3.864 \(\int \frac{e^{-3 \tanh ^{-1}(a+b x)}}{x} \, dx\)

Optimal. Leaf size=103 \[ -\frac{2 (1-a)^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1) \sqrt{1-a^2}}+\frac{4 \sqrt{-a-b x+1}}{(a+1) \sqrt{a+b x+1}}+\sin ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.0853212, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {6163, 98, 157, 53, 619, 216, 93, 208} \[ -\frac{2 (1-a)^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(a+1) \sqrt{1-a^2}}+\frac{4 \sqrt{-a-b x+1}}{(a+1) \sqrt{a+b x+1}}+\sin ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a + b*x])*x),x]

[Out]

(4*Sqrt[1 - a - b*x])/((1 + a)*Sqrt[1 + a + b*x]) + ArcSin[a + b*x] - (2*(1 - a)^2*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 98

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

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

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 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^{-3 \tanh ^{-1}(a+b x)}}{x} \, dx &=\int \frac{(1-a-b x)^{3/2}}{x (1+a+b x)^{3/2}} \, dx\\ &=\frac{4 \sqrt{1-a-b x}}{(1+a) \sqrt{1+a+b x}}+\frac{2 \int \frac{\frac{1}{2} (1-a)^2 b+\frac{1}{2} (1+a) b^2 x}{x \sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{(1+a) b}\\ &=\frac{4 \sqrt{1-a-b x}}{(1+a) \sqrt{1+a+b x}}+\frac{(1-a)^2 \int \frac{1}{x \sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{1+a}+b \int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx\\ &=\frac{4 \sqrt{1-a-b x}}{(1+a) \sqrt{1+a+b x}}+\frac{\left (2 (1-a)^2\right ) \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}+b \int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx\\ &=\frac{4 \sqrt{1-a-b x}}{(1+a) \sqrt{1+a+b x}}-\frac{2 (1-a)^2 \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}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{2 b}\\ &=\frac{4 \sqrt{1-a-b x}}{(1+a) \sqrt{1+a+b x}}+\sin ^{-1}(a+b x)-\frac{2 (1-a)^2 \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 = 0.442148, size = 131, normalized size = 1.27 \[ -\frac{4 (a+b x-1)}{(a+1) \sqrt{-(a+b x-1) (a+b x+1)}}+2 \left (\frac{a-1}{a+1}\right )^{3/2} \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )+\frac{2 \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{\sqrt{b}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a + b*x])*x),x]

[Out]

(-4*(-1 + a + b*x))/((1 + a)*Sqrt[-((-1 + a + b*x)*(1 + a + b*x))]) + (2*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a
- b*x])/(Sqrt[2]*Sqrt[b])])/Sqrt[b] + 2*((-1 + a)/(1 + a))^(3/2)*ArcTan[Sqrt[1 - a - b*x]/(Sqrt[(-1 + a)/(1 +
a)]*Sqrt[1 + a + b*x])]

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Maple [B]  time = 0.066, size = 1062, normalized size = 10.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(1+a)^2/b^2/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)-1/(1+a)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(
1+a)/b))^(3/2)-3/2/(1+a)^2*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x-3/2/(1+a)^2*(-(x+(1+a)/b)^2*b^2+2*b*
(x+(1+a)/b))^(1/2)*a-3/2/(1+a)^2*b/(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))-1/3/(1+a)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)-1/2/(1+a)^3*b*(-(x+(1+a)/b)^2*b^2+2*b*(
x+(1+a)/b))^(1/2)*x-1/2/(1+a)^3*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a-1/2/(1+a)^3*b/(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)/b^3/(x+1/b+a/b)^3*(-(x+(1+a)/b
)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+2/(1+a)/b^2/(x+1/b+a/b)^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(5/2)+2/(1+a)*(-
(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(3/2)+3/(1+a)*b*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*x+3/(1+a)*(-(x+(
1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)*a+3/(1+a)*b/(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))+1/3/(1+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-1/2/(1+a)^3*a*b*(-b^2*x^2-2*a*b*x-a^2
+1)^(1/2)*x-3/2/(1+a)^3*a^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-3/2/(1+a)^3*a*b/(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)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/(1+a)^3*a^3*b/(b^2)^(1/2)*arcta
n((b^2)^(1/2)*(x+a/b)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2))+1/(1+a)^3*(-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)*a^2-1/(1+a)^3*(-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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac{3}{2}}}{{\left (b x + a + 1\right )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.83458, size = 1057, normalized size = 10.26 \begin{align*} \left [\frac{{\left ({\left (a - 1\right )} b x + a^{2} - 1\right )} \sqrt{-\frac{a - 1}{a + 1}} \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{3} +{\left (a^{2} + a\right )} b x + a^{2} - a - 1\right )} \sqrt{-\frac{a - 1}{a + 1}} + 2}{x^{2}}\right ) - 2 \,{\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 8 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \,{\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )}}, \frac{{\left ({\left (a - 1\right )} b x + a^{2} - 1\right )} \sqrt{\frac{a - 1}{a + 1}} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{\frac{a - 1}{a + 1}}}{{\left (a - 1\right )} b^{2} x^{2} + a^{3} + 2 \,{\left (a^{2} - a\right )} b x - a^{2} - a + 1}\right ) -{\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 4 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{{\left (a + 1\right )} b x + a^{2} + 2 \, a + 1}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((a - 1)*b*x + a^2 - 1)*sqrt(-(a - 1)/(a + 1))*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 4*a^
2 - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a^3 + (a^2 + a)*b*x + a^2 - a - 1)*sqrt(-(a - 1)/(a + 1)) + 2)/x^2)
- 2*((a + 1)*b*x + a^2 + 2*a + 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2
- 1)) + 8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a + 1)*b*x + a^2 + 2*a + 1), (((a - 1)*b*x + a^2 - 1)*sqrt((a
- 1)/(a + 1))*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 1)*sqrt((a - 1)/(a + 1))/((a - 1)*b^2*
x^2 + a^3 + 2*(a^2 - a)*b*x - a^2 - a + 1)) - ((a + 1)*b*x + a^2 + 2*a + 1)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a
^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) + 4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a + 1)*b*x + a^2 +
2*a + 1)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.21332, size = 208, normalized size = 2.02 \begin{align*} -\frac{b \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{{\left | b \right |}} + \frac{2 \,{\left (a^{2} b - 2 \, a b + b\right )} \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{8 \, b}{{\left (a{\left | b \right |} +{\left | b \right |}\right )}{\left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left | b \right |} + b}{b^{2} x + a b} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arcsin(-b*x - a)*sgn(b)/abs(b) + 2*(a^2*b - 2*a*b + b)*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))) - 8*b/((a*abs(b) + abs(b))*((sqrt(
-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) + 1))