,x)

[Out]

1/c^2*(-(-a^2*x^2+1)^(1/2)/x-3*a*arctanh(1/(-a^2*x^2+1)^(1/2))+2/3/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1
/2)-11/3/(x-1/a)*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))

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

[Out]

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

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Fricas [A]  time = 1.60348, size = 252, normalized size = 2.55 \begin{align*} \frac{13 \, a^{3} x^{3} - 26 \, a^{2} x^{2} + 13 \, a x + 9 \,{\left (a^{3} x^{3} - 2 \, a^{2} x^{2} + a x\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (14 \, a^{2} x^{2} - 19 \, a x + 3\right )} \sqrt{-a^{2} x^{2} + 1}}{3 \,{\left (a^{2} c^{2} x^{3} - 2 \, a c^{2} x^{2} + c^{2} 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*c*x+c)^2,x, algorithm="fricas")

[Out]

1/3*(13*a^3*x^3 - 26*a^2*x^2 + 13*a*x + 9*(a^3*x^3 - 2*a^2*x^2 + a*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (14*a^
2*x^2 - 19*a*x + 3)*sqrt(-a^2*x^2 + 1))/(a^2*c^2*x^3 - 2*a*c^2*x^2 + c^2*x)

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

[Out]

(Integral(a*x/(a**2*x**4*sqrt(-a**2*x**2 + 1) - 2*a*x**3*sqrt(-a**2*x**2 + 1) + x**2*sqrt(-a**2*x**2 + 1)), x)
+ Integral(1/(a**2*x**4*sqrt(-a**2*x**2 + 1) - 2*a*x**3*sqrt(-a**2*x**2 + 1) + x**2*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}{\sqrt{-a^{2} x^{2} + 1}{\left (a c x - c\right )}^{2} 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*c*x+c)^2,x, algorithm="giac")

[Out]

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