\(\int e^{3 \text {arctanh}(a x)} x (c-a^2 c x^2) \, dx\) [1176]

Optimal result

Integrand size = 21, antiderivative size = 99 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=-\frac {3 c \sqrt {1-a^2 x^2}}{a^2}-\frac {15 c x \sqrt {1-a^2 x^2}}{8 a}-c x^2 \sqrt {1-a^2 x^2}-\frac {1}{4} a c x^3 \sqrt {1-a^2 x^2}+\frac {15 c \arcsin (a x)}{8 a^2} \] Output:

-3*c*(-a^2*x^2+1)^(1/2)/a^2-15/8*c*x*(-a^2*x^2+1)^(1/2)/a-c*x^2*(-a^2*x^2+ 
1)^(1/2)-1/4*a*c*x^3*(-a^2*x^2+1)^(1/2)+15/8*c*arcsin(a*x)/a^2
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.55 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=\frac {-c \sqrt {1-a^2 x^2} \left (24+15 a x+8 a^2 x^2+2 a^3 x^3\right )+15 c \arcsin (a x)}{8 a^2} \] Input:

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

Output:

(-(c*Sqrt[1 - a^2*x^2]*(24 + 15*a*x + 8*a^2*x^2 + 2*a^3*x^3)) + 15*c*ArcSi 
n[a*x])/(8*a^2)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6698, 541, 25, 2340, 27, 533, 27, 455, 223}

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.

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

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 541

\(\displaystyle c \left (-\frac {\int -\frac {x \left (12 x^2 a^4+15 x a^3+4 a^2\right )}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {1}{4} a x^3 \sqrt {1-a^2 x^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {\int \frac {x \left (12 x^2 a^4+15 x a^3+4 a^2\right )}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {1}{4} a x^3 \sqrt {1-a^2 x^2}\right )\)

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

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 223

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

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 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

rule 6698

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a^2*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 + 1)/2, 0] &&  !IntegerQ[p - n/2]
 

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {\left (2 a^{3} x^{3}+8 a^{2} x^{2}+15 a x +24\right ) \left (a^{2} x^{2}-1\right ) c}{8 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {15 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right ) c}{8 a \sqrt {a^{2}}}\)	\(82\)
meijerg	\(\frac {2 c \left (-2 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (-4 a^{2} x^{2}+8\right )}{4 \sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {\pi }}-\frac {c \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {\pi }}+\frac {c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {7}{2}} \left (-14 a^{4} x^{4}-35 a^{2} x^{2}+105\right )}{56 a^{6} \sqrt {-a^{2} x^{2}+1}}-\frac {15 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {7}{2}} \arcsin \left (a x \right )}{8 a^{7}}\right )}{a \sqrt {\pi }\, \sqrt {-a^{2}}}-\frac {2 c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {5}{2}} \left (-5 a^{2} x^{2}+15\right )}{10 a^{4} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \sqrt {\pi }\, \left (-a^{2}\right )^{\frac {5}{2}} \arcsin \left (a x \right )}{2 a^{5}}\right )}{a \sqrt {\pi }\, \sqrt {-a^{2}}}+\frac {3 c \left (\frac {8 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-2 a^{4} x^{4}-8 a^{2} x^{2}+16\right )}{6 \sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {\pi }}-\frac {3 c \left (\frac {\sqrt {\pi }\, x \left (-a^{2}\right )^{\frac {3}{2}}}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\sqrt {\pi }\, \left (-a^{2}\right )^{\frac {3}{2}} \arcsin \left (a x \right )}{a^{3}}\right )}{a \sqrt {\pi }\, \sqrt {-a^{2}}}\)	\(342\)
default	\(-c \left (a^{5} \left (-\frac {x^{5}}{4 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {-\frac {5 x^{3}}{8 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {5 \left (\frac {3 x}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}\right )}{4 a^{2}}}{a^{2}}\right )-\frac {1}{a^{2} \sqrt {-a^{2} x^{2}+1}}-3 a \left (\frac {x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}\right )-2 a^{2} \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )+2 a^{3} \left (-\frac {x^{3}}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {\frac {3 x}{2 a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {3 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}}{a^{2}}\right )+3 a^{4} \left (-\frac {x^{4}}{3 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {-\frac {4 x^{2}}{3 a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {8}{3 a^{4} \sqrt {-a^{2} x^{2}+1}}}{a^{2}}\right )\right )\)	\(368\)

Input:

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

Output:

1/8*(2*a^3*x^3+8*a^2*x^2+15*a*x+24)*(a^2*x^2-1)/a^2/(-a^2*x^2+1)^(1/2)*c+1 
5/8/a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))*c
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.72 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=-\frac {30 \, c \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (2 \, a^{3} c x^{3} + 8 \, a^{2} c x^{2} + 15 \, a c x + 24 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{8 \, a^{2}} \] Input:

