\(\int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx\) [367]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 19, antiderivative size = 129 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=\frac {4 \sqrt {1-a^2 x^2}}{a^4 c^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a^4 c^3 (1-a x)^4}-\frac {14 \left (1-a^2 x^2\right )^{3/2}}{15 a^4 c^3 (1-a x)^3}+\frac {3 \left (1-a^2 x^2\right )^{3/2}}{a^4 c^3 (1-a x)^2}-\frac {4 \arcsin (a x)}{a^4 c^3} \] Output:

4*(-a^2*x^2+1)^(1/2)/a^4/c^3+1/5*(-a^2*x^2+1)^(3/2)/a^4/c^3/(-a*x+1)^4-14/ 
15*(-a^2*x^2+1)^(3/2)/a^4/c^3/(-a*x+1)^3+3*(-a^2*x^2+1)^(3/2)/a^4/c^3/(-a* 
x+1)^2-4*arcsin(a*x)/a^4/c^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=\frac {\frac {\sqrt {1+a x} \left (94-222 a x+149 a^2 x^2-15 a^3 x^3\right )}{(1-a x)^{5/2}}+120 \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )}{15 a^4 c^3} \] Input:

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

Output:

((Sqrt[1 + a*x]*(94 - 222*a*x + 149*a^2*x^2 - 15*a^3*x^3))/(1 - a*x)^(5/2) 
 + 120*ArcSin[Sqrt[1 - a*x]/Sqrt[2]])/(15*a^4*c^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {6678, 27, 581, 25, 2168, 2009}

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.

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 581

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2168

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {-\frac {14 \left (1-a^2 x^2\right )^{3/2}}{15 a (1-a x)^3}+\frac {\left (1-a^2 x^2\right )^{3/2}}{5 a (1-a x)^4}+\frac {8 \sqrt {1-a^2 x^2}}{a (1-a x)}-\frac {4 \arcsin (a x)}{a}}{a^3}-\frac {\left (1-a^2 x^2\right )^{3/2}}{a^4 (1-a x)^2}}{c^3}\)

Input:

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

Output:

(-((1 - a^2*x^2)^(3/2)/(a^4*(1 - a*x)^2)) + ((8*Sqrt[1 - a^2*x^2])/(a*(1 - 
 a*x)) + (1 - a^2*x^2)^(3/2)/(5*a*(1 - a*x)^4) - (14*(1 - a^2*x^2)^(3/2))/ 
(15*a*(1 - a*x)^3) - (4*ArcSin[a*x])/a)/a^3)/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 581
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b*x^ 
2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m 
+ c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p 
)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & 
& IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] 
)
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2168
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Int[ExpandIntegrand[(a + b*x^2)^p, (d + e*x)^m*Pq, x], x] /; FreeQ[{a, b, 
d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq, x] 
+ 2*p + 1, 0] && ILtQ[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 = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.47

method result size
risch \(-\frac {a^{2} x^{2}-1}{a^{4} \sqrt {-a^{2} x^{2}+1}\, c^{3}}-\frac {\frac {4 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{7} \left (x -\frac {1}{a}\right )^{3}}+\frac {31 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{6} \left (x -\frac {1}{a}\right )^{2}}+\frac {104 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{5} \left (x -\frac {1}{a}\right )}}{c^{3}}\) \(190\)
default \(-\frac {-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{4}}+\frac {4 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}+\frac {\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{6}}+\frac {\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a^{5}}+\frac {9 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a^{5} \left (x -\frac {1}{a}\right )}}{c^{3}}\) \(310\)

Input:

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

Output:

-1/a^4*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^3-(4/a^3/(a^2)^(1/2)*arctan((a^2)^ 
(1/2)*x/(-a^2*x^2+1)^(1/2))+2/5/a^7/(x-1/a)^3*(-(x-1/a)^2*a^2-2*a*(x-1/a)) 
^(1/2)+31/15/a^6/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+104/15/a^5/( 
x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))/c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=\frac {94 \, a^{3} x^{3} - 282 \, a^{2} x^{2} + 282 \, a x + 120 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{3} x^{3} - 149 \, a^{2} x^{2} + 222 \, a x - 94\right )} \sqrt {-a^{2} x^{2} + 1} - 94}{15 \, {\left (a^{7} c^{3} x^{3} - 3 \, a^{6} c^{3} x^{2} + 3 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} \] Input:

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

Output:

1/15*(94*a^3*x^3 - 282*a^2*x^2 + 282*a*x + 120*(a^3*x^3 - 3*a^2*x^2 + 3*a* 
x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (15*a^3*x^3 - 149*a^2*x^2 
+ 222*a*x - 94)*sqrt(-a^2*x^2 + 1) - 94)/(a^7*c^3*x^3 - 3*a^6*c^3*x^2 + 3* 
a^5*c^3*x - a^4*c^3)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=-\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{5 \, {\left (a^{7} c^{3} x^{3} - 3 \, a^{6} c^{3} x^{2} + 3 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} - \frac {31 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{6} c^{3} x^{2} - 2 \, a^{5} c^{3} x + a^{4} c^{3}\right )}} - \frac {104 \, \sqrt {-a^{2} x^{2} + 1}}{15 \, {\left (a^{5} c^{3} x - a^{4} c^{3}\right )}} - \frac {4 \, \arcsin \left (a x\right )}{a^{4} c^{3}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{3}} \] Input:

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

Output:

-2/5*sqrt(-a^2*x^2 + 1)/(a^7*c^3*x^3 - 3*a^6*c^3*x^2 + 3*a^5*c^3*x - a^4*c 
^3) - 31/15*sqrt(-a^2*x^2 + 1)/(a^6*c^3*x^2 - 2*a^5*c^3*x + a^4*c^3) - 104 
/15*sqrt(-a^2*x^2 + 1)/(a^5*c^3*x - a^4*c^3) - 4*arcsin(a*x)/(a^4*c^3) + s 
qrt(-a^2*x^2 + 1)/(a^4*c^3)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=-\frac {4 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{a^{3} c^{3} {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a^{4} c^{3}} - \frac {2 \, {\left (\frac {335 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {505 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {285 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {60 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 79\right )}}{15 \, a^{3} c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:

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

Output:

-4*arcsin(a*x)*sgn(a)/(a^3*c^3*abs(a)) + sqrt(-a^2*x^2 + 1)/(a^4*c^3) - 2/ 
15*(335*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 505*(sqrt(-a^2*x^2 + 1)* 
abs(a) + a)^2/(a^4*x^2) + 285*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) 
- 60*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 79)/(a^3*c^3*((sqrt(-a^ 
2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
 

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=\frac {\sqrt {1-a^2\,x^2}}{a^4\,c^3}-\frac {2\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (a^2\,c^3\,\sqrt {-a^2}+3\,a^4\,c^3\,x^2\,\sqrt {-a^2}-a^5\,c^3\,x^3\,\sqrt {-a^2}-3\,a^3\,c^3\,x\,\sqrt {-a^2}\right )}-\frac {104\,\sqrt {1-a^2\,x^2}}{15\,\left (a^2\,c^3\,\sqrt {-a^2}-a^3\,c^3\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {31\,\sqrt {1-a^2\,x^2}}{15\,\left (a^6\,c^3\,x^2-2\,a^5\,c^3\,x+a^4\,c^3\right )}-\frac {4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{a^3\,c^3\,\sqrt {-a^2}} \] Input:

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

Output:

(1 - a^2*x^2)^(1/2)/(a^4*c^3) - (2*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(a 
^2*c^3*(-a^2)^(1/2) + 3*a^4*c^3*x^2*(-a^2)^(1/2) - a^5*c^3*x^3*(-a^2)^(1/2 
) - 3*a^3*c^3*x*(-a^2)^(1/2))) - (104*(1 - a^2*x^2)^(1/2))/(15*(a^2*c^3*(- 
a^2)^(1/2) - a^3*c^3*x*(-a^2)^(1/2))*(-a^2)^(1/2)) - (31*(1 - a^2*x^2)^(1/ 
2))/(15*(a^4*c^3 - 2*a^5*c^3*x + a^6*c^3*x^2)) - (4*asinh(x*(-a^2)^(1/2))) 
/(a^3*c^3*(-a^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\text {arctanh}(a x)} x^3}{(c-a c x)^3} \, dx=\frac {-60 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{2} x^{2}+120 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -60 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-60 \mathit {asin} \left (a x \right ) a^{3} x^{3}+180 \mathit {asin} \left (a x \right ) a^{2} x^{2}-180 \mathit {asin} \left (a x \right ) a x +60 \mathit {asin} \left (a x \right )+15 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-79 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+82 \sqrt {-a^{2} x^{2}+1}\, a x -24 \sqrt {-a^{2} x^{2}+1}-15 a^{4} x^{4}+204 a^{3} x^{3}-283 a^{2} x^{2}+82 a x +24}{15 a^{4} c^{3} \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}+a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

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

Output:

( - 60*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**2*x**2 + 120*sqrt( - a**2*x**2 
+ 1)*asin(a*x)*a*x - 60*sqrt( - a**2*x**2 + 1)*asin(a*x) - 60*asin(a*x)*a* 
*3*x**3 + 180*asin(a*x)*a**2*x**2 - 180*asin(a*x)*a*x + 60*asin(a*x) + 15* 
sqrt( - a**2*x**2 + 1)*a**3*x**3 - 79*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 8 
2*sqrt( - a**2*x**2 + 1)*a*x - 24*sqrt( - a**2*x**2 + 1) - 15*a**4*x**4 + 
204*a**3*x**3 - 283*a**2*x**2 + 82*a*x + 24)/(15*a**4*c**3*(sqrt( - a**2*x 
**2 + 1)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 1)*a*x + sqrt( - a**2*x**2 + 1) 
 + a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))