3.9.68 \(\int \frac {e^{\tanh ^{-1}(1+b x)}}{2+b x} \, dx\) [868]

Optimal. Leaf size=10 \[ \frac {\text {ArcSin}(1+b x)}{b} \]

[Out]

arcsin(b*x+1)/b

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Rubi [A]
time = 0.02, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6298, 55, 633, 222} \begin {gather*} \frac {\text {ArcSin}(b x+1)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[1 + b*x]/(2 + b*x),x]

[Out]

ArcSin[1 + b*x]/b

Rule 55

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 222

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 633

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 6298

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}(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 {\text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(37\) vs. \(2(10)=20\).
time = 0.01, size = 37, normalized size = 3.70 \begin {gather*} -\frac {2 \sqrt {-b x} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2} \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[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] Leaf count of result is larger than twice the leaf count of optimal. \(33\) vs. \(2(10)=20\).
time = 0.05, size = 34, normalized size = 3.40

method	result	size
default	\(\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {1}{b}\right )}{\sqrt {-b^{2} x^{2}-2 b x}}\right )}{\sqrt {b^{2}}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 [A]
time = 0.51, size = 18, normalized size = 1.80 \begin {gather*} -\frac {\arcsin \left (-\frac {b^{2} x + b}{b}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arcsin(-(b^2*x + b)/b)/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (10) = 20\).
time = 0.37, size = 28, normalized size = 2.80 \begin {gather*} -\frac {2 \, \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, b x}}{b x}\right )}{b} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {1 - \left (b x + 1\right )^{2}}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (10) = 20\).
time = 0.41, size = 38, normalized size = 3.80 \begin {gather*} \frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, b x} {\left (x + \frac {1}{b}\right )} - \frac {\arcsin \left (-b x - 1\right ) \mathrm {sgn}\left (b\right )}{2 \, {\left | b \right |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.97, size = 10, normalized size = 1.00 \begin {gather*} \frac {\mathrm {asin}\left (b\,x+1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - (b*x + 1)^2)^(1/2),x)

[Out]

asin(b*x + 1)/b

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