\(\int e^{-3 \text {arctanh}(a+b x)} x \, dx\) [884]

Optimal result

Integrand size = 12, antiderivative size = 119 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=\frac {(1+a) (1-a-b x)^{5/2}}{b^2 \sqrt {1+a+b x}}+\frac {3 (3+2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3+2 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{2 b^2}+\frac {3 (3+2 a) \arcsin (a+b x)}{2 b^2} \] Output:

(1+a)*(-b*x-a+1)^(5/2)/b^2/(b*x+a+1)^(1/2)+3/2*(3+2*a)*(-b*x-a+1)^(1/2)*(b 
*x+a+1)^(1/2)/b^2+1/2*(3+2*a)*(-b*x-a+1)^(3/2)*(b*x+a+1)^(1/2)/b^2+3/2*(3+ 
2*a)*arcsin(b*x+a)/b^2
 

Mathematica [A] (warning: unable to verify)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.32 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=\frac {\sqrt {-b} \left (14-a^3-9 b x-6 b^2 x^2+b^3 x^3-a^2 (14+b x)+a \left (1-20 b x+b^2 x^2\right )\right )+6 (3+2 a) \sqrt {b} \sqrt {1-a^2-2 a b x-b^2 x^2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{2 (-b)^{5/2} \sqrt {-((-1+a+b x) (1+a+b x))}} \] Input:

Integrate[x/E^(3*ArcTanh[a + b*x]),x]
 

Output:

(Sqrt[-b]*(14 - a^3 - 9*b*x - 6*b^2*x^2 + b^3*x^3 - a^2*(14 + b*x) + a*(1 
- 20*b*x + b^2*x^2)) + 6*(3 + 2*a)*Sqrt[b]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^ 
2]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(2*(-b)^(5/2)* 
Sqrt[-((-1 + a + b*x)*(1 + a + b*x))])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6713, 87, 60, 60, 62, 1090, 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.

\(\displaystyle \int x e^{-3 \text {arctanh}(a+b x)} \, dx\)

\(\Big \downarrow \) 6713

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 223

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

Input:

Int[x/E^(3*ArcTanh[a + b*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 60

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

rule 62

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

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1090

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

rule 6713

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.) 
, x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( 
n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.66

method	result	size
risch	\(-\frac {\left (-b x +a +6\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {-\frac {\left (-8 a -8\right ) \sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}{b^{2} \left (x +\frac {a +1}{b}\right )}+\frac {9 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {6 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}}{2 b}\)	\(198\)
default	\(\frac {\frac {\left (-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b \right )^{\frac {5}{2}}}{b \left (x +\frac {a +1}{b}\right )^{2}}+3 b \left (\frac {\left (-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b \right )^{\frac {3}{2}}}{3}+b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a +1}{b}\right )+2 b \right ) \sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}{4 b^{2}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a +1}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}\right )}{2 \sqrt {b^{2}}}\right )\right )}{b^{3}}-\frac {\left (a +1\right ) \left (-\frac {\left (-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b \right )^{\frac {5}{2}}}{b \left (x +\frac {a +1}{b}\right )^{3}}-2 b \left (\frac {\left (-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b \right )^{\frac {5}{2}}}{b \left (x +\frac {a +1}{b}\right )^{2}}+3 b \left (\frac {\left (-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b \right )^{\frac {3}{2}}}{3}+b \left (-\frac {\left (-2 b^{2} \left (x +\frac {a +1}{b}\right )+2 b \right ) \sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}{4 b^{2}}+\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a +1}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}\right )}{2 \sqrt {b^{2}}}\right )\right )\right )\right )}{b^{4}}\)	\(465\)

Input:

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

Output:

-1/2*(-b*x+a+6)*(b^2*x^2+2*a*b*x+a^2-1)/b^2/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2) 
+1/2/b*(-(-8*a-8)/b^2/(x+(a+1)/b)*(-(x+(a+1)/b)^2*b^2+2*(x+(a+1)/b)*b)^(1/ 
2)+9/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+1/b*a)/(-b^2*x^2-2*a*b*x-a^2+1)^(1/ 
2))+6*a/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+1/b*a)/(-b^2*x^2-2*a*b*x-a^2+1)^ 
(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=-\frac {3 \, {\left ({\left (2 \, a + 3\right )} b x + 2 \, a^{2} + 5 \, a + 3\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (b^{2} x^{2} - a^{2} - 5 \, b x - 15 \, a - 14\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left (b^{3} x + {\left (a + 1\right )} b^{2}\right )}} \] Input:

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

Output:

-1/2*(3*((2*a + 3)*b*x + 2*a^2 + 5*a + 3)*arctan(sqrt(-b^2*x^2 - 2*a*b*x - 
 a^2 + 1)*(b*x + a)/(b^2*x^2 + 2*a*b*x + a^2 - 1)) + (b^2*x^2 - a^2 - 5*b* 
x - 15*a - 14)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^3*x + (a + 1)*b^2)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (101) = 202\).

