\(\int \frac {e^{2 \text {arctanh}(a x)} x^m}{(c-a^2 c x^2)^3} \, dx\) [1152]

Optimal result

Integrand size = 25, antiderivative size = 161 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {x^{1+m}}{6 c^3 (1-a x)^3 (1+a x)}+\frac {(4-m) x^{1+m}}{12 c^3 (1-a x)^2 (1+a x)}+\frac {(2-m) (3-m) x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{12 c^3 (1+m)}+\frac {a (2-m) (4-m) x^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{12 c^3 (2+m)} \] Output:

1/6*x^(1+m)/c^3/(-a*x+1)^3/(a*x+1)+1/12*(4-m)*x^(1+m)/c^3/(-a*x+1)^2/(a*x+ 
1)+1/12*(2-m)*(3-m)*x^(1+m)*hypergeom([2, 1/2+1/2*m],[3/2+1/2*m],a^2*x^2)/ 
c^3/(1+m)+1/12*a*(2-m)*(4-m)*x^(2+m)*hypergeom([2, 1+1/2*m],[2+1/2*m],a^2* 
x^2)/c^3/(2+m)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {x^{1+m} \left (-\frac {2 \left (m^2 (-1+a x)^2 (1+a x)+m \left (-7+6 a x+7 a^2 x^2-6 a^3 x^3\right )+2 \left (9-6 a x-5 a^2 x^2+4 a^3 x^3\right )\right )}{(-1+a x)^3 (1+a x)}-\frac {3 (-2+m) \operatorname {Hypergeometric2F1}(1,1+m,2+m,-a x)}{1+m}-\frac {\left (-6+19 m-12 m^2+2 m^3\right ) \operatorname {Hypergeometric2F1}(1,1+m,2+m,a x)}{1+m}\right )}{48 c^3} \] Input:

Integrate[(E^(2*ArcTanh[a*x])*x^m)/(c - a^2*c*x^2)^3,x]
 

Output:

(x^(1 + m)*((-2*(m^2*(-1 + a*x)^2*(1 + a*x) + m*(-7 + 6*a*x + 7*a^2*x^2 - 
6*a^3*x^3) + 2*(9 - 6*a*x - 5*a^2*x^2 + 4*a^3*x^3)))/((-1 + a*x)^3*(1 + a* 
x)) - (3*(-2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/(1 + m) - (( 
-6 + 19*m - 12*m^2 + 2*m^3)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x])/(1 + 
m)))/(48*c^3)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {6700, 114, 25, 27, 168, 27, 168, 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.

\(\displaystyle \int \frac {x^m e^{2 \text {arctanh}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx\)

\(\Big \downarrow \) 6700

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

\(\Big \downarrow \) 114

\(\displaystyle \frac {\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}-\frac {\int -\frac {a x^m (-m+a (3-m) x+5)}{(1-a x)^3 (a x+1)^2}dx}{6 a}}{c^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {a x^m (-m+a (3-m) x+5)}{(1-a x)^3 (a x+1)^2}dx}{6 a}+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \int \frac {x^m (-m+a (3-m) x+5)}{(1-a x)^3 (a x+1)^2}dx+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\frac {1}{6} \left (\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}-\frac {\int -\frac {2 a (2-m) x^m (-m+a (4-m) x+3)}{(1-a x)^2 (a x+1)^2}dx}{4 a}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \int \frac {x^m (-m+a (4-m) x+3)}{(1-a x)^2 (a x+1)^2}dx+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}-\frac {\int \frac {a x^m \left (-2 m^2+7 m-a (7-2 m) (1-m) x+1\right )}{(1-a x) (a x+1)^2}dx}{2 a}\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}-\frac {1}{2} \int \frac {x^m \left (-2 m^2+7 m-a (7-2 m) (1-m) x+1\right )}{(1-a x) (a x+1)^2}dx\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {1}{2} \left (-\frac {\int -\frac {2 a x^m ((1-m) (3-m)-a (4-m) m x)}{(1-a x) (a x+1)}dx}{2 a}-\frac {(4-m) x^{m+1}}{a x+1}\right )+\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {1}{2} \left (\int \frac {x^m ((1-m) (3-m)-a (4-m) m x)}{(1-a x) (a x+1)}dx-\frac {(4-m) x^{m+1}}{a x+1}\right )+\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {1}{2} \left (\frac {1}{2} \left (2 m^2-8 m+3\right ) \int \frac {x^m}{1-a x}dx+\frac {3}{2} \int \frac {x^m}{a x+1}dx-\frac {(4-m) x^{m+1}}{a x+1}\right )+\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

