\(\int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx\) [1099]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=-\frac {6 \sqrt {c-a^2 c x^2}}{5 a^4}-\frac {3 x \sqrt {c-a^2 c x^2}}{4 a^3}-\frac {3 x^2 \sqrt {c-a^2 c x^2}}{5 a^2}-\frac {x^3 \sqrt {c-a^2 c x^2}}{2 a}-\frac {1}{5} x^4 \sqrt {c-a^2 c x^2}+\frac {3 \sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{4 a^4} \] 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 = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=-\frac {\sqrt {c-a^2 c x^2} \left (24+15 a x+12 a^2 x^2+10 a^3 x^3+4 a^4 x^4\right )+15 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )}{20 a^4} \] Input:

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

Output:

-1/20*(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) 
)])/a^4
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.24, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {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.

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

\(\Big \downarrow \) 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*ArcTanh[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]
 
Maple [A] (verified)

Time = 0.30 (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 \(\frac {x^{2} \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 a^{2} c}+\frac {4 \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{5 c \,a^{4}}-\frac {2 \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{a^{3}}-\frac {2 \left (-\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4 a^{2} c}+\frac {\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}}{4 a^{2}}\right )}{a}-\frac {2 \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}-\frac {a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -2 \left (x -\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\right )}{a^{4}}\) \(270\)

Input:

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

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 = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\left [-\frac {2 \, {\left (4 \, a^{4} x^{4} + 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 15 \, a x + 24\right )} \sqrt {-a^{2} c x^{2} + c} - 15 \, \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right )}{40 \, a^{4}}, -\frac {{\left (4 \, a^{4} x^{4} + 10 \, a^{3} x^{3} + 12 \, a^{2} x^{2} + 15 \, a x + 24\right )} \sqrt {-a^{2} c x^{2} + c} + 15 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right )}{20 \, a^{4}}\right ] \] Input:

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

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)*s 
qrt(-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]

\[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=- \int \frac {x^{3} \sqrt {- a^{2} c x^{2} + c}}{a x - 1}\, dx - \int \frac {a x^{4} \sqrt {- a^{2} c x^{2} + c}}{a x - 1}\, dx \] Input:

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

Output:

-Integral(x**3*sqrt(-a**2*c*x**2 + c)/(a*x - 1), x) - Integral(a*x**4*sqrt 
(-a**2*c*x**2 + c)/(a*x - 1), x)
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.78 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {1}{20} \, a {\left (\frac {4 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{2}}{a^{3} c} - \frac {25 \, \sqrt {-a^{2} c x^{2} + c} x}{a^{4}} + \frac {10 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x}{a^{4} c} + \frac {15 \, \sqrt {c} \arcsin \left (a x\right )}{a^{5}} - \frac {40 \, \sqrt {-a^{2} c x^{2} + c}}{a^{5}} + \frac {16 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{a^{5} c}\right )} \] Input:

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

Output:

1/20*a*(4*(-a^2*c*x^2 + c)^(3/2)*x^2/(a^3*c) - 25*sqrt(-a^2*c*x^2 + c)*x/a 
^4 + 10*(-a^2*c*x^2 + c)^(3/2)*x/(a^4*c) + 15*sqrt(c)*arcsin(a*x)/a^5 - 40 
*sqrt(-a^2*c*x^2 + c)/a^5 + 16*(-a^2*c*x^2 + c)^(3/2)/(a^5*c))
 

Giac [F(-2)]

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

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

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.65 \[ \int e^{2 \text {arctanh}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, \left (15 \mathit {asin} \left (a x \right )-4 \sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-10 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-12 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-15 \sqrt {-a^{2} x^{2}+1}\, a x -24 \sqrt {-a^{2} x^{2}+1}+24\right )}{20 a^{4}} \] Input:

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