\(\int e^{3 \text {arctanh}(a x)} (c-\frac {c}{a x})^{9/2} \, dx\) [544]

Optimal result

Integrand size = 24, antiderivative size = 261 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\frac {3 a^4 \left (c-\frac {c}{a x}\right )^{9/2} x^5 \sqrt {1+a x}}{(1-a x)^{9/2}}+\frac {18 a^2 \left (c-\frac {c}{a x}\right )^{9/2} x^3 (1+a x)^{3/2}}{35 (1-a x)^{9/2}}-\frac {156 a^3 \left (c-\frac {c}{a x}\right )^{9/2} x^4 (1+a x)^{3/2}}{35 (1-a x)^{9/2}}+\frac {6 a \left (c-\frac {c}{a x}\right )^{9/2} x^2 (1+a x)^{3/2}}{35 (1-a x)^{5/2}}-\frac {2 \left (c-\frac {c}{a x}\right )^{9/2} x (1+a x)^{3/2}}{7 (1-a x)^{3/2}}+\frac {3 a^{7/2} \left (c-\frac {c}{a x}\right )^{9/2} x^{9/2} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{(1-a x)^{9/2}} \] Output:

3*a^4*(c-c/a/x)^(9/2)*x^5*(a*x+1)^(1/2)/(-a*x+1)^(9/2)+18/35*a^2*(c-c/a/x) 
^(9/2)*x^3*(a*x+1)^(3/2)/(-a*x+1)^(9/2)-156/35*a^3*(c-c/a/x)^(9/2)*x^4*(a* 
x+1)^(3/2)/(-a*x+1)^(9/2)+6/35*a*(c-c/a/x)^(9/2)*x^2*(a*x+1)^(3/2)/(-a*x+1 
)^(5/2)-2/7*(c-c/a/x)^(9/2)*x*(a*x+1)^(3/2)/(-a*x+1)^(3/2)+3*a^(7/2)*(c-c/ 
a/x)^(9/2)*x^(9/2)*arcsinh(a^(1/2)*x^(1/2))/(-a*x+1)^(9/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.33 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=-\frac {c^4 \sqrt {c-\frac {c}{a x}} \left ((1+a x)^{5/2} \left (10-46 a x+35 a^2 x^2\right )+35 a^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},-a x\right )\right )}{35 a^4 x^3 \sqrt {1-a x}} \] Input:

Integrate[E^(3*ArcTanh[a*x])*(c - c/(a*x))^(9/2),x]
 

Output:

-1/35*(c^4*Sqrt[c - c/(a*x)]*((1 + a*x)^(5/2)*(10 - 46*a*x + 35*a^2*x^2) + 
 35*a^2*x^2*Hypergeometric2F1[-3/2, -3/2, -1/2, -(a*x)]))/(a^4*x^3*Sqrt[1 
- a*x])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.61, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6684, 6679, 108, 27, 167, 27, 160, 57, 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.

\(\displaystyle \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx\)

\(\Big \downarrow \) 6684

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

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 167

\(\displaystyle \frac {x^{9/2} \left (c-\frac {c}{a x}\right )^{9/2} \left (-\frac {3}{7} a \left (\frac {2}{5} \int \frac {a (9-17 a x) (1-a x) \sqrt {a x+1}}{2 x^{5/2}}dx-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right )}{(1-a x)^{9/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^{9/2} \left (c-\frac {c}{a x}\right )^{9/2} \left (-\frac {3}{7} a \left (\frac {1}{5} a \int \frac {(9-17 a x) (1-a x) \sqrt {a x+1}}{x^{5/2}}dx-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right )}{(1-a x)^{9/2}}\)

