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

Optimal. Leaf size=41 \[ -\frac{2 (1-a x)}{a c \sqrt{1-a^2 x^2}}-\frac{\sin ^{-1}(a x)}{a c} \]

[Out]

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

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Rubi [A]  time = 0.04145, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {6127, 653, 216} \[ -\frac{2 (1-a x)}{a c \sqrt{1-a^2 x^2}}-\frac{\sin ^{-1}(a x)}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6127

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

Rule 653

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

Mathematica [A]  time = 0.0521901, size = 59, normalized size = 1.44 \[ \frac{2 \left (\sqrt{1-a^2 x^2} \sin ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )+a x-1\right )}{a c \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(-1 + a*x + 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 = 7.1 \begin{align*} -{\frac{3}{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{17}{24\,ac} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{17\,x}{16\,c}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}-{\frac{17}{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}}}}}-{\frac{1}{2\,c{a}^{4} \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}}}} \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)/(-a*c*x+c),x)

[Out]

-3/4/c/a^3/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-17/24/c/a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)-17/16/c*(
-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x-17/16/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/c/a*(-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))-1/2/c/a^4/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2
)

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

[Out]

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

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Fricas [A]  time = 1.67339, size = 138, normalized size = 3.37 \begin{align*} -\frac{2 \,{\left (a x -{\left (a x + 1\right )} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{a x}\right ) + \sqrt{-a^{2} x^{2} + 1} + 1\right )}}{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)/(-a*c*x+c),x, algorithm="fricas")

[Out]

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

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

[Out]

-(Integral(sqrt(-a**2*x**2 + 1)/(a**4*x**4 + 2*a**3*x**3 - 2*a*x - 1), x) + Integral(-a**2*x**2*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 = 1.23199, size = 72, normalized size = 1.76 \begin{align*} -\frac{\arcsin \left (a x\right ) \mathrm{sgn}\left (a\right )}{c{\left | a \right |}} + \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)/(-a*c*x+c),x, algorithm="giac")

[Out]

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