Time = 0.11 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.51 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=-\frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} + 2 \, b^{3} x + 2 \, a b^{2} + b^{2}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{b^{4} x^{2} + 2 \, a b^{3} x + a^{2} b^{2} + 2 \, b^{3} x + 2 \, a b^{2} + b^{2}} + \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{2 \, {\left (b^{3} x + a b^{2} + b^{2}\right )}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{3} x + a b^{2} + b^{2}} + \frac {3 \, a \arcsin \left (b x + a\right )}{b^{2}} + \frac {6 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{3} x + a b^{2} + b^{2}} + \frac {9 \, \arcsin \left (b x + a\right )}{2 \, b^{2}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, b^{2}} \] Input:

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

Output:

-(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a/(b^4*x^2 + 2*a*b^3*x + a^2*b^2 + 2 
*b^3*x + 2*a*b^2 + b^2) - (-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)/(b^4*x^2 + 
2*a*b^3*x + a^2*b^2 + 2*b^3*x + 2*a*b^2 + b^2) + 1/2*(-b^2*x^2 - 2*a*b*x - 
 a^2 + 1)^(3/2)/(b^3*x + a*b^2 + b^2) + 6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 
1)*a/(b^3*x + a*b^2 + b^2) + 3*a*arcsin(b*x + a)/b^2 + 6*sqrt(-b^2*x^2 - 2 
*a*b*x - a^2 + 1)/(b^3*x + a*b^2 + b^2) + 9/2*arcsin(b*x + a)/b^2 + 3/2*sq 
rt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b^2
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=-\frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\frac {x}{b} - \frac {a b^{2} + 6 \, b^{2}}{b^{4}}\right )} - \frac {3 \, {\left (2 \, a + 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} - \frac {8 \, {\left (a + 1\right )}}{b {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} + 1\right )} {\left | b \right |}} \] Input:

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

Output:

-1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x/b - (a*b^2 + 6*b^2)/b^4) - 3/2* 
(2*a + 3)*arcsin(-b*x - a)*sgn(b)/(b*abs(b)) - 8*(a + 1)/(b*((sqrt(-b^2*x^ 
2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) + 1)*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.73 \[ \int e^{-3 \text {arctanh}(a+b x)} x \, dx=\frac {6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right ) a +9 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right )-6 \mathit {asin} \left (b x +a \right ) a^{2}-6 \mathit {asin} \left (b x +a \right ) a b x -15 \mathit {asin} \left (b x +a \right ) a -9 \mathit {asin} \left (b x +a \right ) b x -9 \mathit {asin} \left (b x +a \right )-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}-21 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{2} x^{2}-5 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -18 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-a^{3}-a^{2} b x -8 a^{2}+a \,b^{2} x^{2}-14 a b x +11 a +b^{3} x^{3}-6 b^{2} x^{2}-5 b x +18}{2 b^{2} \left (\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-a -b x -1\right )} \] Input:

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

Output:

(6*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)*a + 9*sqrt( - a** 
2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x) - 6*asin(a + b*x)*a**2 - 6*asin 
(a + b*x)*a*b*x - 15*asin(a + b*x)*a - 9*asin(a + b*x)*b*x - 9*asin(a + b* 
x) - sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a**2 - 21*sqrt( - a**2 - 2*a* 
b*x - b**2*x**2 + 1)*a + sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*b**2*x**2 
 - 5*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*b*x - 18*sqrt( - a**2 - 2*a*b 
*x - b**2*x**2 + 1) - a**3 - a**2*b*x - 8*a**2 + a*b**2*x**2 - 14*a*b*x + 
11*a + b**3*x**3 - 6*b**2*x**2 - 5*b*x + 18)/(2*b**2*(sqrt( - a**2 - 2*a*b 
*x - b**2*x**2 + 1) - a - b*x - 1))