\(\int e^{-2 \text {arctanh}(a x)} (c-a^2 c x^2)^{3/2} \, dx\) [1282]

Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {5}{8} c x \sqrt {c-a^2 c x^2}+\frac {5 \left (c-a^2 c x^2\right )^{3/2}}{12 a}+\frac {\left (c-a^2 c x^2\right )^{5/2}}{4 a c (1+a x)}+\frac {5 c^{3/2} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{8 a} \] Output:

5/8*c*x*(-a^2*c*x^2+c)^(1/2)+5/12*(-a^2*c*x^2+c)^(3/2)/a+1/4*(-a^2*c*x^2+c 
)^(5/2)/a/c/(a*x+1)+5/8*c^(3/2)*arctan(a*c^(1/2)*x/(-a^2*c*x^2+c)^(1/2))/a
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {c \sqrt {c-a^2 c x^2} \left (\sqrt {1+a x} \left (-16+7 a x+25 a^2 x^2-22 a^3 x^3+6 a^4 x^4\right )+30 \sqrt {1-a x} \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )\right )}{24 a \sqrt {1-a x} \sqrt {1-a^2 x^2}} \] Input:

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

Output:

-1/24*(c*Sqrt[c - a^2*c*x^2]*(Sqrt[1 + a*x]*(-16 + 7*a*x + 25*a^2*x^2 - 22 
*a^3*x^3 + 6*a^4*x^4) + 30*Sqrt[1 - a*x]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]]))/( 
a*Sqrt[1 - a*x]*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6692, 469, 455, 211, 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.

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

\(\Big \downarrow \) 6692

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 469

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
((n + p)/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr 
eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 
p + 1, 0] && IntegerQ[2*p]
 

rule 6692

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {\left (6 a^{3} x^{3}-16 a^{2} x^{2}+9 a x +16\right ) \left (a^{2} x^{2}-1\right ) c^{2}}{24 a \sqrt {-c \left (a^{2} x^{2}-1\right )}}+\frac {5 \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right ) c^{2}}{8 \sqrt {a^{2} c}}\)	\(90\)
default	\(-\frac {x \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{4}-\frac {3 c \left (\frac {x \sqrt {-a^{2} c \,x^{2}+c}}{2}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-a^{2} c \,x^{2}+c}}\right )}{2 \sqrt {a^{2} c}}\right )}{4}+\frac {\frac {2 \left (-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}{3}+2 a c \left (-\frac {\left (-2 \left (x +\frac {1}{a}\right ) a^{2} c +2 a c \right ) \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}{4 a^{2} c}+\frac {c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}\right )}{2 \sqrt {a^{2} c}}\right )}{a}\)	\(202\)

Input:

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

Output:

-1/24*(6*a^3*x^3-16*a^2*x^2+9*a*x+16)*(a^2*x^2-1)/a/(-c*(a^2*x^2-1))^(1/2) 
*c^2+5/8/(a^2*c)^(1/2)*arctan((a^2*c)^(1/2)*x/(-a^2*c*x^2+c)^(1/2))*c^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.61 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\left [\frac {15 \, \sqrt {-c} c \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + 2 \, {\left (6 \, a^{3} c x^{3} - 16 \, a^{2} c x^{2} + 9 \, a c x + 16 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{48 \, a}, -\frac {15 \, c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) - {\left (6 \, a^{3} c x^{3} - 16 \, a^{2} c x^{2} + 9 \, a c x + 16 \, c\right )} \sqrt {-a^{2} c x^{2} + c}}{24 \, a}\right ] \] Input:

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

Output:

[1/48*(15*sqrt(-c)*c*log(2*a^2*c*x^2 + 2*sqrt(-a^2*c*x^2 + c)*a*sqrt(-c)*x 
 - c) + 2*(6*a^3*c*x^3 - 16*a^2*c*x^2 + 9*a*c*x + 16*c)*sqrt(-a^2*c*x^2 + 
c))/a, -1/24*(15*c^(3/2)*arctan(sqrt(-a^2*c*x^2 + c)*a*sqrt(c)*x/(a^2*c*x^ 
2 - c)) - (6*a^3*c*x^3 - 16*a^2*c*x^2 + 9*a*c*x + 16*c)*sqrt(-a^2*c*x^2 + 
c))/a]
 

Sympy [A] (verification not implemented)

