\(\int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx\) [358]

Optimal result

Integrand size = 19, antiderivative size = 124 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=-\frac {3 \sqrt {1-a^2 x^2}}{a^4 c^2}-\frac {x \sqrt {1-a^2 x^2}}{2 a^3 c^2}-\frac {6 \sqrt {1-a^2 x^2}}{a^4 c^2 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^4 c^2 (1-a x)^3}+\frac {11 \arcsin (a x)}{2 a^4 c^2} \] Output:

-3*(-a^2*x^2+1)^(1/2)/a^4/c^2-1/2*x*(-a^2*x^2+1)^(1/2)/a^3/c^2-6*(-a^2*x^2 
+1)^(1/2)/a^4/c^2/(-a*x+1)+1/3*(-a^2*x^2+1)^(3/2)/a^4/c^2/(-a*x+1)^3+11/2* 
arcsin(a*x)/a^4/c^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.58 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=-\frac {\frac {\sqrt {1+a x} \left (52-71 a x+12 a^2 x^2+3 a^3 x^3\right )}{(1-a x)^{3/2}}+66 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{6 a^4 c^2} \] Input:

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

Output:

-1/6*((Sqrt[1 + a*x]*(52 - 71*a*x + 12*a^2*x^2 + 3*a^3*x^3))/(1 - a*x)^(3/ 
2) + 66*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(a^4*c^2)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6678, 27, 570, 529, 27, 2166, 27, 676, 223}

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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 570

\(\displaystyle \frac {\int \frac {x^3 (a x+1)^3}{\left (1-a^2 x^2\right )^{5/2}}dx}{c^2}\)

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2166

\(\displaystyle \frac {\int \frac {(a x+1) (a x+5)}{a^3 \sqrt {1-a^2 x^2}}dx+\frac {(a x+1)^3}{3 a^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 \sqrt {1-a^2 x^2}}}{c^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {(a x+1) (a x+5)}{\sqrt {1-a^2 x^2}}dx}{a^3}+\frac {(a x+1)^3}{3 a^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 \sqrt {1-a^2 x^2}}}{c^2}\)

\(\Big \downarrow \) 676

\(\displaystyle \frac {\frac {\frac {11}{2} \int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {6 \sqrt {1-a^2 x^2}}{a}}{a^3}+\frac {(a x+1)^3}{3 a^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 \sqrt {1-a^2 x^2}}}{c^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {(a x+1)^3}{3 a^4 \left (1-a^2 x^2\right )^{3/2}}-\frac {3 (a x+1)^2}{a^4 \sqrt {1-a^2 x^2}}+\frac {-\frac {1}{2} x \sqrt {1-a^2 x^2}-\frac {6 \sqrt {1-a^2 x^2}}{a}+\frac {11 \arcsin (a x)}{2 a}}{a^3}}{c^2}\)

Input:

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

Output:

((1 + a*x)^3/(3*a^4*(1 - a^2*x^2)^(3/2)) - (3*(1 + a*x)^2)/(a^4*Sqrt[1 - a 
^2*x^2]) + ((-6*Sqrt[1 - a^2*x^2])/a - (x*Sqrt[1 - a^2*x^2])/2 + (11*ArcSi 
n[a*x])/(2*a))/a^3)/c^2
 

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 529

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 0]
 

rule 570

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

rule 6678

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.23

method	result	size
risch	\(\frac {\left (a x +6\right ) \left (a^{2} x^{2}-1\right )}{2 a^{4} \sqrt {-a^{2} x^{2}+1}\, c^{2}}+\frac {\frac {11 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{3} \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{6} \left (x -\frac {1}{a}\right )^{2}}+\frac {19 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{5} \left (x -\frac {1}{a}\right )}}{c^{2}}\)	\(153\)
default	\(\frac {\frac {-\frac {x \sqrt {-a^{2} x^{2}+1}}{2 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}}{a}+\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}-\frac {3 \sqrt {-a^{2} x^{2}+1}}{a^{4}}+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a^{5}}+\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{5} \left (x -\frac {1}{a}\right )}}{c^{2}}\)	\(232\)

Input:

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^3/(-a*c*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

1/2*(a*x+6)*(a^2*x^2-1)/a^4/(-a^2*x^2+1)^(1/2)/c^2+(11/2/a^3/(a^2)^(1/2)*a 
rctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2/3/a^6/(x-1/a)^2*(-(x-1/a)^2*a^2- 
2*a*(x-1/a))^(1/2)+19/3/a^5/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))/c^ 
2
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=-\frac {52 \, a^{2} x^{2} - 104 \, a x + 66 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 71 \, a x + 52\right )} \sqrt {-a^{2} x^{2} + 1} + 52}{6 \, {\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}} \] Input:

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

Output:

-1/6*(52*a^2*x^2 - 104*a*x + 66*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^ 
2 + 1) - 1)/(a*x)) + (3*a^3*x^3 + 12*a^2*x^2 - 71*a*x + 52)*sqrt(-a^2*x^2 
+ 1) + 52)/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2)
 

Sympy [F]

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

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

Output:

