\(\int e^{n \text {arctanh}(a+b x)} x^3 \, dx\) [891]

Optimal result

Integrand size = 14, antiderivative size = 245 \[ \int e^{n \text {arctanh}(a+b x)} x^3 \, dx=-\frac {\left (6+30 a^2+2 n+n^2-12 a (1+n)\right ) (1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {2+n}{2}}}{24 b^4}-\frac {x^2 (1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {2+n}{2}}}{4 b^2}-\frac {(6 a-n) (1-a-b x)^{2-\frac {n}{2}} (1+a+b x)^{\frac {2+n}{2}}}{12 b^4}+\frac {2^{-2+\frac {n}{2}} \left (24 a^3-36 a^2 n+12 a \left (2+n^2\right )-n \left (8+n^2\right )\right ) (1-a-b x)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a-b x)\right )}{3 b^4 (2-n)} \] Output:

-1/24*(6+30*a^2+2*n+n^2-12*a*(1+n))*(-b*x-a+1)^(1-1/2*n)*(b*x+a+1)^(1+1/2* 
n)/b^4-1/4*x^2*(-b*x-a+1)^(1-1/2*n)*(b*x+a+1)^(1+1/2*n)/b^2-1/12*(6*a-n)*( 
-b*x-a+1)^(2-1/2*n)*(b*x+a+1)^(1+1/2*n)/b^4+1/3*2^(-2+1/2*n)*(24*a^3-36*a^ 
2*n+12*a*(n^2+2)-n*(n^2+8))*(-b*x-a+1)^(1-1/2*n)*hypergeom([-1/2*n, 1-1/2* 
n],[2-1/2*n],-1/2*b*x-1/2*a+1/2)/b^4/(2-n)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.90 \[ \int e^{n \text {arctanh}(a+b x)} x^3 \, dx=\frac {(1-a-b x)^{1-\frac {n}{2}} \left (b^2 (-2+n) x^2 (1+a+b x)^{1+\frac {n}{2}}-2^{3+\frac {n}{2}} (-6 a+n) \operatorname {Hypergeometric2F1}\left (-2-\frac {n}{2},1-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a-b x)\right )-2^{3+\frac {n}{2}} (1+a) (1+5 a-n) \operatorname {Hypergeometric2F1}\left (-1-\frac {n}{2},1-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a-b x)\right )+2^{1+\frac {n}{2}} (1+a)^2 (2+4 a-n) \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (1-a-b x)\right )\right )}{4 b^4 (2-n)} \] Input:

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

Output:

((1 - a - b*x)^(1 - n/2)*(b^2*(-2 + n)*x^2*(1 + a + b*x)^(1 + n/2) - 2^(3 
+ n/2)*(-6*a + n)*Hypergeometric2F1[-2 - n/2, 1 - n/2, 2 - n/2, (1 - a - b 
*x)/2] - 2^(3 + n/2)*(1 + a)*(1 + 5*a - n)*Hypergeometric2F1[-1 - n/2, 1 - 
 n/2, 2 - n/2, (1 - a - b*x)/2] + 2^(1 + n/2)*(1 + a)^2*(2 + 4*a - n)*Hype 
rgeometric2F1[1 - n/2, -1/2*n, 2 - n/2, (1 - a - b*x)/2]))/(4*b^4*(2 - n))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6713, 111, 25, 164, 79}

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.

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

\(\Big \downarrow \) 6713

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

\(\Big \downarrow \) 111

\(\displaystyle -\frac {\int -x (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \left (2 \left (1-a^2\right )-b (6 a-n) x\right )dx}{4 b^2}-\frac {x^2 (a+b x+1)^{\frac {n+2}{2}} (-a-b x+1)^{1-\frac {n}{2}}}{4 b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int x (-a-b x+1)^{-n/2} (a+b x+1)^{n/2} \left (2 \left (1-a^2\right )-b (6 a-n) x\right )dx}{4 b^2}-\frac {x^2 (-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}}}{4 b^2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {-\frac {\left (24 a^3-36 a^2 n+12 a \left (n^2+2\right )-n \left (n^2+8\right )\right ) \int (-a-b x+1)^{-n/2} (a+b x+1)^{n/2}dx}{6 b}-\frac {(a+b x+1)^{\frac {n+2}{2}} \left (18 a^2-2 b x (6 a-n)-10 a n+n^2+6\right ) (-a-b x+1)^{1-\frac {n}{2}}}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}}}{4 b^2}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\frac {2^{n/2} \left (24 a^3-36 a^2 n+12 a \left (n^2+2\right )-n \left (n^2+8\right )\right ) (-a-b x+1)^{1-\frac {n}{2}} \operatorname {Hypergeometric2F1}\left (1-\frac {n}{2},-\frac {n}{2},2-\frac {n}{2},\frac {1}{2} (-a-b x+1)\right )}{3 b^2 (2-n)}-\frac {(-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}} \left (18 a^2-2 b x (6 a-n)-10 a n+n^2+6\right )}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n+2}{2}}}{4 b^2}\)

Input:

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

Output:

-1/4*(x^2*(1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2))/b^2 + (-1/6*( 
(1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((2 + n)/2)*(6 + 18*a^2 - 10*a*n + n 
^2 - 2*b*(6*a - n)*x))/b^2 + (2^(n/2)*(24*a^3 - 36*a^2*n + 12*a*(2 + n^2) 
- n*(8 + n^2))*(1 - a - b*x)^(1 - n/2)*Hypergeometric2F1[1 - n/2, -1/2*n, 
2 - n/2, (1 - a - b*x)/2])/(3*b^2*(2 - n)))/(4*b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 111

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
 

rule 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 [F]

\[\int {\mathrm e}^{n \,\operatorname {arctanh}\left (b x +a \right )} x^{3}d x\]

Input:

int(exp(n*arctanh(b*x+a))*x^3,x)
 

Output:

int(exp(n*arctanh(b*x+a))*x^3,x)
 

Fricas [F]

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

integrate(exp(n*arctanh(b*x+a))*x^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int e^{n \text {arctanh}(a+b x)} x^3 \, dx=\int x^{3} e^{n \operatorname {atanh}{\left (a + b x \right )}}\, dx \] Input:

integrate(exp(n*atanh(b*x+a))*x**3,x)
 

Output:

Integral(x**3*exp(n*atanh(a + b*x)), x)
 

Maxima [F]

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

integrate(exp(n*arctanh(b*x+a))*x^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(exp(n*arctanh(b*x+a))*x^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int e^{n \text {arctanh}(a+b x)} x^3 \, dx=\int x^3\,{\mathrm {e}}^{n\,\mathrm {atanh}\left (a+b\,x\right )} \,d x \] Input:

int(x^3*exp(n*atanh(a + b*x)),x)
 

Output:

int(x^3*exp(n*atanh(a + b*x)), x)
 

Reduce [F]

\[ \int e^{n \text {arctanh}(a+b x)} x^3 \, dx=\int e^{\mathit {atanh} \left (b x +a \right ) n} x^{3}d x \] Input:

int(exp(n*atanh(b*x+a))*x^3,x)
 

Output:

int(e**(atanh(a + b*x)*n)*x**3,x)