\(\int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^{5/2} \, dx\) [388]

Optimal result

Integrand size = 23, antiderivative size = 214 \[ \int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^{5/2} \, dx=\frac {(9+4 m) \left (1+\frac {1}{a x}\right )^{3/2} x (e x)^m (c-a c x)^{5/2}}{(1+m) (7+2 m) \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {\left (a-\frac {1}{x}\right ) \left (1+\frac {1}{a x}\right )^{3/2} x (e x)^m (c-a c x)^{5/2}}{a (1+m) \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {\left (71+72 m+16 m^2\right ) (e x)^m (c-a c x)^{5/2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {5}{2}-m,-\frac {3}{2}-m,-\frac {1}{a x}\right )}{a (1+m) (5+2 m) (7+2 m) \left (1-\frac {1}{a x}\right )^{5/2}} \] Output:

(9+4*m)*(1+1/a/x)^(3/2)*x*(e*x)^m*(-a*c*x+c)^(5/2)/(1+m)/(7+2*m)/(1-1/a/x) 
^(5/2)-(a-1/x)*(1+1/a/x)^(3/2)*x*(e*x)^m*(-a*c*x+c)^(5/2)/a/(1+m)/(1-1/a/x 
)^(5/2)-(16*m^2+72*m+71)*(e*x)^m*(-a*c*x+c)^(5/2)*hypergeom([-1/2, -5/2-m] 
,[-3/2-m],-1/a/x)/a/(1+m)/(5+2*m)/(7+2*m)/(1-1/a/x)^(5/2)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.62 \[ \int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^{5/2} \, dx=\frac {c^2 x (e x)^m \sqrt {c-a c x} \left ((5+2 m) \sqrt {1+\frac {1}{a x}} (1+a x) (7+2 a x+2 m (1+a x))-a \left (71+72 m+16 m^2\right ) x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {5}{2}-m,-\frac {3}{2}-m,-\frac {1}{a x}\right )\right )}{(1+m) (5+2 m) (7+2 m) \sqrt {1-\frac {1}{a x}}} \] Input:

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

Output:

(c^2*x*(e*x)^m*Sqrt[c - a*c*x]*((5 + 2*m)*Sqrt[1 + 1/(a*x)]*(1 + a*x)*(7 + 
 2*a*x + 2*m*(1 + a*x)) - a*(71 + 72*m + 16*m^2)*x*Hypergeometric2F1[-1/2, 
 -5/2 - m, -3/2 - m, -(1/(a*x))]))/((1 + m)*(5 + 2*m)*(7 + 2*m)*Sqrt[1 - 1 
/(a*x)])
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6730, 27, 101, 27, 88, 74}

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.

\(\displaystyle \int (c-a c x)^{5/2} e^{\coth ^{-1}(a x)} (e x)^m \, dx\)

\(\Big \downarrow \) 6730

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{m+\frac {5}{2}} (c-a c x)^{5/2} (e x)^m \int \frac {\left (a-\frac {1}{x}\right )^2 \sqrt {1+\frac {1}{a x}} \left (\frac {1}{x}\right )^{-m-\frac {9}{2}}}{a^2}d\frac {1}{x}}{\left (1-\frac {1}{a x}\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{m+\frac {5}{2}} (c-a c x)^{5/2} (e x)^m \int \left (a-\frac {1}{x}\right )^2 \sqrt {1+\frac {1}{a x}} \left (\frac {1}{x}\right )^{-m-\frac {9}{2}}d\frac {1}{x}}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\)

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 88

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{m+\frac {5}{2}} (c-a c x)^{5/2} (e x)^m \left (\frac {a \left (-\frac {\left (16 m^2+72 m+71\right ) \int \sqrt {1+\frac {1}{a x}} \left (\frac {1}{x}\right )^{-m-\frac {7}{2}}d\frac {1}{x}}{2 m+7}-\frac {2 a (4 m+9) \left (\frac {1}{a x}+1\right )^{3/2} \left (\frac {1}{x}\right )^{-m-\frac {7}{2}}}{2 m+7}\right )}{2 (m+1)}+\frac {a \left (a-\frac {1}{x}\right ) \left (\frac {1}{a x}+1\right )^{3/2} \left (\frac {1}{x}\right )^{-m-\frac {7}{2}}}{m+1}\right )}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\)

\(\Big \downarrow \) 74

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^{m+\frac {5}{2}} (c-a c x)^{5/2} (e x)^m \left (\frac {a \left (\frac {2 \left (16 m^2+72 m+71\right ) \left (\frac {1}{x}\right )^{-m-\frac {5}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m-\frac {5}{2},-m-\frac {3}{2},-\frac {1}{a x}\right )}{(2 m+5) (2 m+7)}-\frac {2 a (4 m+9) \left (\frac {1}{a x}+1\right )^{3/2} \left (\frac {1}{x}\right )^{-m-\frac {7}{2}}}{2 m+7}\right )}{2 (m+1)}+\frac {a \left (a-\frac {1}{x}\right ) \left (\frac {1}{a x}+1\right )^{3/2} \left (\frac {1}{x}\right )^{-m-\frac {7}{2}}}{m+1}\right )}{a^2 \left (1-\frac {1}{a x}\right )^{5/2}}\)

Input:

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

Output:

-(((x^(-1))^(5/2 + m)*(e*x)^m*(c - a*c*x)^(5/2)*((a*(a - x^(-1))*(1 + 1/(a 
*x))^(3/2)*(x^(-1))^(-7/2 - m))/(1 + m) + (a*((-2*a*(9 + 4*m)*(1 + 1/(a*x) 
)^(3/2)*(x^(-1))^(-7/2 - m))/(7 + 2*m) + (2*(71 + 72*m + 16*m^2)*(x^(-1))^ 
(-5/2 - m)*Hypergeometric2F1[-1/2, -5/2 - m, -3/2 - m, -(1/(a*x))])/((5 + 
2*m)*(7 + 2*m))))/(2*(1 + m))))/(a^2*(1 - 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 74

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x 
)^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] 
/; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] 
 &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))
 

rule 88

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)^Simplify[p + 1], 
 x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimpl 
erQ[p, 1]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

rule 6730

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

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (-a c x +c \right )^{\frac {5}{2}}}{\sqrt {\frac {a x -1}{a x +1}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((a^2*c^2*x^2 - c^2)*sqrt(-a*c*x + c)*(e*x)^m*sqrt((a*x - 1)/(a*x 
+ 1)), x)
 

Sympy [F(-1)]

Timed out. \[ \int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**m*sqrt(c)*c**2*(int((x**m*sqrt(a*x + 1)*sqrt( - a*x + 1)*x**2)/sqrt(a*x 
 - 1),x)*a**2 - 2*int((x**m*sqrt(a*x + 1)*sqrt( - a*x + 1)*x)/sqrt(a*x - 1 
),x)*a + int((x**m*sqrt(a*x + 1)*sqrt( - a*x + 1))/sqrt(a*x - 1),x))