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

Optimal. Leaf size=38 \[ \frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}+\frac {\sin ^{-1}(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6161, 50, 53, 619, 216} \[ \frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}+\frac {\sin ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(-ArcTanh[a + b*x]),x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 6161

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.68 \[ \frac {\sqrt {1-(a+b x)^2}+\sin ^{-1}(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(-ArcTanh[a + b*x]),x]

[Out]

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

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fricas [B]  time = 0.59, size = 77, normalized size = 2.03 \[ \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} - \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{b} \]

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, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 44, normalized size = 1.16 \[ -\frac {\arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b} \]

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, algorithm="giac")

[Out]

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

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maple [B]  time = 0.03, size = 95, normalized size = 2.50 \[ \frac {\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}{b}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1+a}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {1+a}{b}\right )^{2} b^{2}+2 b \left (x +\frac {1+a}{b}\right )}}\right )}{\sqrt {b^{2}}} \]

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)

[Out]

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

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maxima [A]  time = 0.44, size = 37, normalized size = 0.97 \[ \frac {\arcsin \left (b x + a\right )}{b} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b} \]

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, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {1-{\left (a+b\,x\right )}^2}}{a+b\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

[Out]

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

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