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

Output:

-1/8*(30*c*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^3*c*x^3 + 8*a^2*c 
*x^2 + 15*a*c*x + 24*c)*sqrt(-a^2*x^2 + 1))/a^2
 

Sympy [A] (verification not implemented)

Time = 5.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.25 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=\begin {cases} \sqrt {- a^{2} x^{2} + 1} \left (- \frac {a c x^{3}}{4} - c x^{2} - \frac {15 c x}{8 a} - \frac {3 c}{a^{2}}\right ) + \frac {15 c \log {\left (- 2 a^{2} x + 2 \sqrt {- a^{2}} \sqrt {- a^{2} x^{2} + 1} \right )}}{8 a \sqrt {- a^{2}}} & \text {for}\: a^{2} \neq 0 \\\frac {a^{3} c x^{5}}{5} + \frac {3 a^{2} c x^{4}}{4} + a c x^{3} + \frac {c x^{2}}{2} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(-a**2*x**2 + 1)*(-a*c*x**3/4 - c*x**2 - 15*c*x/(8*a) - 3*c 
/a**2) + 15*c*log(-2*a**2*x + 2*sqrt(-a**2)*sqrt(-a**2*x**2 + 1))/(8*a*sqr 
t(-a**2)), Ne(a**2, 0)), (a**3*c*x**5/5 + 3*a**2*c*x**4/4 + a*c*x**3 + c*x 
**2/2, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.27 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=\frac {a^{3} c x^{5}}{4 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {a^{2} c x^{4}}{\sqrt {-a^{2} x^{2} + 1}} + \frac {13 \, a c x^{3}}{8 \, \sqrt {-a^{2} x^{2} + 1}} + \frac {2 \, c x^{2}}{\sqrt {-a^{2} x^{2} + 1}} - \frac {15 \, c x}{8 \, \sqrt {-a^{2} x^{2} + 1} a} + \frac {15 \, c \arcsin \left (a x\right )}{8 \, a^{2}} - \frac {3 \, c}{\sqrt {-a^{2} x^{2} + 1} a^{2}} \] Input:

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

Output:

1/4*a^3*c*x^5/sqrt(-a^2*x^2 + 1) + a^2*c*x^4/sqrt(-a^2*x^2 + 1) + 13/8*a*c 
*x^3/sqrt(-a^2*x^2 + 1) + 2*c*x^2/sqrt(-a^2*x^2 + 1) - 15/8*c*x/(sqrt(-a^2 
*x^2 + 1)*a) + 15/8*c*arcsin(a*x)/a^2 - 3*c/(sqrt(-a^2*x^2 + 1)*a^2)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=-\frac {1}{8} \, \sqrt {-a^{2} x^{2} + 1} {\left ({\left (2 \, {\left (a c x + 4 \, c\right )} x + \frac {15 \, c}{a}\right )} x + \frac {24 \, c}{a^{2}}\right )} + \frac {15 \, c \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{8 \, a {\left | a \right |}} \] Input:

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

Output:

-1/8*sqrt(-a^2*x^2 + 1)*((2*(a*c*x + 4*c)*x + 15*c/a)*x + 24*c/a^2) + 15/8 
*c*arcsin(a*x)*sgn(a)/(a*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=\frac {15\,c\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{8\,a\,\sqrt {-a^2}}-c\,x^2\,\sqrt {1-a^2\,x^2}-\frac {15\,c\,x\,\sqrt {1-a^2\,x^2}}{8\,a}-\frac {a\,c\,x^3\,\sqrt {1-a^2\,x^2}}{4}-\frac {3\,c\,\sqrt {1-a^2\,x^2}}{a^2} \] Input:

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

Output:

(15*c*asinh(x*(-a^2)^(1/2)))/(8*a*(-a^2)^(1/2)) - c*x^2*(1 - a^2*x^2)^(1/2 
) - (15*c*x*(1 - a^2*x^2)^(1/2))/(8*a) - (a*c*x^3*(1 - a^2*x^2)^(1/2))/4 - 
 (3*c*(1 - a^2*x^2)^(1/2))/a^2
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.81 \[ \int e^{3 \text {arctanh}(a x)} x \left (c-a^2 c x^2\right ) \, dx=\frac {c \left (15 \mathit {asin} \left (a x \right )-2 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-8 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-15 \sqrt {-a^{2} x^{2}+1}\, a x -24 \sqrt {-a^{2} x^{2}+1}+24\right )}{8 a^{2}} \] Input:

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

Output:

(c*(15*asin(a*x) - 2*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 8*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))/(8*a**2)