3.877 \(\int e^{n \tanh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=170 \[ -\frac{2^{n/2} \left (6 a^2-6 a n+n^2+2\right ) (-a-b x+1)^{1-\frac{n}{2}} \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{3 b^3 (2-n)}+\frac{(4 a-n) (a+b x+1)^{\frac{n+2}{2}} (-a-b x+1)^{1-\frac{n}{2}}}{6 b^3}-\frac{x (a+b x+1)^{\frac{n+2}{2}} (-a-b x+1)^{1-\frac{n}{2}}}{3 b^2} \]

[Out]

((4*a - n)*(1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/(6*b^3) - (x*(1 - a - b*x)^(1 - n/2)*(1 + a + b*
x)^((2 + n)/2))/(3*b^2) - (2^(n/2)*(2 + 6*a^2 - 6*a*n + n^2)*(1 - a - b*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2
, -n/2, 2 - n/2, (1 - a - b*x)/2])/(3*b^3*(2 - n))

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Rubi [A]  time = 0.158488, antiderivative size = 170, 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, 90, 80, 69} \[ -\frac{2^{n/2} \left (6 a^2-6 a n+n^2+2\right ) (-a-b x+1)^{1-\frac{n}{2}} \, _2F_1\left (1-\frac{n}{2},-\frac{n}{2};2-\frac{n}{2};\frac{1}{2} (-a-b x+1)\right )}{3 b^3 (2-n)}+\frac{(4 a-n) (a+b x+1)^{\frac{n+2}{2}} (-a-b x+1)^{1-\frac{n}{2}}}{6 b^3}-\frac{x (a+b x+1)^{\frac{n+2}{2}} (-a-b x+1)^{1-\frac{n}{2}}}{3 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])*x^2,x]

[Out]

((4*a - n)*(1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/(6*b^3) - (x*(1 - a - b*x)^(1 - n/2)*(1 + a + b*
x)^((2 + n)/2))/(3*b^2) - (2^(n/2)*(2 + 6*a^2 - 6*a*n + n^2)*(1 - a - b*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2
, -n/2, 2 - n/2, (1 - a - b*x)/2])/(3*b^3*(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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

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^2 \, dx &=\int x^2 (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx\\ &=-\frac{x (1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{3 b^2}-\frac{\int (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \left (-1+a^2+b (4 a-n) x\right ) \, dx}{3 b^2}\\ &=\frac{(4 a-n) (1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{6 b^3}-\frac{x (1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{3 b^2}+\frac{\left (2+6 a^2-6 a n+n^2\right ) \int (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx}{6 b^2}\\ &=\frac{(4 a-n) (1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{6 b^3}-\frac{x (1-a-b x)^{1-\frac{n}{2}} (1+a+b x)^{\frac{2+n}{2}}}{3 b^2}-\frac{2^{n/2} \left (2+6 a^2-6 a n+n^2\right ) (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 )}{3 b^3 (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.113065, size = 127, normalized size = 0.75 \[ \frac{(-a-b x+1)^{1-\frac{n}{2}} \left (\frac{2^{\frac{n}{2}+1} \left (6 a^2-6 a n+n^2+2\right ) \text{Hypergeometric2F1}\left (1-\frac{n}{2},-\frac{n}{2},2-\frac{n}{2},\frac{1}{2} (-a-b x+1)\right )}{n-2}+(4 a-n) (a+b x+1)^{\frac{n}{2}+1}-2 b x (a+b x+1)^{\frac{n}{2}+1}\right )}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])*x^2,x]

[Out]

((1 - a - b*x)^(1 - n/2)*((4*a - n)*(1 + a + b*x)^(1 + n/2) - 2*b*x*(1 + a + b*x)^(1 + n/2) + (2^(1 + n/2)*(2
+ 6*a^2 - 6*a*n + n^2)*Hypergeometric2F1[1 - n/2, -n/2, 2 - n/2, (1 - a - b*x)/2])/(-2 + n)))/(6*b^3)

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( bx+a \right ) }}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(b*x+a))*x^2,x)

[Out]

int(exp(n*arctanh(b*x+a))*x^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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^2,x, algorithm="maxima")

[Out]

integrate(x^2*((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^{2} \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^2,x, algorithm="fricas")

[Out]

integral(x^2*((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^{2} 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**2,x)

[Out]

Integral(x**2*exp(n*atanh(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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^2,x, algorithm="giac")

[Out]

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