3.134 \(\int e^{2 \coth ^{-1}(a x)} x^m \, dx\)

Optimal. Leaf size=35 \[ \frac {x^{m+1}}{m+1}-\frac {2 x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6167, 6126, 80, 64} \[ \frac {x^{m+1}}{m+1}-\frac {2 x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCoth[a*x])*x^m,x]

[Out]

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

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 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 6126

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

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int e^{2 \coth ^{-1}(a x)} x^m \, dx &=-\int e^{2 \tanh ^{-1}(a x)} x^m \, dx\\ &=-\int \frac {x^m (1+a x)}{1-a x} \, dx\\ &=\frac {x^{1+m}}{1+m}-2 \int \frac {x^m}{1-a x} \, dx\\ &=\frac {x^{1+m}}{1+m}-\frac {2 x^{1+m} \, _2F_1(1,1+m;2+m;a x)}{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.74 \[ \frac {x^{m+1} (1-2 \, _2F_1(1,m+1;m+2;a x))}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCoth[a*x])*x^m,x]

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a x + 1\right )} x^{m}}{a x - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x + 1)*x^m/(a*x - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{a x - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + 1)*x^m/(a*x - 1), x)

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maple [C]  time = 0.41, size = 106, normalized size = 3.03 \[ \frac {\left (-a \right )^{-m} \left (-\frac {x^{m} \left (-a \right )^{m} \left (-1-m \right )}{\left (1+m \right ) m}-x^{m} \left (-a \right )^{m} \Phi \left (a x , 1, m\right )\right )}{a}-\frac {\left (-a \right )^{-m} \left (-\frac {x^{m} \left (-a \right )^{m} \left (a m x +m +1\right )}{\left (1+m \right ) m}+x^{m} \left (-a \right )^{m} \Phi \left (a x , 1, m\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a)^(-m)/a*(-1/(1+m)*x^m*(-a)^m*(-1-m)/m-x^m*(-a)^m*LerchPhi(a*x,1,m))-(-a)^(-m)/a*(-x^m*(-a)^m*(a*m*x+m+1)/(
1+m)/m+x^m*(-a)^m*LerchPhi(a*x,1,m))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )} x^{m}}{a x - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + 1)*x^m/(a*x - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x^m\,\left (a\,x+1\right )}{a\,x-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(a*x + 1))/(a*x - 1),x)

[Out]

int((x^m*(a*x + 1))/(a*x - 1), x)

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sympy [B]  time = 2.71, size = 100, normalized size = 2.86 \[ - \frac {a m x^{2} x^{m} \Phi \left (a x, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac {2 a x^{2} x^{m} \Phi \left (a x, 1, m + 2\right ) \Gamma \left (m + 2\right )}{\Gamma \left (m + 3\right )} - \frac {m x x^{m} \Phi \left (a x, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} - \frac {x x^{m} \Phi \left (a x, 1, m + 1\right ) \Gamma \left (m + 1\right )}{\Gamma \left (m + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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