\(\Big \downarrow \) 74

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} (2-m) \left (\frac {1}{2} \left (\frac {\left (2 m^2-8 m+3\right ) x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,a x)}{2 (m+1)}+\frac {3 x^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,-a x)}{2 (m+1)}-\frac {(4-m) x^{m+1}}{a x+1}\right )+\frac {(7-2 m) x^{m+1}}{2 (1-a x) (a x+1)}\right )+\frac {(4-m) x^{m+1}}{2 (1-a x)^2 (a x+1)}\right )+\frac {x^{m+1}}{6 (1-a x)^3 (a x+1)}}{c^3}\)

Input:

Int[(E^(2*ArcTanh[a*x])*x^m)/(c - a^2*c*x^2)^3,x]
 

Output:

(x^(1 + m)/(6*(1 - a*x)^3*(1 + a*x)) + (((4 - m)*x^(1 + m))/(2*(1 - a*x)^2 
*(1 + a*x)) + ((2 - m)*(((7 - 2*m)*x^(1 + m))/(2*(1 - a*x)*(1 + a*x)) + (- 
(((4 - m)*x^(1 + m))/(1 + a*x)) + (3*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 
 2 + m, -(a*x)])/(2*(1 + m)) + ((3 - 8*m + 2*m^2)*x^(1 + m)*Hypergeometric 
2F1[1, 1 + m, 2 + m, a*x])/(2*(1 + m)))/2))/2)/6)/c^3
 

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])
 

Maple [F]

\[\int \frac {\left (a x +1\right )^{2} x^{m}}{\left (-a^{2} x^{2}+1\right ) \left (-a^{2} c \,x^{2}+c \right )^{3}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^m/(a^6*c^3*x^6 - 2*a^5*c^3*x^5 - a^4*c^3*x^4 + 4*a^3*c^3*x^3 - 
a^2*c^3*x^2 - 2*a*c^3*x + c^3), x)
 

Sympy [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x^{m}}{a^{6} x^{6} - 2 a^{5} x^{5} - a^{4} x^{4} + 4 a^{3} x^{3} - a^{2} x^{2} - 2 a x + 1}\, dx}{c^{3}} \] Input:

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

Output:

Integral(x**m/(a**6*x**6 - 2*a**5*x**5 - a**4*x**4 + 4*a**3*x**3 - a**2*x* 
*2 - 2*a*x + 1), x)/c**3
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^3} \, dx=-\int \frac {x^m\,{\left (a\,x+1\right )}^2}{{\left (c-a^2\,c\,x^2\right )}^3\,\left (a^2\,x^2-1\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {e^{2 \text {arctanh}(a x)} x^m}{\left (c-a^2 c x^2\right )^3} \, dx=\frac {\int \frac {x^{m}}{a^{6} x^{6}-2 a^{5} x^{5}-a^{4} x^{4}+4 a^{3} x^{3}-a^{2} x^{2}-2 a x +1}d x}{c^{3}} \] Input:

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

Output:

int(x**m/(a**6*x**6 - 2*a**5*x**5 - a**4*x**4 + 4*a**3*x**3 - a**2*x**2 - 
2*a*x + 1),x)/c**3