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

Optimal. Leaf size=114 \[ \frac{2^{n/2} (2 a-n) (-a-b x+1)^{1-\frac{n}{2}} \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{b^2 (2-n)}-\frac{(-a-b x+1)^{1-\frac{n}{2}} (a+b x+1)^{\frac{n+2}{2}}}{2 b^2} \]

[Out]

-((1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/(2*b^2) + (2^(n/2)*(2*a - n)*(1 - a - b*x)^(1 - n/2)*Hype
rgeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(b^2*(2 - n))

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Rubi [A]  time = 0.0687108, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6163, 80, 69} \[ \frac{2^{n/2} (2 a-n) (-a-b x+1)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (-a-b x+1)\right )}{b^2 (2-n)}-\frac{(-a-b x+1)^{1-\frac{n}{2}} (a+b x+1)^{\frac{n+2}{2}}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])*x,x]

[Out]

-((1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/(2*b^2) + (2^(n/2)*(2*a - n)*(1 - a - b*x)^(1 - n/2)*Hype
rgeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(b^2*(2 - n))

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 80

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

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)} x \, dx &=\int x (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx\\ &=-\frac{(1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{2 b^2}-\frac{(2 a-n) \int (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx}{2 b}\\ &=-\frac{(1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{2 b^2}+\frac{2^{n/2} (2 a-n) (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 (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.0403405, size = 96, normalized size = 0.84 \[ \frac{(-a-b x+1)^{1-\frac{n}{2}} \left (\frac{b 2^{\frac{n}{2}+1} (n-2 a) \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{n-2}-b (a+b x+1)^{\frac{n}{2}+1}\right )}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])*x,x]

[Out]

((1 - a - b*x)^(1 - n/2)*(-(b*(1 + a + b*x)^(1 + n/2)) + (2^(1 + n/2)*b*(-2*a + n)*Hypergeometric2F1[1 - n/2,
-n/2, 2 - n/2, (1 - a - b*x)/2])/(-2 + n)))/(2*b^3)

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{{\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 x \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,x, algorithm="maxima")

[Out]

integrate(x*((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 (x \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,x, algorithm="fricas")

[Out]

integral(x*((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 x 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,x)

[Out]

Integral(x*exp(n*atanh(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \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,x, algorithm="giac")

[Out]

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