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

Optimal. Leaf size=132 \[ \frac{a^2 (17 a x+15)}{3 c^2 \sqrt{1-a^2 x^2}}+\frac{4 a^2 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{11 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2} \]

[Out]

(4*a^2*(1 + a*x))/(3*c^2*(1 - a^2*x^2)^(3/2)) + (a^2*(15 + 17*a*x))/(3*c^2*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x
^2]/(2*c^2*x^2) - (3*a*Sqrt[1 - a^2*x^2])/(c^2*x) - (11*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(2*c^2)

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Rubi [A]  time = 0.333784, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {6128, 852, 1805, 1807, 807, 266, 63, 208} \[ \frac{a^2 (17 a x+15)}{3 c^2 \sqrt{1-a^2 x^2}}+\frac{4 a^2 (a x+1)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{11 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*(1 + a*x))/(3*c^2*(1 - a^2*x^2)^(3/2)) + (a^2*(15 + 17*a*x))/(3*c^2*Sqrt[1 - a^2*x^2]) - Sqrt[1 - a^2*x
^2]/(2*c^2*x^2) - (3*a*Sqrt[1 - a^2*x^2])/(c^2*x) - (11*a^2*ArcTanh[Sqrt[1 - a^2*x^2]])/(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 1807

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

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 (c-a c x)^2} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{x^3 (c-a c x)^3} \, dx\\ &=\frac{\int \frac{(c+a c x)^3}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx}{c^5}\\ &=\frac{4 a^2 (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-12 a^2 c^3 x^2-8 a^3 c^3 x^3}{x^3 \left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^5}\\ &=\frac{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}+\frac{\int \frac{3 c^3+9 a c^3 x+15 a^2 c^3 x^2}{x^3 \sqrt{1-a^2 x^2}} \, dx}{3 c^5}\\ &=\frac{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{\int \frac{-18 a c^3-33 a^2 c^3 x}{x^2 \sqrt{1-a^2 x^2}} \, dx}{6 c^5}\\ &=\frac{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}+\frac{\left (11 a^2\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx}{2 c^2}\\ &=\frac{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}+\frac{\left (11 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}-\frac{11 \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{4 a^2 (1+a x)}{3 c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^2 (15+17 a x)}{3 c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{2 c^2 x^2}-\frac{3 a \sqrt{1-a^2 x^2}}{c^2 x}-\frac{11 a^2 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0462549, size = 103, normalized size = 0.78 \[ \frac{52 a^4 x^4-19 a^3 x^3-59 a^2 x^2-33 a^2 x^2 (a x-1) \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+15 a x+3}{6 c^2 x^2 (a x-1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 15*a*x - 59*a^2*x^2 - 19*a^3*x^3 + 52*a^4*x^4 - 33*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 = 0.055, size = 181, normalized size = 1.4 \begin{align*}{\frac{1}{{c}^{2}} \left ( -3\,{\frac{a\sqrt{-{a}^{2}{x}^{2}+1}}{x}}-{\frac{11\,{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+2\,a \left ( 1/3\,{\frac{1}{a}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) } \left ( x-{a}^{-1} \right ) ^{-2}}-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 ) -5\,{a\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*c*x+c)^2,x)

[Out]

1/c^2*(-3*a*(-a^2*x^2+1)^(1/2)/x-11/2*a^2*arctanh(1/(-a^2*x^2+1)^(1/2))+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))-5*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}{\sqrt{-a^{2} x^{2} + 1}{\left (a c x - c\right )}^{2} 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*c*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.58376, size = 285, normalized size = 2.16 \begin{align*} \frac{38 \, a^{4} x^{4} - 76 \, a^{3} x^{3} + 38 \, a^{2} x^{2} + 33 \,{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) -{\left (52 \, a^{3} x^{3} - 71 \, a^{2} x^{2} + 12 \, a x + 3\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a^{2} c^{2} x^{4} - 2 \, 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*c*x+c)^2,x, algorithm="fricas")

[Out]

1/6*(38*a^4*x^4 - 76*a^3*x^3 + 38*a^2*x^2 + 33*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log((sqrt(-a^2*x^2 + 1) - 1)/x)
- (52*a^3*x^3 - 71*a^2*x^2 + 12*a*x + 3)*sqrt(-a^2*x^2 + 1))/(a^2*c^2*x^4 - 2*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 x}{a^{2} x^{5} \sqrt{- a^{2} x^{2} + 1} - 2 a x^{4} \sqrt{- a^{2} x^{2} + 1} + x^{3} \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{1}{a^{2} x^{5} \sqrt{- a^{2} x^{2} + 1} - 2 a x^{4} \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*c*x+c)**2,x)

[Out]

(Integral(a*x/(a**2*x**5*sqrt(-a**2*x**2 + 1) - 2*a*x**4*sqrt(-a**2*x**2 + 1) + x**3*sqrt(-a**2*x**2 + 1)), x)
+ Integral(1/(a**2*x**5*sqrt(-a**2*x**2 + 1) - 2*a*x**4*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}{\sqrt{-a^{2} x^{2} + 1}{\left (a c x - c\right )}^{2} 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*c*x+c)^2,x, algorithm="giac")

[Out]

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