\(\int e^{4 \coth ^{-1}(a x)} (c x)^m \, dx\) [140]

Optimal result

Integrand size = 14, antiderivative size = 60 \[ \int e^{4 \coth ^{-1}(a x)} (c x)^m \, dx=\frac {(c x)^{1+m}}{c (1+m)}+\frac {4 (c x)^{1+m}}{c (1-a x)}-\frac {4 (c x)^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)}{c} \] Output:

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int e^{4 \coth ^{-1}(a x)} (c x)^m \, dx=-\frac {x (c x)^m (5+4 m-a x+4 (1+m) (-1+a x) \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x))}{(1+m) (-1+a x)} \] Input:

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

Output:

-((x*(c*x)^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)))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6717, 6676, 100, 27, 90, 74}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int e^{4 \coth ^{-1}(a x)} (c x)^m \, dx\)

\(\Big \downarrow \) 6717

\(\displaystyle \int e^{4 \text {arctanh}(a x)} (c x)^mdx\)

\(\Big \downarrow \) 6676

\(\displaystyle \int \frac {(a x+1)^2 (c x)^m}{(1-a x)^2}dx\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {4 (c x)^{m+1}}{c (1-a x)}-\frac {\int \frac {a^2 c (c x)^m (4 m+a x+3)}{1-a x}dx}{a^2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 (c x)^{m+1}}{c (1-a x)}-\int \frac {(c x)^m (4 m+a x+3)}{1-a x}dx\)

\(\Big \downarrow \) 90

\(\displaystyle -4 (m+1) \int \frac {(c x)^m}{1-a x}dx+\frac {4 (c x)^{m+1}}{c (1-a x)}+\frac {(c x)^{m+1}}{c (m+1)}\)

\(\Big \downarrow \) 74

\(\displaystyle -\frac {4 (c x)^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,a x)}{c}+\frac {4 (c x)^{m+1}}{c (1-a x)}+\frac {(c x)^{m+1}}{c (m+1)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(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 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> 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] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(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 6676

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

rule 6717

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

Maple [C] (verified)

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

Time = 0.34 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.85

method	result	size
meijerg	\(-\frac {\left (-a \right )^{-m} \left (x c \right )^{m} x^{-m} \left (\frac {x^{m} \left (-a \right )^{m} \left (a^{2} m \,x^{2}+a m x +2 a x -m^{2}-3 m -2\right )}{\left (1+m \right ) m \left (-a x +1\right )}+x^{m} \left (-a \right )^{m} \left (2+m \right ) \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}+\frac {2 \left (-a \right )^{-m} \left (x c \right )^{m} x^{-m} \left (-\frac {x^{m} \left (-a \right )^{m} \left (a x -m -1\right )}{m \left (-a x +1\right )}-x^{m} \left (-a \right )^{m} \left (1+m \right ) \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}-\frac {\left (-a \right )^{-m} \left (x c \right )^{m} x^{-m} \left (\frac {x^{m} \left (-a \right )^{m} \left (-1-m \right )}{\left (1+m \right ) \left (-a x +1\right )}+x^{m} \left (-a \right )^{m} m \operatorname {LerchPhi}\left (a x , 1, m\right )\right )}{a}\)	\(231\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int e^{4 \coth ^{-1}(a x)} (c x)^m \, dx=\frac {c^{m} \left (x^{m} a^{2} m^{2} x^{2}-x^{m} a^{2} m \,x^{2}+3 x^{m} a \,m^{2} x +x^{m} a m x -4 x^{m} a x +4 x^{m} m^{2}+8 x^{m} m +4 x^{m}+4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) a \,m^{4} x +4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) a \,m^{3} x -4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) a \,m^{2} x -4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) a m x -4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) m^{4}-4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) m^{3}+4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) m^{2}+4 \left (\int \frac {x^{m}}{a^{2} m \,x^{3}-a^{2} x^{3}-2 a m \,x^{2}+2 a \,x^{2}+m x -x}d x \right ) m \right )}{a m \left (a \,m^{2} x -a x -m^{2}+1\right )} \] Input:

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

Output:

(c**m*(x**m*a**2*m**2*x**2 - x**m*a**2*m*x**2 + 3*x**m*a*m**2*x + x**m*a*m 
*x - 4*x**m*a*x + 4*x**m*m**2 + 8*x**m*m + 4*x**m + 4*int(x**m/(a**2*m*x** 
3 - a**2*x**3 - 2*a*m*x**2 + 2*a*x**2 + m*x - x),x)*a*m**4*x + 4*int(x**m/ 
(a**2*m*x**3 - a**2*x**3 - 2*a*m*x**2 + 2*a*x**2 + m*x - x),x)*a*m**3*x - 
4*int(x**m/(a**2*m*x**3 - a**2*x**3 - 2*a*m*x**2 + 2*a*x**2 + m*x - x),x)* 
a*m**2*x - 4*int(x**m/(a**2*m*x**3 - a**2*x**3 - 2*a*m*x**2 + 2*a*x**2 + m 
*x - x),x)*a*m*x - 4*int(x**m/(a**2*m*x**3 - a**2*x**3 - 2*a*m*x**2 + 2*a* 
x**2 + m*x - x),x)*m**4 - 4*int(x**m/(a**2*m*x**3 - a**2*x**3 - 2*a*m*x**2 
 + 2*a*x**2 + m*x - x),x)*m**3 + 4*int(x**m/(a**2*m*x**3 - a**2*x**3 - 2*a 
*m*x**2 + 2*a*x**2 + m*x - x),x)*m**2 + 4*int(x**m/(a**2*m*x**3 - a**2*x** 
3 - 2*a*m*x**2 + 2*a*x**2 + m*x - x),x)*m))/(a*m*(a*m**2*x - a*x - m**2 + 
1))