(Integral(x**3/(a**2*x**2*sqrt(-a**2*x**2 + 1) - 2*a*x*sqrt(-a**2*x**2 + 1 
) + sqrt(-a**2*x**2 + 1)), x) + Integral(a*x**4/(a**2*x**2*sqrt(-a**2*x**2 
 + 1) - 2*a*x*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x))/c**2
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{6} c^{2} x^{2} - 2 \, a^{5} c^{2} x + a^{4} c^{2}\right )}} + \frac {19 \, \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{5} c^{2} x - a^{4} c^{2}\right )}} - \frac {\sqrt {-a^{2} x^{2} + 1} x}{2 \, a^{3} c^{2}} + \frac {11 \, \arcsin \left (a x\right )}{2 \, a^{4} c^{2}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{2}} \] Input:

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

Output:

2/3*sqrt(-a^2*x^2 + 1)/(a^6*c^2*x^2 - 2*a^5*c^2*x + a^4*c^2) + 19/3*sqrt(- 
a^2*x^2 + 1)/(a^5*c^2*x - a^4*c^2) - 1/2*sqrt(-a^2*x^2 + 1)*x/(a^3*c^2) + 
11/2*arcsin(a*x)/(a^4*c^2) - 3*sqrt(-a^2*x^2 + 1)/(a^4*c^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=-\frac {4 \, a^{3} c^{6} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} + 132 \, a^{3} c^{6} \arctan \left (\sqrt {-\frac {2 \, c}{a c x - c} - 1}\right ) - 72 \, a^{3} c^{6} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - \frac {3 \, {\left (7 \, a^{3} c^{6} {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} + 5 \, a^{3} c^{6} \sqrt {-\frac {2 \, c}{a c x - c} - 1}\right )} {\left (a c x - c\right )}^{2}}{c^{2}}}{12 \, a^{6} c^{8} {\left | a \right |} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )} \] Input:

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

Output:

-1/12*(4*a^3*c^6*(-2*c/(a*c*x - c) - 1)^(3/2) + 132*a^3*c^6*arctan(sqrt(-2 
*c/(a*c*x - c) - 1)) - 72*a^3*c^6*sqrt(-2*c/(a*c*x - c) - 1) - 3*(7*a^3*c^ 
6*(-2*c/(a*c*x - c) - 1)^(3/2) + 5*a^3*c^6*sqrt(-2*c/(a*c*x - c) - 1))*(a* 
c*x - c)^2/c^2)/(a^6*c^8*abs(a)*sgn(1/(a*c*x - c))*sgn(a)*sgn(c))
 

Mupad [B] (verification not implemented)

Time = 14.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.34 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=\frac {2\,\sqrt {1-a^2\,x^2}}{3\,\left (a^6\,c^2\,x^2-2\,a^5\,c^2\,x+a^4\,c^2\right )}+\frac {19\,\sqrt {1-a^2\,x^2}}{3\,\left (a^2\,c^2\,\sqrt {-a^2}-a^3\,c^2\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {3\,\sqrt {1-a^2\,x^2}}{a^4\,c^2}-\frac {x\,\sqrt {1-a^2\,x^2}}{2\,a^3\,c^2}+\frac {11\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^3\,c^2\,\sqrt {-a^2}} \] Input:

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

Output:

(2*(1 - a^2*x^2)^(1/2))/(3*(a^4*c^2 - 2*a^5*c^2*x + a^6*c^2*x^2)) + (19*(1 
 - a^2*x^2)^(1/2))/(3*(a^2*c^2*(-a^2)^(1/2) - a^3*c^2*x*(-a^2)^(1/2))*(-a^ 
2)^(1/2)) - (3*(1 - a^2*x^2)^(1/2))/(a^4*c^2) - (x*(1 - a^2*x^2)^(1/2))/(2 
*a^3*c^2) + (11*asinh(x*(-a^2)^(1/2)))/(2*a^3*c^2*(-a^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^2} \, dx=\frac {33 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -33 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )+33 \mathit {asin} \left (a x \right ) a^{2} x^{2}-66 \mathit {asin} \left (a x \right ) a x +33 \mathit {asin} \left (a x \right )-3 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+41 \sqrt {-a^{2} x^{2}+1}\, a x -22 \sqrt {-a^{2} x^{2}+1}+3 a^{4} x^{4}+15 a^{3} x^{3}-89 a^{2} x^{2}+41 a x +22}{6 a^{4} c^{2} \left (\sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+a^{2} x^{2}-2 a x +1\right )} \] Input:

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

Output:

(33*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x - 33*sqrt( - a**2*x**2 + 1)*asin( 
a*x) + 33*asin(a*x)*a**2*x**2 - 66*asin(a*x)*a*x + 33*asin(a*x) - 3*sqrt( 
- a**2*x**2 + 1)*a**3*x**3 - 12*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 41*sqrt 
( - a**2*x**2 + 1)*a*x - 22*sqrt( - a**2*x**2 + 1) + 3*a**4*x**4 + 15*a**3 
*x**3 - 89*a**2*x**2 + 41*a*x + 22)/(6*a**4*c**2*(sqrt( - a**2*x**2 + 1)*a 
*x - sqrt( - a**2*x**2 + 1) + a**2*x**2 - 2*a*x + 1))