Time = 2.14 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.28 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=a^{2} c \left (\begin {cases} \left (\frac {x^{3}}{4} - \frac {x}{8 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} + \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{8 a^{2}} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{3}}{3} & \text {otherwise} \end {cases}\right ) - 2 a c \left (\begin {cases} \left (\frac {x^{2}}{3} - \frac {1}{3 a^{2}}\right ) \sqrt {- a^{2} c x^{2} + c} & \text {for}\: a^{2} c \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} \frac {c \left (\begin {cases} \frac {\log {\left (- 2 a^{2} c x + 2 \sqrt {- a^{2} c} \sqrt {- a^{2} c x^{2} + c} \right )}}{\sqrt {- a^{2} c}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- a^{2} c x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {- a^{2} c x^{2} + c}}{2} & \text {for}\: a^{2} c \neq 0 \\\sqrt {c} x & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**2*c*Piecewise(((x**3/4 - x/(8*a**2))*sqrt(-a**2*c*x**2 + c) + c*Piecewi 
se((log(-2*a**2*c*x + 2*sqrt(-a**2*c)*sqrt(-a**2*c*x**2 + c))/sqrt(-a**2*c 
), Ne(c, 0)), (x*log(x)/sqrt(-a**2*c*x**2), True))/(8*a**2), Ne(a**2*c, 0) 
), (sqrt(c)*x**3/3, True)) - 2*a*c*Piecewise(((x**2/3 - 1/(3*a**2))*sqrt(- 
a**2*c*x**2 + c), Ne(a**2*c, 0)), (sqrt(c)*x**2/2, True)) + c*Piecewise((c 
*Piecewise((log(-2*a**2*c*x + 2*sqrt(-a**2*c)*sqrt(-a**2*c*x**2 + c))/sqrt 
(-a**2*c), Ne(c, 0)), (x*log(x)/sqrt(-a**2*c*x**2), True))/2 + x*sqrt(-a** 
2*c*x**2 + c)/2, Ne(a**2*c, 0)), (sqrt(c)*x, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {1}{4} \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x + \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c x - \frac {3}{8} \, \sqrt {-a^{2} c x^{2} + c} c x - \frac {c^{3} \arcsin \left (a x + 2\right )}{a \left (-c\right )^{\frac {3}{2}}} - \frac {3 \, c^{\frac {3}{2}} \arcsin \left (a x\right )}{8 \, a} + \frac {2 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}}}{3 \, a} + \frac {2 \, \sqrt {a^{2} c x^{2} + 4 \, a c x + 3 \, c} c}{a} \] Input:

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

Output:

-1/4*(-a^2*c*x^2 + c)^(3/2)*x + sqrt(a^2*c*x^2 + 4*a*c*x + 3*c)*c*x - 3/8* 
sqrt(-a^2*c*x^2 + c)*c*x - c^3*arcsin(a*x + 2)/(a*(-c)^(3/2)) - 3/8*c^(3/2 
)*arcsin(a*x)/a + 2/3*(-a^2*c*x^2 + c)^(3/2)/a + 2*sqrt(a^2*c*x^2 + 4*a*c* 
x + 3*c)*c/a
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (92) = 184\).

Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.00 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {{\left (240 \, a^{5} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-c + \frac {2 \, c}{a x + 1}}}{\sqrt {c}}\right ) \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) - \frac {{\left (15 \, a^{5} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{3} c^{2} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) + 73 \, a^{5} {\left (c - \frac {2 \, c}{a x + 1}\right )}^{2} c^{3} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) + 15 \, a^{5} c^{5} \sqrt {-c + \frac {2 \, c}{a x + 1}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right ) + 55 \, a^{5} c^{4} {\left (-c + \frac {2 \, c}{a x + 1}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{a x + 1}\right ) \mathrm {sgn}\left (a\right )\right )} {\left (a x + 1\right )}^{4}}{c^{4}}\right )} {\left | a \right |}}{192 \, a^{7}} \] Input:

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

Output:

-1/192*(240*a^5*c^(3/2)*arctan(sqrt(-c + 2*c/(a*x + 1))/sqrt(c))*sgn(1/(a* 
x + 1))*sgn(a) - (15*a^5*(c - 2*c/(a*x + 1))^3*c^2*sqrt(-c + 2*c/(a*x + 1) 
)*sgn(1/(a*x + 1))*sgn(a) + 73*a^5*(c - 2*c/(a*x + 1))^2*c^3*sqrt(-c + 2*c 
/(a*x + 1))*sgn(1/(a*x + 1))*sgn(a) + 15*a^5*c^5*sqrt(-c + 2*c/(a*x + 1))* 
sgn(1/(a*x + 1))*sgn(a) + 55*a^5*c^4*(-c + 2*c/(a*x + 1))^(3/2)*sgn(1/(a*x 
 + 1))*sgn(a))*(a*x + 1)^4/c^4)*abs(a)/a^7
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.73 \[ \int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {\sqrt {c}\, c \left (15 \mathit {asin} \left (a x \right )+6 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-16 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+9 \sqrt {-a^{2} x^{2}+1}\, a x +16 \sqrt {-a^{2} x^{2}+1}-16\right )}{24 a} \] Input:

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

Output:

(sqrt(c)*c*(15*asin(a*x) + 6*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 16*sqrt( - 
 a**2*x**2 + 1)*a**2*x**2 + 9*sqrt( - a**2*x**2 + 1)*a*x + 16*sqrt( - a**2 
*x**2 + 1) - 16))/(24*a)