\(\int e^{\text {arcsinh}(a+b x)} x^3 \, dx\) [1]

Optimal result

Integrand size = 12, antiderivative size = 165 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {e^{-3 \text {arcsinh}(a+b x)}}{48 b^4}+\frac {3 a e^{-2 \text {arcsinh}(a+b x)}}{16 b^4}-\frac {\left (1-6 a^2\right ) e^{-\text {arcsinh}(a+b x)}}{8 b^4}+\frac {a \left (3-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}}{16 b^4}-\frac {\left (1-6 a^2\right ) e^{3 \text {arcsinh}(a+b x)}}{24 b^4}-\frac {3 a e^{4 \text {arcsinh}(a+b x)}}{32 b^4}+\frac {e^{5 \text {arcsinh}(a+b x)}}{80 b^4}+\frac {a \left (3-4 a^2\right ) \text {arcsinh}(a+b x)}{8 b^4} \] Output:

1/48/b^4/(b*x+a+(1+(b*x+a)^2)^(1/2))^3+3/16*a/b^4/(b*x+a+(1+(b*x+a)^2)^(1/ 
2))^2-1/8*(-6*a^2+1)/b^4/(b*x+a+(1+(b*x+a)^2)^(1/2))+1/16*a*(-4*a^2+3)*(b* 
x+a+(1+(b*x+a)^2)^(1/2))^2/b^4-1/24*(-6*a^2+1)*(b*x+a+(1+(b*x+a)^2)^(1/2)) 
^3/b^4-3/32*a*(b*x+a+(1+(b*x+a)^2)^(1/2))^4/b^4+1/80*(b*x+a+(1+(b*x+a)^2)^ 
(1/2))^5/b^4+1/8*a*(-4*a^2+3)*arcsinh(b*x+a)/b^4
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.72 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {30 a b^4 x^4+24 b^5 x^5-\sqrt {1+a^2+2 a b x+b^2 x^2} \left (16-83 a^2+6 a^4+a \left (29-6 a^2\right ) b x+2 \left (-4+3 a^2\right ) b^2 x^2-6 a b^3 x^3-24 b^4 x^4\right )+15 a \left (3-4 a^2\right ) \text {arcsinh}(a+b x)}{120 b^4} \] Input:

Integrate[E^ArcSinh[a + b*x]*x^3,x]
 

Output:

(30*a*b^4*x^4 + 24*b^5*x^5 - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(16 - 83*a^ 
2 + 6*a^4 + a*(29 - 6*a^2)*b*x + 2*(-4 + 3*a^2)*b^2*x^2 - 6*a*b^3*x^3 - 24 
*b^4*x^4) + 15*a*(3 - 4*a^2)*ArcSinh[a + b*x])/(120*b^4)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6288, 25, 2720, 27, 2159, 2009}

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.

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

\(\Big \downarrow \) 6288

\(\displaystyle \frac {\int -e^{\text {arcsinh}(a+b x)} \left (\frac {a}{b}-\frac {a+b x}{b}\right )^3 \sqrt {(a+b x)^2+1}d\text {arcsinh}(a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int e^{\text {arcsinh}(a+b x)} \left (\frac {a}{b}-\frac {a+b x}{b}\right )^3 \sqrt {(a+b x)^2+1}d\text {arcsinh}(a+b x)}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {\int \frac {e^{-4 \text {arcsinh}(a+b x)} \left (2 e^{\text {arcsinh}(a+b x)} a-e^{2 \text {arcsinh}(a+b x)}+1\right )^3 \left (1+e^{2 \text {arcsinh}(a+b x)}\right )}{16 b^3}de^{\text {arcsinh}(a+b x)}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int e^{-4 \text {arcsinh}(a+b x)} \left (2 e^{\text {arcsinh}(a+b x)} a-e^{2 \text {arcsinh}(a+b x)}+1\right )^3 \left (1+e^{2 \text {arcsinh}(a+b x)}\right )de^{\text {arcsinh}(a+b x)}}{16 b^4}\)

