3.849 \(\int \frac{e^{-\tanh ^{-1}(a+b x)}}{x^3} \, dx\)
Optimal. Leaf size=162 \[ -\frac{(1-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(1-a) (a+1)^2 \sqrt{1-a^2}}-\frac{(-a-b x+1)^{3/2} \sqrt{a+b x+1}}{2 \left (1-a^2\right ) x^2}+\frac{(1-2 a) b \sqrt{-a-b x+1} \sqrt{a+b x+1}}{2 (1-a) (a+1)^2 x} \]
[Out]
((1 - 2*a)*b*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*(1 - a)*(1 + a)^2*x) - ((1 - a - b*x)^(3/2)*Sqrt[1 + a +
b*x])/(2*(1 - a^2)*x^2) - ((1 - 2*a)*b^2*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x
])])/((1 - a)*(1 + a)^2*Sqrt[1 - a^2])
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Rubi [A] time = 0.0979128, antiderivative size = 162, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used =
{6163, 96, 94, 93, 208} \[ -\frac{(1-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{(1-a) (a+1)^2 \sqrt{1-a^2}}-\frac{(-a-b x+1)^{3/2} \sqrt{a+b x+1}}{2 \left (1-a^2\right ) x^2}+\frac{(1-2 a) b \sqrt{-a-b x+1} \sqrt{a+b x+1}}{2 (1-a) (a+1)^2 x} \]
Antiderivative was successfully verified.
[In]
Int[1/(E^ArcTanh[a + b*x]*x^3),x]
[Out]
((1 - 2*a)*b*Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/(2*(1 - a)*(1 + a)^2*x) - ((1 - a - b*x)^(3/2)*Sqrt[1 + a +
b*x])/(2*(1 - a^2)*x^2) - ((1 - 2*a)*b^2*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x
])])/((1 - a)*(1 + a)^2*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 96
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])
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^3} \, dx &=\int \frac{\sqrt{1-a-b x}}{x^3 \sqrt{1+a+b x}} \, dx\\ &=-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x^2}-\frac{((1-2 a) b) \int \frac{\sqrt{1-a-b x}}{x^2 \sqrt{1+a+b x}} \, dx}{2 \left (1-a^2\right )}\\ &=\frac{(1-2 a) b \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 (1-a) (1+a)^2 x}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x^2}+\frac{\left ((1-2 a) b^2\right ) \int \frac{1}{x \sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{2 (1-a) (1+a)^2}\\ &=\frac{(1-2 a) b \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 (1-a) (1+a)^2 x}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x^2}+\frac{\left ((1-2 a) b^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) (1+a)^2}\\ &=\frac{(1-2 a) b \sqrt{1-a-b x} \sqrt{1+a+b x}}{2 (1-a) (1+a)^2 x}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x^2}-\frac{(1-2 a) b^2 \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{1-a-b x}}\right )}{(1-a) (1+a)^2 \sqrt{1-a^2}}\\ \end{align*}
Mathematica [A] time = 0.124291, size = 117, normalized size = 0.72 \[ \frac{(2 a-1) b^2 \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )}{(a-1)^{3/2} (a+1)^{5/2}}-\frac{\left (a^2-a b x+2 b x-1\right ) \sqrt{-a^2-2 a b x-b^2 x^2+1}}{2 (a-1) (a+1)^2 x^2} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(E^ArcTanh[a + b*x]*x^3),x]
[Out]
-((-1 + a^2 + 2*b*x - a*b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2])/(2*(-1 + a)*(1 + a)^2*x^2) + ((-1 + 2*a)*b^2*A
rcTan[Sqrt[1 - a - b*x]/(Sqrt[(-1 + a)/(1 + a)]*Sqrt[1 + a + b*x])])/((-1 + a)^(3/2)*(1 + a)^(5/2))
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Maple [B] time = 0.049, size = 1116, normalized size = 6.9 \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)*(1-(b*x+a)^2)^(1/2)/x^3,x)
[Out]
1/(1+a)^2*b/(-a^2+1)/x*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)+2/(1+a)^2*b^2*a/(-a^2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-
1/(1+a)^2*b^3*a^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^
2*a/(-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)^2*b^3/(-a^
2+1)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x+1/(1+a)^2*b^3/(-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)^3*b^2*(-(x+(1+a)/b)^2*b^2+2*b*(x+(1+a)/b))^(1/2)-1/(1+a)^3*b^3/(b^2)^(1/2)*arcta
n((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)^3*b^2*(-b^2*x^2-2*a*b*x-a^2+
1)^(1/2)-1/(1+a)^3*b^3*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)^3*b^2*
(-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)-1/2/(1+a)/(-a^2+1)/x^2
