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

Optimal. Leaf size=94 \[ -\frac{(a x+1)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 (a x+1)}{3 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{2 \sin ^{-1}(a x)}{a c^3} \]

[Out]

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

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Rubi [A]  time = 0.259897, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6131, 6128, 852, 1635, 641, 216} \[ -\frac{(a x+1)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 (a x+1)}{3 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{2 \sin ^{-1}(a x)}{a c^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]

[Out]

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

Rule 6131

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

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 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{-\tanh ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^3} \, dx &=-\frac{a^3 \int \frac{e^{-\tanh ^{-1}(a x)} x^3}{(1-a x)^3} \, dx}{c^3}\\ &=-\frac{a^3 \int \frac{x^3}{(1-a x)^2 \sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac{a^3 \int \frac{x^3 (1+a x)^2}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^3}\\ &=-\frac{(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{a^3 \int \frac{(1+a x) \left (\frac{2}{a^3}+\frac{3 x}{a^2}+\frac{3 x^2}{a}\right )}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^3}\\ &=-\frac{(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 (1+a x)}{3 a c^3 \sqrt{1-a^2 x^2}}-\frac{a^3 \int \frac{\frac{6}{a^3}+\frac{3 x}{a^2}}{\sqrt{1-a^2 x^2}} \, dx}{3 c^3}\\ &=-\frac{(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 (1+a x)}{3 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c^3}\\ &=-\frac{(1+a x)^2}{3 a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac{8 (1+a x)}{3 a c^3 \sqrt{1-a^2 x^2}}+\frac{\sqrt{1-a^2 x^2}}{a c^3}-\frac{2 \sin ^{-1}(a x)}{a c^3}\\ \end{align*}

Mathematica [A]  time = 0.125562, size = 53, normalized size = 0.56 \[ \frac{\frac{\sqrt{1-a^2 x^2} \left (3 a^2 x^2-14 a x+10\right )}{(a x-1)^2}-6 \sin ^{-1}(a x)}{3 a c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]

[Out]

((Sqrt[1 - a^2*x^2]*(10 - 14*a*x + 3*a^2*x^2))/(-1 + a*x)^2 - 6*ArcSin[a*x])/(3*a*c^3)

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Maple [B]  time = 0.049, size = 242, normalized size = 2.6 \begin{align*}{\frac{1}{6\,{a}^{4}{c}^{3}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-3}}+{\frac{5}{4\,{a}^{3}{c}^{3}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}+{\frac{17}{8\,a{c}^{3}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{17}{8\,{c}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}}+{\frac{1}{8\,a{c}^{3}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}+{\frac{1}{8\,{c}^{3}}\arctan \left ({x\sqrt{{a}^{2}}{\frac{1}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^4/c^3/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+5/4/a^3/c^3/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3
/2)+17/8/a/c^3*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-17/8/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*
a*(x-1/a))^(1/2))+1/8/a/c^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/8/c^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-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{\sqrt{-a^{2} x^{2} + 1}}{{\left (a x + 1\right )}{\left (c - \frac{c}{a x}\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.18325, size = 244, normalized size = 2.6 \begin{align*} \frac{10 \, a^{2} x^{2} - 20 \, a x + 12 \,{\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) +{\left (3 \, a^{2} x^{2} - 14 \, a x + 10\right )} \sqrt{-a^{2} x^{2} + 1} + 10}{3 \,{\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,x, algorithm="fricas")

[Out]

1/3*(10*a^2*x^2 - 20*a*x + 12*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^2*x^2 - 14*a
*x + 10)*sqrt(-a^2*x^2 + 1) + 10)/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21157, size = 170, normalized size = 1.81 \begin{align*} -\frac{2 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{3}{\left | a \right |}} + \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{3}} - \frac{2 \,{\left (\frac{15 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}}{a^{2} x} - \frac{6 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} - 7\right )}}{3 \, c^{3}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} - 1\right )}^{3}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,x, algorithm="giac")

[Out]

-2*arcsin(a*x)*sgn(a)/(c^3*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^3) - 2/3*(15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2
*x) - 6*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) - 7)/(c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^3*
abs(a))