\(\int e^{4 \coth ^{-1}(a x)} x^m \, dx\) [132]

Optimal result

Integrand size = 12, antiderivative size = 45 \[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\frac {x^{1+m}}{1+m}+\frac {4 x^{1+m}}{1-a x}-4 x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6302, 6261, 91, 81, 66} \[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=-4 x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,a x)+\frac {4 x^{m+1}}{1-a x}+\frac {x^{m+1}}{m+1} \]

[In]

Int[E^(4*ArcCoth[a*x])*x^m,x]

[Out]

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 81

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 91

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

Rule 6261

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 6302

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*} \text {integral}& = \int e^{4 \text {arctanh}(a x)} x^m \, dx \\ & = \int \frac {x^m (1+a x)^2}{(1-a x)^2} \, dx \\ & = \frac {4 x^{1+m}}{1-a x}-\frac {\int \frac {x^m \left (a^2 (3+4 m)+a^3 x\right )}{1-a x} \, dx}{a^2} \\ & = \frac {x^{1+m}}{1+m}+\frac {4 x^{1+m}}{1-a x}-(4 (1+m)) \int \frac {x^m}{1-a x} \, dx \\ & = \frac {x^{1+m}}{1+m}+\frac {4 x^{1+m}}{1-a x}-4 x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\frac {x^{1+m} (-5-4 m+a x-4 (1+m) (-1+a x) \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x))}{(1+m) (-1+a x)} \]

[In]

Integrate[E^(4*ArcCoth[a*x])*x^m,x]

[Out]

(x^(1 + m)*(-5 - 4*m + a*x - 4*(1 + m)*(-1 + a*x)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x]))/((1 + m)*(-1 + a*x
))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5.

Time = 0.66 (sec) , antiderivative size = 201, normalized size of antiderivative = 4.47

method	result	size
meijerg	\(-\frac {\left (-a \right )^{-m} \left (\frac {x^{m} \left (-a \right )^{m} \left (a^{2} m \,x^{2}+a m x +2 a x -m^{2}-3 m -2\right )}{\left (1+m \right ) m \left (-a x +1\right )}+x^{m} \left (-a \right )^{m} \left (2+m \right ) \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}+\frac {2 \left (-a \right )^{-m} \left (-\frac {x^{m} \left (-a \right )^{m} \left (a x -m -1\right )}{m \left (-a x +1\right )}-x^{m} \left (-a \right )^{m} \left (1+m \right ) \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}-\frac {\left (-a \right )^{-m} \left (\frac {x^{m} \left (-a \right )^{m} \left (-m -1\right )}{\left (1+m \right ) \left (-a x +1\right )}+x^{m} \left (-a \right )^{m} m \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}\)	\(201\)

[In]

int(1/(a*x-1)^2*(a*x+1)^2*x^m,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\int { \frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (a x - 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\int \frac {x^{m} \left (a x + 1\right )^{2}}{\left (a x - 1\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\int { \frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (a x - 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\int { \frac {{\left (a x + 1\right )}^{2} x^{m}}{{\left (a x - 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int e^{4 \coth ^{-1}(a x)} x^m \, dx=\int \frac {x^m\,{\left (a\,x+1\right )}^2}{{\left (a\,x-1\right )}^2} \,d x \]

[In]

int((x^m*(a*x + 1)^2)/(a*x - 1)^2,x)

[Out]

int((x^m*(a*x + 1)^2)/(a*x - 1)^2, x)