3.868 \(\int \frac{e^{\tanh ^{-1}(1+b x)}}{2+b x} \, dx\)

Optimal. Leaf size=10 \[ \frac{\sin ^{-1}(b x+1)}{b} \]

[Out]

ArcSin[1 + b*x]/b

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Rubi [A]  time = 0.0353483, antiderivative size = 10, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6163, 53, 619, 216} \[ \frac{\sin ^{-1}(b x+1)}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[1 + b*x]/(2 + b*x),x]

[Out]

ArcSin[1 + b*x]/b

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 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]

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(1+b x)}}{2+b x} \, dx &=\int \frac{1}{\sqrt{-b x} \sqrt{2+b x}} \, dx\\ &=\int \frac{1}{\sqrt{-2 b x-b^2 x^2}} \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 b^2}}} \, dx,x,-2 b-2 b^2 x\right )}{2 b^2}\\ &=\frac{\sin ^{-1}(1+b x)}{b}\\ \end{align*}

Mathematica [B]  time = 0.0120201, size = 37, normalized size = 3.7 \[ -\frac{2 \sqrt{-b x} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2} \sqrt{x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[1 + b*x]/(2 + b*x),x]

[Out]

(-2*Sqrt[-(b*x)]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(b^(3/2)*Sqrt[x])

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Maple [B]  time = 0.032, size = 34, normalized size = 3.4 \begin{align*}{\arctan \left ({(x+{b}^{-1})\sqrt{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,bx}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.4805, size = 58, normalized size = 5.8 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, b x}}{b x}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*arctan(sqrt(-b^2*x^2 - 2*b*x)/(b*x))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(1 - (b*x + 1)**2), x)

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Giac [A]  time = 1.20334, size = 20, normalized size = 2. \begin{align*} -\frac{\arcsin \left (-b x - 1\right ) \mathrm{sgn}\left (b\right )}{{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(-b*x - 1)*sgn(b)/abs(b)