*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-1/2/(1+a)*a*b/(-a^2+1)^2/x*(-b^2*x^2-2*a*b*x-a^2+1)^(3/2)-1/(1+a)*a^2*b^2/(-a^
2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/2/(1+a)*a^3*b^3/(-a^2+1)^2/(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))+1/2/(1+a)*a^2*b^2/(-a^2+1)^(3/2)*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)-1/2/(1+a)*a*b^3/(-a^2+1)^2*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*x-1/2/(1+a)*a*b^3/(-a^2+1)^2
/(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))-1/2/(1+a)*b^2/(-a^2+1)*(-b^2*x^2-2*a*b
*x-a^2+1)^(1/2)+1/2/(1+a)*b^3/(-a^2+1)*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)
)+1/2/(1+a)*b^2/(-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{\sqrt{-{\left (b x + a\right )}^{2} + 1}}{{\left (b x + a + 1\right )} x^{3}}\,{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^3,x, algorithm="maxima")
[Out]
integrate(sqrt(-(b*x + a)^2 + 1)/((b*x + a + 1)*x^3), x)
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Fricas [A] time = 1.7765, size = 821, normalized size = 5.07 \begin{align*} \left [-\frac{\sqrt{-a^{2} + 1}{\left (2 \, a - 1\right )} b^{2} x^{2} \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 \,{\left (a^{4} -{\left (a^{3} - 2 \, a^{2} - a + 2\right )} b x - 2 \, a^{2} + 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{4 \,{\left (a^{5} + a^{4} - 2 \, a^{3} - 2 \, a^{2} + a + 1\right )} x^{2}}, \frac{\sqrt{a^{2} - 1}{\left (2 \, a - 1\right )} b^{2} x^{2} \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 ) -{\left (a^{4} -{\left (a^{3} - 2 \, a^{2} - a + 2\right )} b x - 2 \, a^{2} + 1\right )} \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \,{\left (a^{5} + a^{4} - 2 \, a^{3} - 2 \, a^{2} + a + 1\right )} x^{2}}\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^3,x, algorithm="fricas")
[Out]
[-1/4*(sqrt(-a^2 + 1)*(2*a - 1)*b^2*x^2*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*(a^4 - (a^3 - 2*a^2 - a + 2)*b*x -
2*a^2 + 1)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/((a^5 + a^4 - 2*a^3 - 2*a^2 + a + 1)*x^2), 1/2*(sqrt(a^2 - 1)*(
2*a - 1)*b^2*x^2*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)) - (a^4 - (a^3 - 2*a^2 - a + 2)*b*x - 2*a^2 + 1)*sqrt(-b^2*x^2 - 2*a*b*x
- a^2 + 1))/((a^5 + a^4 - 2*a^3 - 2*a^2 + a + 1)*x^2)]
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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^{3} \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**3,x)
[Out]
Integral(sqrt(-(a + b*x - 1)*(a + b*x + 1))/(x**3*(a + b*x + 1)), x)
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Giac [B] time = 1.27263, size = 1013, normalized size = 6.25 \begin{align*} \text{result too large to display} \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^3,x, algorithm="giac")
[Out]
(2*a*b^3 - b^3)*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))/((
a^3*abs(b) + a^2*abs(b) - a*abs(b) - abs(b))*sqrt(a^2 - 1)) + (2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) +
b)^2*a^4*b^3/(b^2*x + a*b)^2 + 2*a^4*b^3 - 5*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^3*b^3/(b^2*x +
a*b) - 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^3*b^3/(b^2*x + a*b)^2 - 3*(sqrt(-b^2*x^2 - 2*a*b*
x - a^2 + 1)*abs(b) + b)^3*a^3*b^3/(b^2*x + a*b)^3 - 2*a^3*b^3 + 6*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b)
+ b)*a^2*b^3/(b^2*x + a*b) + 3*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^2*b^3/(b^2*x + a*b)^2 + 2*(
sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a^2*b^3/(b^2*x + a*b)^3 - a^2*b^3 + 2*(sqrt(-b^2*x^2 - 2*a*b*
x - a^2 + 1)*abs(b) + b)*a*b^3/(b^2*x + a*b) - 4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a*b^3/(b^2*
x + a*b)^2 + 2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^3*a*b^3/(b^2*x + a*b)^3 - 2*(sqrt(-b^2*x^2 - 2*
a*b*x - a^2 + 1)*abs(b) + b)^2*b^3/(b^2*x + a*b)^2)/((a^5*abs(b) + a^4*abs(b) - a^3*abs(b) - a^2*abs(b))*((sqr
t(-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)*ab
s(b) + b)/(b^2*x + a*b))^2)