\(\Big \downarrow \) 2159

\(\displaystyle -\frac {\int \left (6 e^{-3 \text {arcsinh}(a+b x)} a+2 \left (4 a^2-3\right ) e^{-\text {arcsinh}(a+b x)} a+2 \left (4 a^2-3\right ) e^{\text {arcsinh}(a+b x)} a+6 e^{3 \text {arcsinh}(a+b x)} a+e^{-4 \text {arcsinh}(a+b x)}+2 \left (6 a^2-1\right ) e^{-2 \text {arcsinh}(a+b x)}-2 \left (6 a^2-1\right ) e^{2 \text {arcsinh}(a+b x)}-e^{4 \text {arcsinh}(a+b x)}\right )de^{\text {arcsinh}(a+b x)}}{16 b^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\left (3-4 a^2\right ) a e^{2 \text {arcsinh}(a+b x)}+2 \left (1-6 a^2\right ) e^{-\text {arcsinh}(a+b x)}+\frac {2}{3} \left (1-6 a^2\right ) e^{3 \text {arcsinh}(a+b x)}-2 \left (3-4 a^2\right ) a \log \left (e^{\text {arcsinh}(a+b x)}\right )-3 a e^{-2 \text {arcsinh}(a+b x)}+\frac {3}{2} a e^{4 \text {arcsinh}(a+b x)}-\frac {1}{3} e^{-3 \text {arcsinh}(a+b x)}-\frac {1}{5} e^{5 \text {arcsinh}(a+b x)}}{16 b^4}\)

Input:

Int[E^ArcSinh[a + b*x]*x^3,x]
 

Output:

-1/16*(-1/3*1/E^(3*ArcSinh[a + b*x]) - (3*a)/E^(2*ArcSinh[a + b*x]) + (2*( 
1 - 6*a^2))/E^ArcSinh[a + b*x] - a*(3 - 4*a^2)*E^(2*ArcSinh[a + b*x]) + (2 
*(1 - 6*a^2)*E^(3*ArcSinh[a + b*x]))/3 + (3*a*E^(4*ArcSinh[a + b*x]))/2 - 
E^(5*ArcSinh[a + b*x])/5 - 2*a*(3 - 4*a^2)*Log[E^ArcSinh[a + b*x]])/b^4
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2159

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x 
], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 6288

Int[(f_)^(ArcSinh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.))*(x_)^(m_.), x_Symbol] :> 
 Simp[1/b   Subst[Int[(-a/b + Sinh[x]/b)^m*f^(c*x^n)*Cosh[x], x], x, ArcSin 
h[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(463\) vs. \(2(205)=410\).

Time = 0.21 (sec) , antiderivative size = 464, normalized size of antiderivative = 2.81

method	result	size
default	\(\frac {x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{5 b^{2}}-\frac {7 a \left (\frac {x \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{4 b^{2}}-\frac {5 a \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )}{4 b}-\frac {\left (a^{2}+1\right ) \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{4 b^{2}}\right )}{5 b}-\frac {2 \left (a^{2}+1\right ) \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )}{5 b^{2}}+\frac {b \,x^{5}}{5}+\frac {a \,x^{4}}{4}\)	\(464\)

Input:

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

Output:

1/5*x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/b^2-7/5*a/b*(1/4*x*(b^2*x^2+2*a*b*x+ 
a^2+1)^(3/2)/b^2-5/4*a/b*(1/3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/b^2-a/b*(1/4*( 
2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/8*(4*b^2*(a^2+1)-4*a^2* 
b^2)/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^( 
1/2)))-1/4*(a^2+1)/b^2*(1/4*(2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1 
/2)+1/8*(4*b^2*(a^2+1)-4*a^2*b^2)/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+ 
2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)))-2/5*(a^2+1)/b^2*(1/3*(b^2*x^2+2*a*b*x+ 
a^2+1)^(3/2)/b^2-a/b*(1/4*(2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2 
)+1/8*(4*b^2*(a^2+1)-4*a^2*b^2)/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2* 
a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)))+1/5*b*x^5+1/4*a*x^4
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.84 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {24 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 15 \, {\left (4 \, a^{3} - 3 \, a\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (24 \, b^{4} x^{4} + 6 \, a b^{3} x^{3} - 2 \, {\left (3 \, a^{2} - 4\right )} b^{2} x^{2} - 6 \, a^{4} + {\left (6 \, a^{3} - 29 \, a\right )} b x + 83 \, a^{2} - 16\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{120 \, b^{4}} \] Input:

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

Output:

1/120*(24*b^5*x^5 + 30*a*b^4*x^4 + 15*(4*a^3 - 3*a)*log(-b*x - a + sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1)) + (24*b^4*x^4 + 6*a*b^3*x^3 - 2*(3*a^2 - 4)*b^ 
2*x^2 - 6*a^4 + (6*a^3 - 29*a)*b*x + 83*a^2 - 16)*sqrt(b^2*x^2 + 2*a*b*x + 
 a^2 + 1))/b^4
 

Sympy [A] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.76 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {a x^{4}}{4} + \frac {b x^{5}}{5} + \begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (- \frac {5 a \left (\frac {1}{5} - \frac {3 a^{2}}{20}\right )}{3 b} - \frac {a \left (3 a^{2} + 3\right )}{20 b}\right )}{2 b} - \frac {\left (\frac {1}{5} - \frac {3 a^{2}}{20}\right ) \left (2 a^{2} + 2\right )}{3 b^{2}}\right )}{b} - \frac {\left (a^{2} + 1\right ) \left (- \frac {5 a \left (\frac {1}{5} - \frac {3 a^{2}}{20}\right )}{3 b} - \frac {a \left (3 a^{2} + 3\right )}{20 b}\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \left (\frac {a x^{3}}{20 b} + \frac {x^{4}}{5} + \frac {x^{2} \cdot \left (\frac {1}{5} - \frac {3 a^{2}}{20}\right )}{3 b^{2}} + \frac {x \left (- \frac {5 a \left (\frac {1}{5} - \frac {3 a^{2}}{20}\right )}{3 b} - \frac {a \left (3 a^{2} + 3\right )}{20 b}\right )}{2 b^{2}} + \frac {- \frac {3 a \left (- \frac {5 a \left (\frac {1}{5} - \frac {3 a^{2}}{20}\right )}{3 b} - \frac {a \left (3 a^{2} + 3\right )}{20 b}\right )}{2 b} - \frac {\left (\frac {1}{5} - \frac {3 a^{2}}{20}\right ) \left (2 a^{2} + 2\right )}{3 b^{2}}}{b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {\left (- 3 a^{2} - 3\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}} \cdot \left (3 a^{4} + 6 a^{2} + 3\right )}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \left (- a^{6} - 3 a^{4} - 3 a^{2} - 1\right )}{3}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2} + 1}}{4} & \text {otherwise} \end {cases} \] Input:

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

Output:

a*x**4/4 + b*x**5/5 + Piecewise(((-a*(-3*a*(-5*a*(1/5 - 3*a**2/20)/(3*b) - 
 a*(3*a**2 + 3)/(20*b))/(2*b) - (1/5 - 3*a**2/20)*(2*a**2 + 2)/(3*b**2))/b 
 - (a**2 + 1)*(-5*a*(1/5 - 3*a**2/20)/(3*b) - a*(3*a**2 + 3)/(20*b))/(2*b* 
*2))*log(2*a*b + 2*b**2*x + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*sqrt(b* 
*2))/sqrt(b**2) + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*(a*x**3/(20*b) + x* 
*4/5 + x**2*(1/5 - 3*a**2/20)/(3*b**2) + x*(-5*a*(1/5 - 3*a**2/20)/(3*b) - 
 a*(3*a**2 + 3)/(20*b))/(2*b**2) + (-3*a*(-5*a*(1/5 - 3*a**2/20)/(3*b) - a 
*(3*a**2 + 3)/(20*b))/(2*b) - (1/5 - 3*a**2/20)*(2*a**2 + 2)/(3*b**2))/b** 
2), Ne(b**2, 0)), (((-3*a**2 - 3)*(a**2 + 2*a*b*x + 1)**(7/2)/7 + (a**2 + 
2*a*b*x + 1)**(9/2)/9 + (a**2 + 2*a*b*x + 1)**(5/2)*(3*a**4 + 6*a**2 + 3)/ 
5 + (a**2 + 2*a*b*x + 1)**(3/2)*(-a**6 - 3*a**4 - 3*a**2 - 1)/3)/(8*a**4*b 
**4), Ne(a*b, 0)), (x**4*sqrt(a**2 + 1)/4, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (205) = 410\).

