\(\int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx\) [215]

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

Optimal result

Integrand size = 18, antiderivative size = 129 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {\sqrt {1-a^2 x^2}}{7 a c^5 (1-a x)^4}+\frac {3 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)^2}+\frac {2 \sqrt {1-a^2 x^2}}{35 a c^5 (1-a x)} \] Output:

1/7*(-a^2*x^2+1)^(1/2)/a/c^5/(-a*x+1)^4+3/35*(-a^2*x^2+1)^(1/2)/a/c^5/(-a* 
x+1)^3+2/35*(-a^2*x^2+1)^(1/2)/a/c^5/(-a*x+1)^2+2/35*(-a^2*x^2+1)^(1/2)/a/ 
c^5/(-a*x+1)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.40 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=-\frac {\sqrt {1+a x} \left (-12+13 a x-8 a^2 x^2+2 a^3 x^3\right )}{35 a c^5 (1-a x)^{7/2}} \] Input:

Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^5),x]
 

Output:

-1/35*(Sqrt[1 + a*x]*(-12 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(a*c^5*(1 - a 
*x)^(7/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6677, 27, 461, 461, 461, 460}

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.

\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx\)

\(\Big \downarrow \) 6677

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {1}{(1-a x)^4 \sqrt {1-a^2 x^2}}dx}{c^5}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {3}{7} \int \frac {1}{(1-a x)^3 \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{7 a (1-a x)^4}}{c^5}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \int \frac {1}{(1-a x)^2 \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}\right )+\frac {\sqrt {1-a^2 x^2}}{7 a (1-a x)^4}}{c^5}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (\frac {1}{3} \int \frac {1}{(1-a x) \sqrt {1-a^2 x^2}}dx+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}\right )+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}\right )+\frac {\sqrt {1-a^2 x^2}}{7 a (1-a x)^4}}{c^5}\)

\(\Big \downarrow \) 460

\(\displaystyle \frac {\frac {3}{7} \left (\frac {2}{5} \left (\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)}+\frac {\sqrt {1-a^2 x^2}}{3 a (1-a x)^2}\right )+\frac {\sqrt {1-a^2 x^2}}{5 a (1-a x)^3}\right )+\frac {\sqrt {1-a^2 x^2}}{7 a (1-a x)^4}}{c^5}\)

Input:

Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^5),x]
 

Output:

(Sqrt[1 - a^2*x^2]/(7*a*(1 - a*x)^4) + (3*(Sqrt[1 - a^2*x^2]/(5*a*(1 - a*x 
)^3) + (2*(Sqrt[1 - a^2*x^2]/(3*a*(1 - a*x)^2) + Sqrt[1 - a^2*x^2]/(3*a*(1 
 - a*x))))/5))/7)/c^5
 

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 460
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, 
 p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
 

rule 461
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl 
ify[n + 2*p + 2]/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x 
], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp 
lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
 

rule 6677
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S 
imp[c^n   Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, 
 d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
 
Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.39

method result size
gosper \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right )}{35 \left (a x -1\right )^{4} a \,c^{5}}\) \(50\)
trager \(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right )}{35 \left (a x -1\right )^{4} a \,c^{5}}\) \(50\)
orering \(\frac {\left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x -12\right ) \left (a x -1\right ) \sqrt {-a^{2} x^{2}+1}}{35 a \left (-a c x +c \right )^{5}}\) \(54\)
default \(-\frac {\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{7 a \left (x -\frac {1}{a}\right )^{5}}-\frac {2 a \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {1}{a}\right )^{4}}-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{a}\right )^{3}}\right )}{7}}{2 a^{5}}-\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {1}{a}\right )^{4}}-\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{a}\right )^{3}}}{4 a^{4}}+\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{24 a^{4} \left (x -\frac {1}{a}\right )^{3}}-\frac {\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{16 a^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{32 a}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{32 a}}{c^{5}}\) \(525\)

Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

-1/35*(-a^2*x^2+1)^(1/2)*(2*a^3*x^3-8*a^2*x^2+13*a*x-12)/(a*x-1)^4/a/c^5
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {12 \, a^{4} x^{4} - 48 \, a^{3} x^{3} + 72 \, a^{2} x^{2} - 48 \, a x - {\left (2 \, a^{3} x^{3} - 8 \, a^{2} x^{2} + 13 \, a x - 12\right )} \sqrt {-a^{2} x^{2} + 1} + 12}{35 \, {\left (a^{5} c^{5} x^{4} - 4 \, a^{4} c^{5} x^{3} + 6 \, a^{3} c^{5} x^{2} - 4 \, a^{2} c^{5} x + a c^{5}\right )}} \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="fricas")
 

