\(\int e^{3 \text {arctanh}(a+b x)} x \, dx\) [874]

Optimal result

Integrand size = 12, antiderivative size = 121 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=\frac {3 (3-2 a) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^2}+\frac {(3-2 a) \sqrt {1-a-b x} (1+a+b x)^{3/2}}{2 b^2}+\frac {(1-a) (1+a+b x)^{5/2}}{b^2 \sqrt {1-a-b x}}-\frac {3 (3-2 a) \arcsin (a+b x)}{2 b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.17 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=\frac {\frac {\sqrt {b} \sqrt {1+a+b x} \left (14-15 a+a^2-5 b x-b^2 x^2\right )}{\sqrt {1-a-b x}}+12 a \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )+18 \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{2 b^{5/2}} \] Input:

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

Output:

((Sqrt[b]*Sqrt[1 + a + b*x]*(14 - 15*a + a^2 - 5*b*x - b^2*x^2))/Sqrt[1 - 
a - b*x] + 12*a*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqr 
t[b])] + 18*Sqrt[-b]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b] 
)])/(2*b^(5/2))
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.08, 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

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

((1 - a)*(1 + a + b*x)^(5/2))/(b^2*Sqrt[1 - a - b*x]) - ((3 - 2*a)*(-1/2*( 
Sqrt[1 - a - b*x]*(1 + a + b*x)^(3/2))/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 = 0.47 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.63

method	result	size
risch	\(\frac {\left (-b x +a -6\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{2 b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {\frac {\left (8 a -8\right ) \sqrt {-\left (x +\frac {-1+a}{b}\right )^{2} b^{2}-2 \left (x +\frac {-1+a}{b}\right ) b}}{b^{2} \left (x +\frac {-1+a}{b}\right )}-\frac {9 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {6 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}}{2 b}\)	\(197\)
default	\(\text {Expression too large to display}\)	\(1118\)

Input:

int((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)*x,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+(-1+a)/b)*(-(x+(-1+a)/b)^2*b^2-2*(x+(-1+a)/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 = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.08 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=-\frac {3 \, {\left ({\left (2 \, a - 3\right )} b x + 2 \, a^{2} - 5 \, a + 3\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (b^{2} x^{2} - a^{2} + 5 \, b x + 15 \, a - 14\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left (b^{3} x + {\left (a - 1\right )} b^{2}\right )}} \] Input:

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

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1137 vs. \(2 (102) = 204\).

Time = 0.12 (sec) , antiderivative size = 1137, normalized size of antiderivative = 9.40 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=\text {Too large to display} \] Input:

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

Output:

15*a^4*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) 
- 31/2*(a^2 - 1)*a^2*b*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b* 
x - a^2 + 1)) - 1/2*b*x^3/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) + 5/2*(a^2 - 
1)*a^3/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 6* 
(a^2*b + 2*a*b + b)*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b 
*x - a^2 + 1)) - 18*(a*b^2 + b^2)*a^3*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b 
^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 3/2*(a^2 - 1)^2*b*x/((a^2*b^2 - (a^2 - 1) 
*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - (a^3 + 3*a^2 + 3*a + 1)*a*b*x/ 
((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 5/2*a*x^2 
/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) - 3/2*(a^2 - 1)^2*a/((a^2*b^2 - (a^2 - 
 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - (a^3 + 3*a^2 + 3*a + 1)*a^2 
/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 3*(a^2*b 
 + 2*a*b + b)*(a^2 - 1)*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b 
*x - a^2 + 1)) + 15*(a*b^2 + b^2)*(a^2 - 1)*a*x/((a^2*b^2 - (a^2 - 1)*b^2) 
*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 15/2*a^2*arcsin(-(b^2*x + a*b)/sq 
rt(a^2*b^2 - (a^2 - 1)*b^2))/b^2 - 3*(a*b^2 + b^2)*(a^2 - 1)*a^2/((a^2*b^2 
 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) + 3*(a^2*b + 2*a 
*b + b)*(a^2 - 1)*a/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a 
^2 + 1)*b) - 3*(a*b^2 + b^2)*x^2/(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^2) 
- 3/2*(a^2 - 1)*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 - (a^2 - 1)*b^2))/b^...
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=\frac {1}{2} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\frac {x}{b} - \frac {a b^{2} - 6 \, b^{2}}{b^{4}}\right )} - \frac {3 \, {\left (2 \, a - 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b {\left | b \right |}} - \frac {8 \, {\left (a - 1\right )}}{b {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \] Input:

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.65 \[ \int e^{3 \text {arctanh}(a+b x)} x \, dx=\frac {6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right ) a -9 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right )+6 \mathit {asin} \left (b x +a \right ) a^{2}+6 \mathit {asin} \left (b x +a \right ) a b x -15 \mathit {asin} \left (b x +a \right ) a -9 \mathit {asin} \left (b x +a \right ) b x +9 \mathit {asin} \left (b x +a \right )-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}+21 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{2} x^{2}+5 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -18 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+a^{3}+a^{2} b x -8 a^{2}-a \,b^{2} x^{2}-14 a b x -11 a -b^{3} x^{3}-6 b^{2} x^{2}+5 b x +18}{2 b^{2} \left (\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+a +b x -1\right )} \] Input:

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