\(\int e^{\text {arctanh}(a x)} (c-\frac {c}{a x})^{5/2} \, dx\) [544]

Optimal result

Integrand size = 22, antiderivative size = 171 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx=-\frac {3 a^2 \left (c-\frac {c}{a x}\right )^{5/2} x^3 \sqrt {1+a x}}{(1-a x)^{5/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{5/2} x (1+a x)^{3/2}}{3 (1-a x)^{5/2}}+\frac {4 a \left (c-\frac {c}{a x}\right )^{5/2} x^2 (1+a x)^{3/2}}{(1-a x)^{5/2}}-\frac {3 a^{3/2} \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.51 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx=\frac {c^2 \sqrt {c-\frac {c}{a x}} \left (\sqrt {1+a x} \left (-2+10 a x+3 a^2 x^2\right )-9 a^{3/2} x^{3/2} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )\right )}{3 a^2 x \sqrt {1-a x}} \] Input:

Integrate[E^ArcTanh[a*x]*(c - c/(a*x))^(5/2),x]
 

Output:

(c^2*Sqrt[c - c/(a*x)]*(Sqrt[1 + a*x]*(-2 + 10*a*x + 3*a^2*x^2) - 9*a^(3/2 
)*x^(3/2)*ArcSinh[Sqrt[a]*Sqrt[x]]))/(3*a^2*x*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.63, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6684, 6679, 100, 27, 87, 60, 63, 222}

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.

\(\displaystyle \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx\)

\(\Big \downarrow \) 6684

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 222

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

Input:

Int[E^ArcTanh[a*x]*(c - c/(a*x))^(5/2),x]
 

Output:

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

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 222

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

rule 6679

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

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 = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{2} \sqrt {-a^{2} x^{2}+1}\, \left (6 a^{\frac {5}{2}} x^{2} \sqrt {-x \left (a x +1\right )}+9 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}}\right ) a^{2} x^{2}+20 a^{\frac {3}{2}} x \sqrt {-x \left (a x +1\right )}-4 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}\right )}{6 x \,a^{\frac {5}{2}} \left (a x -1\right ) \sqrt {-x \left (a x +1\right )}}\)	\(136\)
risch	\(\frac {\left (3 a^{3} x^{3}+13 a^{2} x^{2}+8 a x -2\right ) c^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{3 x \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}\, a^{2}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, \left (x +\frac {1}{2 a}\right )}{\sqrt {-a^{2} c \,x^{2}-a c x}}\right ) c^{2} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{2 \sqrt {a^{2} c}\, \sqrt {-a^{2} x^{2}+1}}\)	\(192\)

Input:

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

Output:

-1/6*(c*(a*x-1)/a/x)^(1/2)/x*c^2/a^(5/2)*(-a^2*x^2+1)^(1/2)*(6*a^(5/2)*x^2 
*(-x*(a*x+1))^(1/2)+9*arctan(1/2/a^(1/2)*(2*a*x+1)/(-x*(a*x+1))^(1/2))*a^2 
*x^2+20*a^(3/2)*x*(-x*(a*x+1))^(1/2)-4*a^(1/2)*(-x*(a*x+1))^(1/2))/(a*x-1) 
/(-x*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.93 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx=\left [\frac {9 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x - 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) - 4 \, {\left (3 \, a^{2} c^{2} x^{2} + 10 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{12 \, {\left (a^{3} x^{2} - a^{2} x\right )}}, \frac {9 \, {\left (a^{2} c^{2} x^{2} - a c^{2} x\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (3 \, a^{2} c^{2} x^{2} + 10 \, a c^{2} x - 2 \, c^{2}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{6 \, {\left (a^{3} x^{2} - a^{2} x\right )}}\right ] \] Input:

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

Output:

[1/12*(9*(a^2*c^2*x^2 - a*c^2*x)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4* 
(2*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c) 
/(a*x - 1)) - 4*(3*a^2*c^2*x^2 + 10*a*c^2*x - 2*c^2)*sqrt(-a^2*x^2 + 1)*sq 
rt((a*c*x - c)/(a*x)))/(a^3*x^2 - a^2*x), 1/6*(9*(a^2*c^2*x^2 - a*c^2*x)*s 
qrt(c)*arctan(2*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2* 
a^2*c*x^2 - a*c*x - c)) - 2*(3*a^2*c^2*x^2 + 10*a*c^2*x - 2*c^2)*sqrt(-a^2 
*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*x^2 - a^2*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.52 \[ \int e^{\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{5/2} \, dx=\frac {\sqrt {c}\, c^{2} \left (9 \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i}{a x +1}\right ) a^{2} x^{2}-3 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{2} i \,x^{2}-10 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a i x +2 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i \right )}{3 a^{3} x^{2}} \] Input:

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

Output:

(sqrt(c)*c**2*(9*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i)/(a*x + 1))*a**2*x* 
*2 - 3*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a**2*i*x**2 - 10*sqrt(x)*sqrt(a)*sqrt 
(a*x + 1)*a*i*x + 2*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i))/(3*a**3*x**2)