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

Optimal. Leaf size=135 \[ \frac{2 (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \text{Hypergeometric2F1}\left (1,-\frac{n}{2},1-\frac{n}{2},\frac{(a+1) (-a-b x+1)}{(1-a) (a+b x+1)}\right )}{n}-\frac{2^{\frac{n}{2}+1} (-a-b x+1)^{-n/2} \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.0704511, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6163, 105, 69, 131} \[ \frac{2 (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \, _2F_1\left (1,-\frac{n}{2};1-\frac{n}{2};\frac{(a+1) (-a-b x+1)}{(1-a) (a+b x+1)}\right )}{n}-\frac{2^{\frac{n}{2}+1} (-a-b x+1)^{-n/2} \, _2F_1\left (-\frac{n}{2},-\frac{n}{2};1-\frac{n}{2};\frac{1}{2} (-a-b x+1)\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])/x,x]

[Out]

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

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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]))

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

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

Mathematica [A]  time = 0.0354214, size = 111, normalized size = 0.82 \[ \frac{2 (-a-b x+1)^{-n/2} \left ((a+b x+1)^{n/2} \text{Hypergeometric2F1}\left (1,-\frac{n}{2},1-\frac{n}{2},\frac{(a+1) (a+b x-1)}{(a-1) (a+b x+1)}\right )-2^{n/2} \text{Hypergeometric2F1}\left (-\frac{n}{2},-\frac{n}{2},1-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])/x,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(n*atanh(a + b*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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