3.9.75 \(\int e^{n \tanh ^{-1}(a+b x)} x^m \, dx\) [875]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 109, normalized size of antiderivative = 1.00, 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 = {6298, 140, 138} \begin {gather*} \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} \end {gather*}

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, -1/2*n, 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 138

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

Rule 140

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 6298

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]

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*}

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Mathematica [F]
time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int e^{n \tanh ^{-1}(a+b x)} x^m \, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{n \arctanh \left (b x +a \right )} x^{m}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m} e^{n \operatorname {atanh}{\left (a + b x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^m\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*exp(n*atanh(a + b*x)),x)

[Out]

int(x^m*exp(n*atanh(a + b*x)), x)

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