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

Optimal. Leaf size=63 \[ -\frac{1-a x}{a c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{\sin ^{-1}(a x)}{a c^2} \]

[Out]

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

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Rubi [A]  time = 0.124952, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6131, 6128, 797, 641, 216, 637} \[ -\frac{1-a x}{a c^2 \sqrt{1-a^2 x^2}}-\frac{\sqrt{1-a^2 x^2}}{a c^2}-\frac{\sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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 797

Int[(x_)^2*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/c, Int[(f + g*x)*(a + c*x^2)^(p
+ 1), x], x] - Dist[a/c, Int[(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && EqQ[a*g^2 + f^2*
c, 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]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

Mathematica [A]  time = 0.134099, size = 46, normalized size = 0.73 \[ -\frac{\sqrt{1-a^2 x^2} (a x+2)+(a x+1) \sin ^{-1}(a x)}{a c^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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Maple [B]  time = 0.053, size = 336, normalized size = 5.3 \begin{align*} -{\frac{1}{4\,{a}^{4}{c}^{2} \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{3}{4\,{a}^{3}{c}^{2} \left ( x+{a}^{-1} \right ) ^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{37}{48\,a{c}^{2}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{37\,x}{32\,{c}^{2}}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{37}{32\,{c}^{2}}\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}^{3}{c}^{2}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}} \left ( x-{a}^{-1} \right ) ^{-2}}-{\frac{5}{48\,a{c}^{2}} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{5\,x}{32\,{c}^{2}}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}+{\frac{5}{32\,{c}^{2}}\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)^3*(-a^2*x^2+1)^(3/2)/(c-c/a/x)^2,x)

[Out]

-1/4/a^4/c^2/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-3/4/a^3/c^2/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(
5/2)-37/48/a/c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-37/32/c^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-37/32/c^2/(
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^3/c^2/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*
(x-1/a))^(5/2)-5/48/a/c^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+5/32/c^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x+5/3
2/c^2/(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{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3}{\left (c - \frac{c}{a x}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33273, size = 159, normalized size = 2.52 \begin{align*} -\frac{2 \, a x - 2 \,{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt{-a^{2} x^{2} + 1}{\left (a x + 2\right )} + 2}{a^{2} c^{2} x + a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17544, size = 99, normalized size = 1.57 \begin{align*} -\frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c^{2}{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c^{2}} + \frac{2}{c^{2}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(a*x)*sgn(a)/(c^2*abs(a)) - sqrt(-a^2*x^2 + 1)/(a*c^2) + 2/(c^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x
) + 1)*abs(a))