3.848 \(\int \frac{e^{-\tanh ^{-1}(a+b x)}}{x^2} \, dx\)

Optimal. Leaf size=94 \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1) \sqrt{1-a^2}}-\frac{\sqrt{-a-b x+1} \sqrt{a+b x+1}}{(a+1) x} \]

[Out]

-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/((1 + a)*x)) + (2*b*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 +
a]*Sqrt[1 - a - b*x])])/((1 + a)*Sqrt[1 - a^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0567481, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6163, 94, 93, 208} \[ \frac{2 b \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1) \sqrt{1-a^2}}-\frac{\sqrt{-a-b x+1} \sqrt{a+b x+1}}{(a+1) x} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a + b*x]*x^2),x]

[Out]

-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/((1 + a)*x)) + (2*b*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 +
a]*Sqrt[1 - a - b*x])])/((1 + a)*Sqrt[1 - a^2])

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*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+b x)}}{x^2} \, dx &=\int \frac{\sqrt{1-a-b x}}{x^2 \sqrt{1+a+b x}} \, dx\\ &=-\frac{\sqrt{1-a-b x} \sqrt{1+a+b x}}{(1+a) x}-\frac{b \int \frac{1}{x \sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{1+a}\\ &=-\frac{\sqrt{1-a-b x} \sqrt{1+a+b x}}{(1+a) x}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{-1-a-(-1+a) x^2} \, dx,x,\frac{\sqrt{1+a+b x}}{\sqrt{1-a-b x}}\right )}{1+a}\\ &=-\frac{\sqrt{1-a-b x} \sqrt{1+a+b x}}{(1+a) x}+\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{1-a-b x}}\right )}{(1+a) \sqrt{1-a^2}}\\ \end{align*}

Mathematica [A]  time = 0.0619592, size = 89, normalized size = 0.95 \[ -\frac{\sqrt{-a^2-2 a b x-b^2 x^2+1}}{a x+x}-\frac{2 b \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )}{\sqrt{a-1} (a+1)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a + b*x]*x^2),x]

[Out]

-(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]/(x + a*x)) - (2*b*ArcTan[Sqrt[1 - a - b*x]/(Sqrt[(-1 + a)/(1 + a)]*Sqrt[1
+ a + b*x])])/(Sqrt[-1 + a]*(1 + a)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.083, size = 565, normalized size = 6. \begin{align*} -{\frac{1}{ \left ( 1+a \right ) \left ( -{a}^{2}+1 \right ) x} \left ( -{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-2\,{\frac{ab\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}{ \left ( 1+a \right ) \left ( -{a}^{2}+1 \right ) }}+{\frac{{a}^{2}{b}^{2}}{ \left ( 1+a \right ) \left ( -{a}^{2}+1 \right ) }\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{ab}{1+a}\ln \left ({\frac{1}{x} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ){\frac{1}{\sqrt{-{a}^{2}+1}}}}-{\frac{{b}^{2}x}{ \left ( 1+a \right ) \left ( -{a}^{2}+1 \right ) }\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}-{\frac{{b}^{2}}{ \left ( 1+a \right ) \left ( -{a}^{2}+1 \right ) }\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{b}{ \left ( 1+a \right ) ^{2}}\sqrt{- \left ( x+{\frac{1+a}{b}} \right ) ^{2}{b}^{2}+2\,b \left ( x+{\frac{1+a}{b}} \right ) }}+{\frac{{b}^{2}}{ \left ( 1+a \right ) ^{2}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{1+a}{b}}-{b}^{-1} \right ){\frac{1}{\sqrt{- \left ( x+{\frac{1+a}{b}} \right ) ^{2}{b}^{2}+2\,b \left ( x+{\frac{1+a}{b}} \right ) }}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-{\frac{b}{ \left ( 1+a \right ) ^{2}}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}+{\frac{a{b}^{2}}{ \left ( 1+a \right ) ^{2}}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{b}{ \left ( 1+a \right ) ^{2}}\sqrt{-{a}^{2}+1}\ln \left ({\frac{1}{x} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(1+a)/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-2/(1+a)*a*b/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/(1+a)
*a^2*b^2/(-a^2+1)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))+1/(1+a)*a*b/(-a^2+1)^
(1/2)*ln((-2*a^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-1/(1+a)*b^2/(-a^2+1)*(-b^2*x^2-
2*a*b*x-a^2+1)^(1/2)*x-1/(1+a)*b^2/(-a^2+1)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1
/2))+1/(1+a)^2*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)+1/(1+a)^2*b^2/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+(1
+a)/b-1/b)/(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2))-1/(1+a)^2*b*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/(1+a)^2*b^
2*a/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))+1/(1+a)^2*b*(-a^2+1)^(1/2)*ln((-2*a
^2+2-2*x*a*b+2*(-a^2+1)^(1/2)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{{\left (b x + a + 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.70272, size = 633, normalized size = 6.73 \begin{align*} \left [-\frac{\sqrt{-a^{2} + 1} b x \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) + 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{2} - 1\right )}}{2 \,{\left (a^{3} + a^{2} - a - 1\right )} x}, -\frac{\sqrt{a^{2} - 1} b x \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) + \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{2} - 1\right )}}{{\left (a^{3} + a^{2} - a - 1\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(-a^2 + 1)*b*x*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2
+ 1)*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1) - 4*a^2 + 2)/x^2) + 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1))/((a
^3 + a^2 - a - 1)*x), -(sqrt(a^2 - 1)*b*x*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt(a^2
- 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1)) + sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^2 - 1))
/((a^3 + a^2 - a - 1)*x)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(a + b*x - 1)*(a + b*x + 1))/(x**2*(a + b*x + 1)), x)

________________________________________________________________________________________

Giac [B]  time = 1.27779, size = 301, normalized size = 3.2 \begin{align*} -\frac{2 \, b^{2} \arctan \left (\frac{\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt{a^{2} - 1}}\right )}{\sqrt{a^{2} - 1}{\left (a{\left | b \right |} +{\left | b \right |}\right )}} - \frac{2 \,{\left (a b^{2} - \frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b}\right )}}{{\left (a^{2}{\left | b \right |} + a{\left | b \right |}\right )}{\left (\frac{{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac{2 \,{\left (\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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