Output:

1/35*(12*a^4*x^4 - 48*a^3*x^3 + 72*a^2*x^2 - 48*a*x - (2*a^3*x^3 - 8*a^2*x 
^2 + 13*a*x - 12)*sqrt(-a^2*x^2 + 1) + 12)/(a^5*c^5*x^4 - 4*a^4*c^5*x^3 + 
6*a^3*c^5*x^2 - 4*a^2*c^5*x + a*c^5)
 

Sympy [F]

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

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**5,x)
 

Output:

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

Maxima [F]

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

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="maxima")
 

Output:

-integrate(sqrt(-a^2*x^2 + 1)/((a*c*x - c)^5*(a*x + 1)), x)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {1}{280} \, c^{2} {\left (\frac {{\left (5 \, {\left (\frac {2 \, c}{a c x - c} + 1\right )}^{3} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 21 \, {\left (\frac {2 \, c}{a c x - c} + 1\right )}^{2} \sqrt {-\frac {2 \, c}{a c x - c} - 1} - 35 \, {\left (-\frac {2 \, c}{a c x - c} - 1\right )}^{\frac {3}{2}} - 35 \, \sqrt {-\frac {2 \, c}{a c x - c} - 1}\right )} \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )}{a^{2} c^{7}} + \frac {16 i \, \mathrm {sgn}\left (\frac {1}{a c x - c}\right ) \mathrm {sgn}\left (a\right ) \mathrm {sgn}\left (c\right )}{a^{2} c^{7}}\right )} {\left | a \right |} \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x, algorithm="giac")
 

Output:

1/280*c^2*((5*(2*c/(a*c*x - c) + 1)^3*sqrt(-2*c/(a*c*x - c) - 1) - 21*(2*c 
/(a*c*x - c) + 1)^2*sqrt(-2*c/(a*c*x - c) - 1) - 35*(-2*c/(a*c*x - c) - 1) 
^(3/2) - 35*sqrt(-2*c/(a*c*x - c) - 1))*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/( 
a^2*c^7) + 16*I*sgn(1/(a*c*x - c))*sgn(a)*sgn(c)/(a^2*c^7))*abs(a)
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=-\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {3\,a^4}{35\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^3}-\frac {2\,a^4}{35\,c^5\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}+\frac {a^5}{7\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^4\,\sqrt {-a^2}}+\frac {2\,a^7}{35\,c^5\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^2\,{\left (-a^2\right )}^{3/2}}\right )}{a^4\,\sqrt {-a^2}} \] Input:

int((1 - a^2*x^2)^(1/2)/((c - a*c*x)^5*(a*x + 1)),x)
 

Output:

-((1 - a^2*x^2)^(1/2)*((3*a^4)/(35*c^5*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)^3 
) - (2*a^4)/(35*c^5*(x*(-a^2)^(1/2) - (-a^2)^(1/2)/a)) + a^5/(7*c^5*(x*(-a 
^2)^(1/2) - (-a^2)^(1/2)/a)^4*(-a^2)^(1/2)) + (2*a^7)/(35*c^5*(x*(-a^2)^(1 
/2) - (-a^2)^(1/2)/a)^2*(-a^2)^(3/2))))/(a^4*(-a^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-\text {arctanh}(a x)}}{(c-a c x)^5} \, dx=\frac {\sqrt {-a^{2} x^{2}+1}\, \left (-2 a^{3} x^{3}+8 a^{2} x^{2}-13 a x +12\right )}{35 a \,c^{5} \left (a^{4} x^{4}-4 a^{3} x^{3}+6 a^{2} x^{2}-4 a x +1\right )} \] Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^5,x)
 

Output:

(sqrt( - a**2*x**2 + 1)*( - 2*a**3*x**3 + 8*a**2*x**2 - 13*a*x + 12))/(35* 
a*c**5*(a**4*x**4 - 4*a**3*x**3 + 6*a**2*x**2 - 4*a*x + 1))