3.12.29 \(\int \frac {e^{2 \text {arctanh}(a x)} x^m}{(c-a^2 c x^2)^2} \, dx\) [1129]

3.12.29.1 Optimal result
3.12.29.2 Mathematica [A] (verified)
3.12.29.3 Rubi [A] (verified)
3.12.29.4 Maple [F]
3.12.29.5 Fricas [F]
3.12.29.6 Sympy [F]
3.12.29.7 Maxima [F]
3.12.29.8 Giac [F]
3.12.29.9 Mupad [F(-1)]

3.12.29.1 Optimal result

Integrand size = 25, antiderivative size = 113 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {x^{1+m}}{4 c^2 (1-a x)^2}+\frac {(2-m) x^{1+m}}{4 c^2 (1-a x)}+\frac {x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,-a x)}{8 c^2 (1+m)}+\frac {\left (1-4 m+2 m^2\right ) x^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)}{8 c^2 (1+m)} \]

output
1/4*x^(1+m)/c^2/(-a*x+1)^2+1/4*(2-m)*x^(1+m)/c^2/(-a*x+1)+1/8*x^(1+m)*hype 
rgeom([1, 1+m],[2+m],-a*x)/c^2/(1+m)+1/8*(2*m^2-4*m+1)*x^(1+m)*hypergeom([ 
1, 1+m],[2+m],a*x)/c^2/(1+m)
 
3.12.29.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^2} \, dx=\frac {x^{1+m} \left (2 (1+m) (3-2 a x+m (-1+a x))+(-1+a x)^2 \operatorname {Hypergeometric2F1}(1,1+m,2+m,-a x)+\left (1-4 m+2 m^2\right ) (-1+a x)^2 \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)\right )}{8 c^2 (1+m) (-1+a x)^2} \]

input
Integrate[(E^(2*ArcTanh[a*x])*x^m)/(c - a^2*c*x^2)^2,x]
 
output
(x^(1 + m)*(2*(1 + m)*(3 - 2*a*x + m*(-1 + a*x)) + (-1 + a*x)^2*Hypergeome 
tric2F1[1, 1 + m, 2 + m, -(a*x)] + (1 - 4*m + 2*m^2)*(-1 + a*x)^2*Hypergeo 
metric2F1[1, 1 + m, 2 + m, a*x]))/(8*c^2*(1 + m)*(-1 + a*x)^2)
 
3.12.29.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {6700, 114, 25, 27, 168, 27, 174, 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.

\(\displaystyle \int \frac {x^m e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 74

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

input
Int[(E^(2*ArcTanh[a*x])*x^m)/(c - a^2*c*x^2)^2,x]
 
output
(x^(1 + m)/(4*(1 - a*x)^2) + (((2 - m)*x^(1 + m))/(1 - a*x) + (x^(1 + m)*H 
ypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/(2*(1 + m)) + ((1 - 4*m + 2*m^2 
)*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x])/(2*(1 + m)))/4)/c^2
 

3.12.29.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 114
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 168
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

rule 6700
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], 
 x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || 
 GtQ[c, 0])
 
3.12.29.4 Maple [F]

\[\int \frac {\left (a x +1\right )^{2} x^{m}}{\left (-a^{2} x^{2}+1\right ) \left (-a^{2} c \,x^{2}+c \right )^{2}}d x\]

input
int((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^2,x)
 
output
int((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^2,x)
 
3.12.29.5 Fricas [F]

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

input
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^2,x, algorithm="fricas 
")
 
output
integral(-x^m/(a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 2*a*c^2*x - c^2), x)
 
3.12.29.6 Sympy [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^2} \, dx=- \frac {\int \frac {x^{m}}{a^{4} x^{4} - 2 a^{3} x^{3} + 2 a x - 1}\, dx}{c^{2}} \]

input
integrate((a*x+1)**2/(-a**2*x**2+1)*x**m/(-a**2*c*x**2+c)**2,x)
 
output
-Integral(x**m/(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1), x)/c**2
 
3.12.29.7 Maxima [F]

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

input
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^2,x, algorithm="maxima 
")
 
output
-integrate((a*x + 1)^2*x^m/((a^2*c*x^2 - c)^2*(a^2*x^2 - 1)), x)
 
3.12.29.8 Giac [F]

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

input
integrate((a*x+1)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^2,x, algorithm="giac")
 
output
integrate(-(a*x + 1)^2*x^m/((a^2*c*x^2 - c)^2*(a^2*x^2 - 1)), x)
 
3.12.29.9 Mupad [F(-1)]

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

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