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

Optimal. Leaf size=39 \[ \frac {\sin ^{-1}(a+b x)}{b}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b} \]

[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 = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6161, 50, 53, 619, 216} \[ \frac {\sin ^{-1}(a+b x)}{b}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b} \]

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 = 28, normalized size = 0.72 \[ \frac {\sin ^{-1}(a+b x)-\sqrt {1-(a+b x)^2}}{b} \]

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 = 0.65, size = 76, normalized size = 1.95 \[ -\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} + \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.21, size = 36, normalized size = 0.92 \[ -\frac {\arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} - \frac {\sqrt {-{\left (b x + a\right )}^{2} + 1}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*x + a)^2 + 1)/b

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maple [A]  time = 0.04, size = 71, normalized size = 1.82 \[ -\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{b}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 65, normalized size = 1.67 \[ -\frac {\arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{b} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.42, size = 101, normalized size = 2.59 \[ \frac {\mathrm {asin}\left (a+b\,x\right )}{b}-\frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{b}-\frac {a\,\ln \left (\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}-\frac {x\,b^2+a\,b}{\sqrt {-b^2}}\right )}{\sqrt {-b^2}}+\frac {a\,\mathrm {asin}\left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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