\(\int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx\) [655]

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {6 \sqrt {c-a^2 c x^2}}{5 a^4}+\frac {3 x \sqrt {c-a^2 c x^2}}{4 a^3}+\frac {3 x^2 \sqrt {c-a^2 c x^2}}{5 a^2}+\frac {x^3 \sqrt {c-a^2 c x^2}}{2 a}+\frac {1}{5} x^4 \sqrt {c-a^2 c x^2}-\frac {3 \sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (24+15 a x+12 a^2 x^2+10 a^3 x^3+4 a^4 x^4\right )+15 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{20 a^4} \] Input:

Integrate[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a^2*c*x^2],x]
 

Output:

(Sqrt[c - a^2*c*x^2]*(24 + 15*a*x + 12*a^2*x^2 + 10*a^3*x^3 + 4*a^4*x^4) + 
 15*Sqrt[c]*ArcTan[(a*x*Sqrt[c - a^2*c*x^2])/(Sqrt[c]*(-1 + a^2*x^2))])/(2 
0*a^4)
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.25, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {6717, 6701, 541, 25, 27, 533, 27, 533, 27, 533, 27, 455, 224, 216}

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.

\(\displaystyle \int x^3 \sqrt {c-a^2 c x^2} e^{2 \coth ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 6717

\(\displaystyle -\int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2}dx\)

\(\Big \downarrow \) 6701

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

\(\Big \downarrow \) 541

\(\displaystyle -c \left (-\frac {\int -\frac {a^2 c x^3 (10 a x+9)}{\sqrt {c-a^2 c x^2}}dx}{5 a^2 c}-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -c \left (\frac {\int \frac {a^2 c x^3 (10 a x+9)}{\sqrt {c-a^2 c x^2}}dx}{5 a^2 c}-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -c \left (\frac {1}{5} \int \frac {x^3 (10 a x+9)}{\sqrt {c-a^2 c x^2}}dx-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 533

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {\int \frac {6 a c x^2 (6 a x+5)}{\sqrt {c-a^2 c x^2}}dx}{4 a^2 c}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \int \frac {x^2 (6 a x+5)}{\sqrt {c-a^2 c x^2}}dx}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \left (\frac {\frac {\int \frac {a c (8 a x+5)}{\sqrt {c-a^2 c x^2}}dx}{2 a^2 c}-\frac {5 x \sqrt {c-a^2 c x^2}}{2 a c}}{a}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a c}\right )}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \left (\frac {\frac {\int \frac {8 a x+5}{\sqrt {c-a^2 c x^2}}dx}{2 a}-\frac {5 x \sqrt {c-a^2 c x^2}}{2 a c}}{a}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a c}\right )}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 455

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \left (\frac {\frac {5 \int \frac {1}{\sqrt {c-a^2 c x^2}}dx-\frac {8 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {5 x \sqrt {c-a^2 c x^2}}{2 a c}}{a}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a c}\right )}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \left (\frac {\frac {5 \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}-\frac {8 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {5 x \sqrt {c-a^2 c x^2}}{2 a c}}{a}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a c}\right )}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle -c \left (\frac {1}{5} \left (\frac {3 \left (\frac {\frac {\frac {5 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{a \sqrt {c}}-\frac {8 \sqrt {c-a^2 c x^2}}{a c}}{2 a}-\frac {5 x \sqrt {c-a^2 c x^2}}{2 a c}}{a}-\frac {2 x^2 \sqrt {c-a^2 c x^2}}{a c}\right )}{2 a}-\frac {5 x^3 \sqrt {c-a^2 c x^2}}{2 a c}\right )-\frac {x^4 \sqrt {c-a^2 c x^2}}{5 c}\right )\)

Input:

Int[E^(2*ArcCoth[a*x])*x^3*Sqrt[c - a^2*c*x^2],x]
 

Output:

-(c*(-1/5*(x^4*Sqrt[c - a^2*c*x^2])/c + ((-5*x^3*Sqrt[c - a^2*c*x^2])/(2*a 
*c) + (3*((-2*x^2*Sqrt[c - a^2*c*x^2])/(a*c) + ((-5*x*Sqrt[c - a^2*c*x^2]) 
/(2*a*c) + ((-8*Sqrt[c - a^2*c*x^2])/(a*c) + (5*ArcTan[(a*Sqrt[c]*x)/Sqrt[ 
c - a^2*c*x^2]])/(a*Sqrt[c]))/(2*a))/a))/(2*a))/5))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

