\(\int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx\) [623]

Optimal result

Integrand size = 25, antiderivative size = 172 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=-\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{9 x^4 \sqrt {1-a x}}+\frac {4 a \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{21 x^3 \sqrt {1-a x}}-\frac {16 a^2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{105 x^2 \sqrt {1-a x}}+\frac {32 a^3 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2}}{315 x \sqrt {1-a x}} \] Output:

-2/9*(c-c/a/x)^(1/2)*(a*x+1)^(3/2)/x^4/(-a*x+1)^(1/2)+4/21*a*(c-c/a/x)^(1/ 
2)*(a*x+1)^(3/2)/x^3/(-a*x+1)^(1/2)-16/105*a^2*(c-c/a/x)^(1/2)*(a*x+1)^(3/ 
2)/x^2/(-a*x+1)^(1/2)+32/315*a^3*(c-c/a/x)^(1/2)*(a*x+1)^(3/2)/x/(-a*x+1)^ 
(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.37 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {2 \sqrt {c-\frac {c}{a x}} (1+a x)^{3/2} \left (-35+30 a x-24 a^2 x^2+16 a^3 x^3\right )}{315 x^4 \sqrt {1-a x}} \] Input:

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

Output:

(2*Sqrt[c - c/(a*x)]*(1 + a*x)^(3/2)*(-35 + 30*a*x - 24*a^2*x^2 + 16*a^3*x 
^3))/(315*x^4*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.68, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6684, 6678, 516, 55, 55, 55, 48}

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.

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

\(\Big \downarrow \) 6684

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {e^{\text {arctanh}(a x)} \sqrt {1-a x}}{x^{11/2}}dx}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 6678

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \int \frac {\sqrt {1-a^2 x^2}}{x^{11/2} \sqrt {1-a x}}dx}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 516

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

\(\Big \downarrow \) 55

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {2}{3} a \int \frac {\sqrt {a x+1}}{x^{9/2}}dx-\frac {2 (a x+1)^{3/2}}{9 x^{9/2}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {2}{3} a \left (-\frac {4}{7} a \int \frac {\sqrt {a x+1}}{x^{7/2}}dx-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}\right )-\frac {2 (a x+1)^{3/2}}{9 x^{9/2}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {\sqrt {x} \sqrt {c-\frac {c}{a x}} \left (-\frac {2}{3} a \left (-\frac {4}{7} a \left (-\frac {2}{5} a \int \frac {\sqrt {a x+1}}{x^{5/2}}dx-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}\right )-\frac {2 (a x+1)^{3/2}}{9 x^{9/2}}\right )}{\sqrt {1-a x}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {\sqrt {x} \left (-\frac {2 (a x+1)^{3/2}}{9 x^{9/2}}-\frac {2}{3} a \left (-\frac {2 (a x+1)^{3/2}}{7 x^{7/2}}-\frac {4}{7} a \left (\frac {4 a (a x+1)^{3/2}}{15 x^{3/2}}-\frac {2 (a x+1)^{3/2}}{5 x^{5/2}}\right )\right )\right ) \sqrt {c-\frac {c}{a x}}}{\sqrt {1-a x}}\)

Input:

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

Output:

(Sqrt[c - c/(a*x)]*Sqrt[x]*((-2*(1 + a*x)^(3/2))/(9*x^(9/2)) - (2*a*((-2*( 
1 + a*x)^(3/2))/(7*x^(7/2)) - (4*a*((-2*(1 + a*x)^(3/2))/(5*x^(5/2)) + (4* 
a*(1 + a*x)^(3/2))/(15*x^(3/2))))/7))/3))/Sqrt[1 - a*x]
 

Defintions of rubi rules used

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 516

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free 
Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || 
(GtQ[a, 0] && GtQ[c, 0] &&  !IntegerQ[n]))
 

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]
 

rule 6684

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] 
 :> Simp[x^p*((c + d/x)^p/(1 + c*(x/d))^p)   Int[u*(1 + c*(x/d))^p*(E^(n*Ar 
cTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 
 0] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.35

method	result	size
orering	\(\frac {2 \left (16 a^{3} x^{3}-24 a^{2} x^{2}+30 a x -35\right ) \left (a x +1\right )^{2} \sqrt {c -\frac {c}{a x}}}{315 x^{4} \sqrt {-a^{2} x^{2}+1}}\)	\(60\)
gosper	\(\frac {2 \left (a x +1\right )^{2} \left (16 a^{3} x^{3}-24 a^{2} x^{2}+30 a x -35\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}}{315 x^{4} \sqrt {-a^{2} x^{2}+1}}\)	\(62\)
default	\(-\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (16 a^{3} x^{3}-24 a^{2} x^{2}+30 a x -35\right )}{315 x^{4} \left (a x -1\right )}\)	\(67\)
risch	\(\frac {2 \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (16 a^{5} x^{5}+8 a^{4} x^{4}-2 a^{3} x^{3}+a^{2} x^{2}-40 a x -35\right )}{315 \sqrt {-a^{2} x^{2}+1}\, x^{4} \sqrt {-\left (a x +1\right ) a c x}}\)	\(105\)

Input:

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

Output:

2/315*(16*a^3*x^3-24*a^2*x^2+30*a*x-35)/x^4*(a*x+1)^2/(-a^2*x^2+1)^(1/2)*( 
c-c/a/x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.43 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=-\frac {2 \, {\left (16 \, a^{4} x^{4} - 8 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 5 \, a x - 35\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{315 \, {\left (a x^{5} - x^{4}\right )}} \] Input:

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

Output:

-2/315*(16*a^4*x^4 - 8*a^3*x^3 + 6*a^2*x^2 - 5*a*x - 35)*sqrt(-a^2*x^2 + 1 
)*sqrt((a*c*x - c)/(a*x))/(a*x^5 - x^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 23.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.40 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=-\frac {\sqrt {c-\frac {c}{a\,x}}\,\left (-\frac {32\,a^5\,x^5}{315}-\frac {16\,a^4\,x^4}{315}+\frac {4\,a^3\,x^3}{315}-\frac {2\,a^2\,x^2}{315}+\frac {16\,a\,x}{63}+\frac {2}{9}\right )}{x^4\,\sqrt {1-a^2\,x^2}} \] Input:

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

Output:

-((c - c/(a*x))^(1/2)*((16*a*x)/63 - (2*a^2*x^2)/315 + (4*a^3*x^3)/315 - ( 
16*a^4*x^4)/315 - (32*a^5*x^5)/315 + 2/9))/(x^4*(1 - a^2*x^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.30 \[ \int \frac {e^{\text {arctanh}(a x)} \sqrt {c-\frac {c}{a x}}}{x^5} \, dx=\frac {2 \sqrt {x}\, \sqrt {c}\, \sqrt {a}\, \sqrt {a x +1}\, i \left (16 a^{4} x^{4}-8 a^{3} x^{3}+6 a^{2} x^{2}-5 a x -35\right )}{315 a \,x^{5}} \] Input:

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

Output:

(2*sqrt(x)*sqrt(c)*sqrt(a)*sqrt(a*x + 1)*i*(16*a**4*x**4 - 8*a**3*x**3 + 6 
*a**2*x**2 - 5*a*x - 35))/(315*a*x**5)