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

Optimal. Leaf size=65 \[ -\frac{(1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{a c}-\frac{2 \sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.104094, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6131, 6128, 789, 641, 216} \[ -\frac{(1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{a c}-\frac{2 \sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 789

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g + e*f)*
(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(p + 1)), x] - Dist[(e*(m*(d*g + e*f) + 2*e*f*(p + 1)))/(2*c*d*(p + 1)
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && 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^{-3 \tanh ^{-1}(a x)}}{c-\frac{c}{a x}} \, dx &=-\frac{a \int \frac{e^{-3 \tanh ^{-1}(a x)} x}{1-a x} \, dx}{c}\\ &=-\frac{a \int \frac{x (1-a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{c}\\ &=-\frac{(1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{2 \int \frac{1-a x}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=-\frac{(1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{a c}-\frac{2 \int \frac{1}{\sqrt{1-a^2 x^2}} \, dx}{c}\\ &=-\frac{(1-a x)^2}{a c \sqrt{1-a^2 x^2}}-\frac{2 \sqrt{1-a^2 x^2}}{a c}-\frac{2 \sin ^{-1}(a x)}{a c}\\ \end{align*}

Mathematica [A]  time = 0.0329864, size = 67, normalized size = 1.03 \[ \frac{a^2 x^2+4 \sqrt{1-a^2 x^2} \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )+2 a x-3}{a c \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.052, size = 292, normalized size = 4.5 \begin{align*} -{\frac{1}{2\,{a}^{4}c \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{5}{4\,{a}^{3}c \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{31}{24\,ac} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{31\,x}{16\,c}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{31}{16\,c}\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}{24\,ac} \left ( -{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{x}{16\,c}\sqrt{-{a}^{2} \left ( x-{a}^{-1} \right ) ^{2}-2\,a \left ( x-{a}^{-1} \right ) }}-{\frac{1}{16\,c}\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),x)

[Out]

-1/2/a^4/c/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-5/4/a^3/c/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)
-31/24/c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-31/16/c*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-31/16/c/(a^2)^(1/2)
*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))+1/24/a/c*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-1/16/c*(
-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x-1/16/c/(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 )}}\,{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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.21634, size = 154, normalized size = 2.37 \begin{align*} -\frac{3 \, a x - 4 \,{\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 + 3\right )} + 3}{a^{2} c x + a c} \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),x, algorithm="fricas")

[Out]

-(3*a*x - 4*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + sqrt(-a^2*x^2 + 1)*(a*x + 3) + 3)/(a^2*c*x + a*
c)

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

[Out]

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

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Giac [A]  time = 1.24643, size = 99, normalized size = 1.52 \begin{align*} -\frac{2 \, \arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \right |}} - \frac{\sqrt{-a^{2} x^{2} + 1}}{a c} + \frac{4}{c{\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),x, algorithm="giac")

[Out]

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