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

rule 224

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

rule 455

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

rule 533

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

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 6701

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

rule 6717

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2)   Int[ 
u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63

method	result	size
risch	\(-\frac {\left (4 a^{4} x^{4}+10 a^{3} x^{3}+12 a^{2} x^{2}+15 a x +24\right ) \left (a^{2} x^{2}-1\right ) c}{20 a^{4} \sqrt {-c \left (a^{2} x^{2}-1\right )}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c}{4 a^{3} \sqrt {a^{2} c}}\)	\(97\)
default	\(-\frac {x^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 a^{2} c}-\frac {4 \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,a^{4}}+\frac {x \sqrt {-a^{2} c \,x^{2}+c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{\sqrt {a^{2} c}}}{a^{3}}+\frac {-\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{2 a^{2} c}+\frac {\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}}{2 a^{2}}}{a}+\frac {2 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {2 a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}}{a^{4}}\)	\(270\)

Input:

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

Output:

-1/20*(4*a^4*x^4+10*a^3*x^3+12*a^2*x^2+15*a*x+24)*(a^2*x^2-1)/a^4/(-c*(a^2 
*x^2-1))^(1/2)*c-3/4/a^3/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+ 
c)^(1/2))*c
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\left [\frac {2 \, {\left (4 \, a^{4} x^{4} + 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 15 \, a x + 24\right )} \sqrt {-a^{2} c x^{2} + c} + 15 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{40 \, a^{4}}, \frac {{\left (4 \, a^{4} x^{4} + 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 15 \, a x + 24\right )} \sqrt {-a^{2} c x^{2} + c} + 15 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{20 \, a^{4}}\right ] \] Input:

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

Output:

[1/40*(2*(4*a^4*x^4 + 10*a^3*x^3 + 12*a^2*x^2 + 15*a*x + 24)*sqrt(-a^2*c*x 
^2 + c) + 15*sqrt(-c)*log(2*a^2*c*x^2 - 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)* 
x - c))/a^4, 1/20*((4*a^4*x^4 + 10*a^3*x^3 + 12*a^2*x^2 + 15*a*x + 24)*sqr 
t(-a^2*c*x^2 + c) + 15*sqrt(c)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^ 
2*c*x^2 - c)))/a^4]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.76 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=-\frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{2}}{5 \, a^{2} c} + \frac {5 \, \sqrt {-a^{2} c x^{2} + c} x}{4 \, a^{3}} - \frac {{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{2 \, a^{3} c} - \frac {3 \, \sqrt {c} \arcsin \left (a x\right )}{4 \, a^{4}} + \frac {2 \, \sqrt {-a^{2} c x^{2} + c}}{a^{4}} - \frac {4 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{5 \, a^{4} c} \] Input:

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

Output:

-1/5*(-a^2*c*x^2 + c)^(3/2)*x^2/(a^2*c) + 5/4*sqrt(-a^2*c*x^2 + c)*x/a^3 - 
 1/2*(-a^2*c*x^2 + c)^(3/2)*x/(a^3*c) - 3/4*sqrt(c)*arcsin(a*x)/a^4 + 2*sq 
rt(-a^2*c*x^2 + c)/a^4 - 4/5*(-a^2*c*x^2 + c)^(3/2)/(a^4*c)
 

Giac [F(-2)]

Exception generated. \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int e^{2 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, \left (-15 \mathit {asin} \left (a x \right )+4 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}+10 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+15 \sqrt {-a^{2} x^{2}+1}\, a x +24 \sqrt {-a^{2} x^{2}+1}-24\right )}{20 a^{4}} \] Input:

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

Output:

(sqrt(c)*( - 15*asin(a*x) + 4*sqrt( - a**2*x**2 + 1)*a**4*x**4 + 10*sqrt( 
- a**2*x**2 + 1)*a**3*x**3 + 12*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 15*sqrt 
( - a**2*x**2 + 1)*a*x + 24*sqrt( - a**2*x**2 + 1) - 24))/(20*a**4)