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

Optimal. Leaf size=109 \[ \frac{x^{m+1} (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \left (1-\frac{b x}{1-a}\right )^{n/2} \left (\frac{b x}{a+1}+1\right )^{-n/2} F_1\left (m+1;\frac{n}{2},-\frac{n}{2};m+2;\frac{b x}{1-a},-\frac{b x}{a+1}\right )}{m+1} \]

[Out]

(x^(1 + m)*(1 + a + b*x)^(n/2)*(1 - (b*x)/(1 - a))^(n/2)*AppellF1[1 + m, n/2, -n/2, 2 + m, (b*x)/(1 - a), -((b
*x)/(1 + a))])/((1 + m)*(1 - a - b*x)^(n/2)*(1 + (b*x)/(1 + a))^(n/2))

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Rubi [A]  time = 0.0759803, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {6163, 135, 133} \[ \frac{x^{m+1} (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \left (1-\frac{b x}{1-a}\right )^{n/2} \left (\frac{b x}{a+1}+1\right )^{-n/2} F_1\left (m+1;\frac{n}{2},-\frac{n}{2};m+2;\frac{b x}{1-a},-\frac{b x}{a+1}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])*x^m,x]

[Out]

(x^(1 + m)*(1 + a + b*x)^(n/2)*(1 - (b*x)/(1 - a))^(n/2)*AppellF1[1 + m, n/2, -n/2, 2 + m, (b*x)/(1 - a), -((b
*x)/(1 + a))])/((1 + m)*(1 - a - b*x)^(n/2)*(1 + (b*x)/(1 + a))^(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 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int e^{n \tanh ^{-1}(a+b x)} x^m \, dx &=\int x^m (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \, dx\\ &=\left ((1-a-b x)^{-n/2} \left (1-\frac{b x}{1-a}\right )^{n/2}\right ) \int x^m (1+a+b x)^{n/2} \left (1-\frac{b x}{1-a}\right )^{-n/2} \, dx\\ &=\left ((1-a-b x)^{-n/2} (1+a+b x)^{n/2} \left (1-\frac{b x}{1-a}\right )^{n/2} \left (1+\frac{b x}{1+a}\right )^{-n/2}\right ) \int x^m \left (1-\frac{b x}{1-a}\right )^{-n/2} \left (1+\frac{b x}{1+a}\right )^{n/2} \, dx\\ &=\frac{x^{1+m} (1-a-b x)^{-n/2} (1+a+b x)^{n/2} \left (1-\frac{b x}{1-a}\right )^{n/2} \left (1+\frac{b x}{1+a}\right )^{-n/2} F_1\left (1+m;\frac{n}{2},-\frac{n}{2};2+m;\frac{b x}{1-a},-\frac{b x}{1+a}\right )}{1+m}\\ \end{align*}

Mathematica [F]  time = 0.790792, size = 0, normalized size = 0. \[ \int e^{n \tanh ^{-1}(a+b x)} x^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n{\it Artanh} \left ( bx+a \right ) }}{x}^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

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