\(\Big \downarrow \) 160

\(\displaystyle \frac {x^{9/2} \left (c-\frac {c}{a x}\right )^{9/2} \left (-\frac {3}{7} a \left (\frac {1}{5} a \left (-\frac {35}{2} a \int \frac {\sqrt {a x+1}}{x^{3/2}}dx-\frac {(6-17 a x) (a x+1)^{3/2}}{x^{3/2}}\right )-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right )}{(1-a x)^{9/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {x^{9/2} \left (c-\frac {c}{a x}\right )^{9/2} \left (-\frac {3}{7} a \left (\frac {1}{5} a \left (-\frac {35}{2} a \left (a \int \frac {1}{\sqrt {x} \sqrt {a x+1}}dx-\frac {2 \sqrt {a x+1}}{\sqrt {x}}\right )-\frac {(6-17 a x) (a x+1)^{3/2}}{x^{3/2}}\right )-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right )}{(1-a x)^{9/2}}\)

\(\Big \downarrow \) 63

\(\displaystyle \frac {x^{9/2} \left (c-\frac {c}{a x}\right )^{9/2} \left (-\frac {3}{7} a \left (\frac {1}{5} a \left (-\frac {35}{2} a \left (2 a \int \frac {1}{\sqrt {a x+1}}d\sqrt {x}-\frac {2 \sqrt {a x+1}}{\sqrt {x}}\right )-\frac {(6-17 a x) (a x+1)^{3/2}}{x^{3/2}}\right )-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right )}{(1-a x)^{9/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {x^{9/2} \left (-\frac {3}{7} a \left (\frac {1}{5} a \left (-\frac {35}{2} a \left (2 \sqrt {a} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )-\frac {2 \sqrt {a x+1}}{\sqrt {x}}\right )-\frac {(6-17 a x) (a x+1)^{3/2}}{x^{3/2}}\right )-\frac {2 (1-a x)^2 (a x+1)^{3/2}}{5 x^{5/2}}\right )-\frac {2 (a x+1)^{3/2} (1-a x)^3}{7 x^{7/2}}\right ) \left (c-\frac {c}{a x}\right )^{9/2}}{(1-a x)^{9/2}}\)

Input:

Int[E^(3*ArcTanh[a*x])*(c - c/(a*x))^(9/2),x]
 

Output:

((c - c/(a*x))^(9/2)*x^(9/2)*((-2*(1 - a*x)^3*(1 + a*x)^(3/2))/(7*x^(7/2)) 
 - (3*a*((-2*(1 - a*x)^2*(1 + a*x)^(3/2))/(5*x^(5/2)) + (a*(-(((6 - 17*a*x 
)*(1 + a*x)^(3/2))/x^(3/2)) - (35*a*((-2*Sqrt[1 + a*x])/Sqrt[x] + 2*Sqrt[a 
]*ArcSinh[Sqrt[a]*Sqrt[x]]))/2))/5))/7))/(1 - a*x)^(9/2)
 

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 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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 108

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

rule 160

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

rule 167

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
 

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 = 0.20 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, c^{4} \sqrt {-a^{2} x^{2}+1}\, \left (70 a^{\frac {9}{2}} \sqrt {-x \left (a x +1\right )}\, x^{4}+105 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}}\right ) a^{4} x^{4}+328 a^{\frac {7}{2}} x^{3} \sqrt {-x \left (a x +1\right )}-24 a^{\frac {5}{2}} x^{2} \sqrt {-x \left (a x +1\right )}-52 a^{\frac {3}{2}} x \sqrt {-x \left (a x +1\right )}+20 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}\right )}{70 x^{3} a^{\frac {9}{2}} \left (a x -1\right ) \sqrt {-x \left (a x +1\right )}}\)	\(172\)
risch	\(-\frac {\left (35 a^{5} x^{5}+199 a^{4} x^{4}+152 a^{3} x^{3}-38 a^{2} x^{2}-16 a x +10\right ) c^{4} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{35 x^{3} \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {-a^{2} x^{2}+1}\, a^{4}}+\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, \left (x +\frac {1}{2 a}\right )}{\sqrt {-a^{2} c \,x^{2}-a c x}}\right ) c^{4} \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}}{2 \sqrt {a^{2} c}\, \sqrt {-a^{2} x^{2}+1}}\)	\(208\)

Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(9/2),x,method=_RETURNVERBOSE)
 

Output:

1/70*(c*(a*x-1)/a/x)^(1/2)/x^3*c^4/a^(9/2)*(-a^2*x^2+1)^(1/2)*(70*a^(9/2)* 
(-x*(a*x+1))^(1/2)*x^4+105*arctan(1/2/a^(1/2)*(2*a*x+1)/(-x*(a*x+1))^(1/2) 
)*a^4*x^4+328*a^(7/2)*x^3*(-x*(a*x+1))^(1/2)-24*a^(5/2)*x^2*(-x*(a*x+1))^( 
1/2)-52*a^(3/2)*x*(-x*(a*x+1))^(1/2)+20*a^(1/2)*(-x*(a*x+1))^(1/2))/(a*x-1 
)/(-x*(a*x+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.48 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\left [\frac {105 \, {\left (a^{4} c^{4} x^{4} - a^{3} c^{4} x^{3}\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \, {\left (35 \, a^{4} c^{4} x^{4} + 164 \, a^{3} c^{4} x^{3} - 12 \, a^{2} c^{4} x^{2} - 26 \, a c^{4} x + 10 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{140 \, {\left (a^{5} x^{4} - a^{4} x^{3}\right )}}, -\frac {105 \, {\left (a^{4} c^{4} x^{4} - a^{3} c^{4} x^{3}\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 2 \, {\left (35 \, a^{4} c^{4} x^{4} + 164 \, a^{3} c^{4} x^{3} - 12 \, a^{2} c^{4} x^{2} - 26 \, a c^{4} x + 10 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{70 \, {\left (a^{5} x^{4} - a^{4} x^{3}\right )}}\right ] \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(9/2),x, algorithm="frica 
s")
 

Output:

[1/140*(105*(a^4*c^4*x^4 - a^3*c^4*x^3)*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*(35*a^4*c^4*x^4 + 164*a^3*c^4*x^3 - 12*a^2*c^4*x^2 
- 26*a*c^4*x + 10*c^4)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^5*x^ 
4 - a^4*x^3), -1/70*(105*(a^4*c^4*x^4 - a^3*c^4*x^3)*sqrt(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*(35*a^4*c^4*x^4 + 164*a^3*c^4*x^3 - 12*a^2*c^4*x^2 - 26*a*c^4*x + 
10*c^4)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^5*x^4 - a^4*x^3)]
 

Sympy [F]

\[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\int \frac {\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{\frac {9}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(c-c/a/x)**(9/2),x)
 

Output:

Integral((-c*(-1 + 1/(a*x)))**(9/2)*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**( 
3/2), x)
 

Maxima [F]

\[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{\frac {9}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(9/2),x, algorithm="maxim 
a")
 

Output:

integrate((a*x + 1)^3*(c - c/(a*x))^(9/2)/(-a^2*x^2 + 1)^(3/2), x)
 

Giac [F]

\[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\int { \frac {{\left (a x + 1\right )}^{3} {\left (c - \frac {c}{a x}\right )}^{\frac {9}{2}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(9/2),x, algorithm="giac" 
)
 

Output:

integrate((a*x + 1)^3*(c - c/(a*x))^(9/2)/(-a^2*x^2 + 1)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\int \frac {{\left (c-\frac {c}{a\,x}\right )}^{9/2}\,{\left (a\,x+1\right )}^3}{{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \] Input:

int(((c - c/(a*x))^(9/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
 

Output:

int(((c - c/(a*x))^(9/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.49 \[ \int e^{3 \text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^{9/2} \, dx=\frac {\sqrt {c}\, c^{4} \left (-105 \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i}{a x +1}\right ) a^{4} x^{4}+35 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{4} i \,x^{4}+164 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{3} i \,x^{3}-12 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a^{2} i \,x^{2}-26 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a i x +10 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i \right )}{35 a^{5} x^{4}} \] Input:

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(c-c/a/x)^(9/2),x)
 

Output:

(sqrt(c)*c**4*( - 105*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i)/(a*x + 1))*a* 
*4*x**4 + 35*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a**4*i*x**4 + 164*sqrt(x)*sqrt( 
a)*sqrt(a*x + 1)*a**3*i*x**3 - 12*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a**2*i*x** 
2 - 26*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*a*i*x + 10*sqrt(x)*sqrt(a)*sqrt(a*x + 
 1)*i))/(35*a**5*x**4)