3.147 \(\int e^{-3 \tanh ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=150 \[ -\frac{3 x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}+\frac{4 x^{m+1} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{m+2}-\frac{4 a x^{m+2} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.813987, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6124, 6742, 364, 850, 808} \[ -\frac{3 x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{4 x^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2}-\frac{4 a x^{m+2} \, _2F_1\left (\frac{3}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Int[x^m/E^(3*ArcTanh[a*x]),x]

[Out]

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

Rule 6124

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/
2)*Sqrt[1 - a^2*x^2])), x] /; FreeQ[{a, m}, x] && IntegerQ[(n - 1)/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 850

Int[((x_)^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Int[x^n*(a/d + (c*x)/e)*(a + c*x
^2)^(p - 1), x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ( !IntegerQ[n] ||
!IntegerQ[2*p] || IGtQ[n, 2] || (GtQ[p, 0] && NeQ[n, 2]))

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} x^m \, dx &=\int \frac{x^m (1-a x)^2}{(1+a x) \sqrt{1-a^2 x^2}} \, dx\\ &=\int \left (-\frac{3 x^m}{\sqrt{1-a^2 x^2}}+\frac{a x^{1+m}}{\sqrt{1-a^2 x^2}}+\frac{4 x^m}{(1+a x) \sqrt{1-a^2 x^2}}\right ) \, dx\\ &=-\left (3 \int \frac{x^m}{\sqrt{1-a^2 x^2}} \, dx\right )+4 \int \frac{x^m}{(1+a x) \sqrt{1-a^2 x^2}} \, dx+a \int \frac{x^{1+m}}{\sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{1+m}+\frac{a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{2+m}+4 \int \frac{x^m (1-a x)}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{1+m}+\frac{a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{2+m}+4 \int \frac{x^m}{\left (1-a^2 x^2\right )^{3/2}} \, dx-(4 a) \int \frac{x^{1+m}}{\left (1-a^2 x^2\right )^{3/2}} \, dx\\ &=-\frac{3 x^{1+m} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{1+m}+\frac{a x^{2+m} \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{2+m}+\frac{4 x^{1+m} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{1+m}-\frac{4 a x^{2+m} \, _2F_1\left (\frac{3}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{2+m}\\ \end{align*}

Mathematica [C]  time = 0.0517339, size = 55, normalized size = 0.37 \[ -\frac{x^{m+1} \left (F_1\left (m+1;-\frac{1}{2},\frac{1}{2};m+2;a x,-a x\right )-2 F_1\left (m+1;-\frac{1}{2},\frac{3}{2};m+2;a x,-a x\right )\right )}{m+1} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^m/E^(3*ArcTanh[a*x]),x]

[Out]

-((x^(1 + m)*(AppellF1[1 + m, -1/2, 1/2, 2 + m, a*x, -(a*x)] - 2*AppellF1[1 + m, -1/2, 3/2, 2 + m, a*x, -(a*x)
]))/(1 + m))

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Maple [F]  time = 0.332, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( ax+1 \right ) ^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

int(x^m/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{m}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*x^m/(a*x + 1)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*(a*x - 1)*x^m/(a^2*x^2 + 2*a*x + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(a*x+1)**3*(-a**2*x**2+1)**(3/2),x)

[Out]

Integral(x**m*(-(a*x - 1)*(a*x + 1))**(3/2)/(a*x + 1)**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{m}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*x^m/(a*x + 1)^3, x)