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

Optimal. Leaf size=68 \[ -\frac {2 (1-a) \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}}-\sin ^{-1}(a+b x) \]

[Out]

-arcsin(b*x+a)-2*(1-a)*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(-a^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6163, 105, 53, 619, 216, 93, 208} \[ -\frac {2 (1-a) \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}}-\sin ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a + b*x]*x),x]

[Out]

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

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 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 105

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

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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a+b x)}}{x} \, dx &=\int \frac {\sqrt {1-a-b x}}{x \sqrt {1+a+b x}} \, dx\\ &=-\left ((-1+a) \int \frac {1}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx\right )-b \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx\\ &=(2 (1-a)) \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 )-b \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx\\ &=-\frac {2 (1-a) \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{\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}\\ &=-\sin ^{-1}(a+b x)-\frac {2 (1-a) \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{\sqrt {1-a^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 106, normalized size = 1.56 \[ \frac {2 \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {-b}}\right )}{\sqrt {b}}-\frac {2 \sqrt {a-1} \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )}{\sqrt {-a-1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a + b*x]*x),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.61, size = 303, normalized size = 4.46 \[ \left [\frac {1}{2} \, \sqrt {-\frac {a - 1}{a + 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} + a\right )} b x + a^{2} - a - 1\right )} \sqrt {-\frac {a - 1}{a + 1}} + 2}{x^{2}}\right ) + \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ), -\sqrt {\frac {a - 1}{a + 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a - 1}{a + 1}}}{{\left (a - 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} - a\right )} b x - a^{2} - a + 1}\right ) + \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )\right ] \]

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,x, algorithm="fricas")

[Out]

[1/2*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) + 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)), -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
)) + 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))]

________________________________________________________________________________________

giac [A]  time = 0.41, size = 89, normalized size = 1.31 \[ \frac {b \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} - \frac {2 \, {\left (a b - b\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left | b \right |}} \]

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,x, algorithm="giac")

[Out]

b*arcsin(-b*x - a)*sgn(b)/abs(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)*abs(b))

________________________________________________________________________________________

maple [B]  time = 0.04, size = 249, normalized size = 3.66 \[ \frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{1+a}-\frac {a b \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\left (1+a \right ) \sqrt {b^{2}}}-\frac {\sqrt {-a^{2}+1}\, \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{1+a}-\frac {\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{1+a}-\frac {b \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{\left (1+a \right ) \sqrt {b^{2}}} \]

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,x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is a-1 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-{\left (a+b\,x\right )}^2}}{x\,\left (a+b\,x+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{x \left (a + b x + 1\right )}\, dx \]

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,x)

[Out]

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

________________________________________________________________________________________