Time = 0.11 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.98 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{5 \, b^{2}} - \frac {{\left (a^{2} + 1\right )} a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a x}{20 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a x}{5 \, b^{3}} + \frac {{\left (a^{2} + 1\right )}^{2} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{5 \, b^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{12 \, b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a^{2}}{5 \, b^{4}} + \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{40 \, b^{6}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} {\left (a^{2} + 1\right )}}{15 \, b^{4}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{40 \, b^{5}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{40 \, b^{6}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{40 \, b^{6}} \] Input:

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

Output:

1/5*b*x^5 + 1/4*a*x^4 + 1/5*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x^2/b^2 - 
1/5*(a^2 + 1)*a^3*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^ 
2))/b^4 - 7/20*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a*x/b^3 + 1/5*sqrt(b^2* 
x^2 + 2*a*b*x + a^2 + 1)*(a^2 + 1)*a*x/b^3 + 1/5*(a^2 + 1)^2*a*arcsinh(2*( 
b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^4 + 7/12*(b^2*x^2 + 2*a 
*b*x + a^2 + 1)^(3/2)*a^2/b^4 + 1/5*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 
 + 1)*a^2/b^4 + 7/40*(5*a^2*b^2 - (a^2 + 1)*b^2)*a^3*arcsinh(2*(b^2*x + a* 
b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^6 - 2/15*(b^2*x^2 + 2*a*b*x + a^2 
 + 1)^(3/2)*(a^2 + 1)/b^4 - 7/40*(5*a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)*a*x/b^5 - 7/40*(5*a^2*b^2 - (a^2 + 1)*b^2)*(a^2 + 1)* 
a*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^6 - 7/40*( 
5*a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2/b^6
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.05 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} + \frac {1}{120} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, {\left (3 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2} b^{5} - 4 \, b^{5}}{b^{7}}\right )} x + \frac {6 \, a^{3} b^{4} - 29 \, a b^{4}}{b^{7}}\right )} x - \frac {6 \, a^{4} b^{3} - 83 \, a^{2} b^{3} + 16 \, b^{3}}{b^{7}}\right )} + \frac {{\left (4 \, a^{3} - 3 \, a\right )} \log \left ({\left | -a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |} \right |}\right )}{8 \, b^{3} {\left | b \right |}} \] Input:

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

Output:

1/5*b*x^5 + 1/4*a*x^4 + 1/120*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*(3*(4* 
x + a/b)*x - (3*a^2*b^5 - 4*b^5)/b^7)*x + (6*a^3*b^4 - 29*a*b^4)/b^7)*x - 
(6*a^4*b^3 - 83*a^2*b^3 + 16*b^3)/b^7) + 1/8*(4*a^3 - 3*a)*log(abs(-a*b - 
(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b)))/(b^3*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.82 \[ \int e^{\text {arcsinh}(a+b x)} x^3 \, dx=\frac {-6 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{4}+6 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{3} b x -6 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{2} b^{2} x^{2}+83 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{2}+6 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a \,b^{3} x^{3}-29 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a b x +24 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{4} x^{4}+8 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{2} x^{2}-16 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}-60 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{3}+45 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a +30 a \,b^{4} x^{4}+24 b^{5} x^{5}}{120 b^{4}} \] Input:

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

Output:

( - 6*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**4 + 6*sqrt(a**2 + 2*a*b*x + 
b**2*x**2 + 1)*a**3*b*x - 6*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2*b**2 
*x**2 + 83*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**2 + 6*sqrt(a**2 + 2*a*b 
*x + b**2*x**2 + 1)*a*b**3*x**3 - 29*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)* 
a*b*x + 24*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*b**4*x**4 + 8*sqrt(a**2 + 
2*a*b*x + b**2*x**2 + 1)*b**2*x**2 - 16*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 
1) - 60*log(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a**3 + 45*log( 
sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a + 30*a*b**4*x**4 + 24*b* 
*5*x**5)/(120*b**4)