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

Optimal. Leaf size=71 \[ -\frac{2^{\frac{n}{2}+1} (-a-b x+1)^{1-\frac{n}{2}} \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{b (2-n)} \]

[Out]

-((2^(1 + n/2)*(1 - a - b*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(b*(2 - n))
)

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Rubi [A]  time = 0.0141212, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6161, 69} \[ -\frac{2^{\frac{n}{2}+1} (-a-b x+1)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (-a-b x+1)\right )}{b (2-n)} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x]),x]

[Out]

-((2^(1 + n/2)*(1 - a - b*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(b*(2 - n))
)

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]

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int e^{n \tanh ^{-1}(a+b x)} \, dx &=\int (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx\\ &=-\frac{2^{1+\frac{n}{2}} (1-a-b x)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (1-a-b x)\right )}{b (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.0406865, size = 50, normalized size = 0.7 \[ \frac{4 e^{(n+2) \tanh ^{-1}(a+b x)} \text{Hypergeometric2F1}\left (2,\frac{n}{2}+1,\frac{n}{2}+2,-e^{2 \tanh ^{-1}(a+b x)}\right )}{b (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(n*ArcTanh[a + b*x]),x]

[Out]

(4*E^((2 + n)*ArcTanh[a + b*x])*Hypergeometric2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcTanh[a + b*x])])/(b*(2 + n))

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( bx+a \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(b*x+a)),x)

[Out]

int(exp(n*arctanh(b*x+a)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (\frac{b x + a + 1}{b x + a - 1}\right )^{\frac{1}{2} \, n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{n \operatorname{atanh}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(b*x+a)),x)

[Out]

Integral(exp(n*atanh(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a)),x, algorithm="giac")

[Out]

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