\(\int e^{\text {arcsinh}(a+b x)} x^2 \, dx\) [2]

Optimal result

Integrand size = 12, antiderivative size = 115 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=-\frac {e^{-2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{-\text {arcsinh}(a+b x)}}{2 b^3}-\frac {\left (1-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}}{16 b^3}-\frac {a e^{3 \text {arcsinh}(a+b x)}}{6 b^3}+\frac {e^{4 \text {arcsinh}(a+b x)}}{32 b^3}-\frac {\left (1-4 a^2\right ) \text {arcsinh}(a+b x)}{8 b^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {8 a b^3 x^3+6 b^4 x^4+\sqrt {1+a^2+2 a b x+b^2 x^2} \left (2 a^3+3 b x-2 a^2 b x+6 b^3 x^3+a \left (-13+2 b^2 x^2\right )\right )+3 (-1+2 a) (1+2 a) \text {arcsinh}(a+b x)}{24 b^3} \] Input:

Integrate[E^ArcSinh[a + b*x]*x^2,x]
 

Output:

(8*a*b^3*x^3 + 6*b^4*x^4 + Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(2*a^3 + 3*b* 
x - 2*a^2*b*x + 6*b^3*x^3 + a*(-13 + 2*b^2*x^2)) + 3*(-1 + 2*a)*(1 + 2*a)* 
ArcSinh[a + b*x])/(24*b^3)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6288, 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.

\(\displaystyle \int x^2 e^{\text {arcsinh}(a+b x)} \, dx\)

\(\Big \downarrow \) 6288

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

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int \frac {e^{-3 \text {arcsinh}(a+b x)} \left (2 e^{\text {arcsinh}(a+b x)} a-e^{2 \text {arcsinh}(a+b x)}+1\right )^2 \left (1+e^{2 \text {arcsinh}(a+b x)}\right )}{8 b^2}de^{\text {arcsinh}(a+b x)}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int e^{-3 \text {arcsinh}(a+b x)} \left (2 e^{\text {arcsinh}(a+b x)} a-e^{2 \text {arcsinh}(a+b x)}+1\right )^2 \left (1+e^{2 \text {arcsinh}(a+b x)}\right )de^{\text {arcsinh}(a+b x)}}{8 b^3}\)

\(\Big \downarrow \) 2159

\(\displaystyle \frac {\int \left (4 e^{-2 \text {arcsinh}(a+b x)} a-4 e^{2 \text {arcsinh}(a+b x)} a+e^{-3 \text {arcsinh}(a+b x)}+\left (4 a^2-1\right ) e^{-\text {arcsinh}(a+b x)}+\left (4 a^2-1\right ) e^{\text {arcsinh}(a+b x)}+e^{3 \text {arcsinh}(a+b x)}\right )de^{\text {arcsinh}(a+b x)}}{8 b^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{2} \left (1-4 a^2\right ) e^{2 \text {arcsinh}(a+b x)}-\left (1-4 a^2\right ) \log \left (e^{\text {arcsinh}(a+b x)}\right )-4 a e^{-\text {arcsinh}(a+b x)}-\frac {4}{3} a e^{3 \text {arcsinh}(a+b x)}-\frac {1}{2} e^{-2 \text {arcsinh}(a+b x)}+\frac {1}{4} e^{4 \text {arcsinh}(a+b x)}}{8 b^3}\)

Input:

Int[E^ArcSinh[a + b*x]*x^2,x]
 

Output:

(-1/2*1/E^(2*ArcSinh[a + b*x]) - (4*a)/E^ArcSinh[a + b*x] - ((1 - 4*a^2)*E 
^(2*ArcSinh[a + b*x]))/2 - (4*a*E^(3*ArcSinh[a + b*x]))/3 + E^(4*ArcSinh[a 
 + b*x])/4 - (1 - 4*a^2)*Log[E^ArcSinh[a + b*x]])/(8*b^3)
 

Defintions of rubi rules used

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. \(287\) vs. \(2(143)=286\).

Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.50

method	result	size
default	\(\frac {x \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{4 b^{2}}-\frac {5 a \left (\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )}{4 b}-\frac {\left (a^{2}+1\right ) \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{4 b^{2}}+\frac {b \,x^{4}}{4}+\frac {a \,x^{3}}{3}\)	\(288\)

Input:

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

Output:

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))+1/4*b*x^4+1/3*a* 
x^3
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {6 \, b^{4} x^{4} + 8 \, a b^{3} x^{3} - 3 \, {\left (4 \, a^{2} - 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (6 \, b^{3} x^{3} + 2 \, a b^{2} x^{2} + 2 \, a^{3} - {\left (2 \, a^{2} - 3\right )} b x - 13 \, a\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{24 \, b^{3}} \] Input:

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

Output:

1/24*(6*b^4*x^4 + 8*a*b^3*x^3 - 3*(4*a^2 - 1)*log(-b*x - a + sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1)) + (6*b^3*x^3 + 2*a*b^2*x^2 + 2*a^3 - (2*a^2 - 3)*b*x 
 - 13*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/b^3
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 1.06 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.54 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {a x^{3}}{3} + \frac {b x^{4}}{4} + \begin {cases} \frac {\left (- \frac {a \left (- \frac {3 a \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b} - \frac {a \left (2 a^{2} + 2\right )}{12 b}\right )}{b} - \frac {\left (\frac {1}{4} - \frac {a^{2}}{6}\right ) \left (a^{2} + 1\right )}{2 b^{2}}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \left (\frac {a x^{2}}{12 b} + \frac {x^{3}}{4} + \frac {x \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b^{2}} + \frac {- \frac {3 a \left (\frac {1}{4} - \frac {a^{2}}{6}\right )}{2 b} - \frac {a \left (2 a^{2} + 2\right )}{12 b}}{b^{2}}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {\left (- 2 a^{2} - 2\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}} \left (a^{4} + 2 a^{2} + 1\right )}{3}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2} + 1}}{3} & \text {otherwise} \end {cases} \] Input:

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

Output:

a*x**3/3 + b*x**4/4 + Piecewise(((-a*(-3*a*(1/4 - a**2/6)/(2*b) - a*(2*a** 
2 + 2)/(12*b))/b - (1/4 - a**2/6)*(a**2 + 1)/(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**2/(12*b) + x**3/4 + x*(1/4 - a**2/6)/ 
(2*b**2) + (-3*a*(1/4 - a**2/6)/(2*b) - a*(2*a**2 + 2)/(12*b))/b**2), Ne(b 
**2, 0)), (((-2*a**2 - 2)*(a**2 + 2*a*b*x + 1)**(5/2)/5 + (a**2 + 2*a*b*x 
+ 1)**(7/2)/7 + (a**2 + 2*a*b*x + 1)**(3/2)*(a**4 + 2*a**2 + 1)/3)/(4*a**3 
*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2 + 1)/3, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.37 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a}{12 \, b^{3}} - \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x}{8 \, b^{4}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{8 \, b^{5}} + \frac {{\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{8 \, b^{5}} \] Input:

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

Output:

1/4*b*x^4 + 1/3*a*x^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*x/b^2 - 5/ 
12*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)*a/b^3 - 1/8*(5*a^2*b^2 - (a^2 + 1)* 
b^2)*a^2*arcsinh(2*(b^2*x + a*b)/sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^5 + 
 1/8*(5*a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*x/b^4 + 
 1/8*(5*a^2*b^2 - (a^2 + 1)*b^2)*(a^2 + 1)*arcsinh(2*(b^2*x + a*b)/sqrt(-4 
*a^2*b^2 + 4*(a^2 + 1)*b^2))/b^5 + 1/8*(5*a^2*b^2 - (a^2 + 1)*b^2)*sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1)*a/b^5
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.23 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} + \frac {1}{24} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, {\left (3 \, x + \frac {a}{b}\right )} x - \frac {2 \, a^{2} b^{3} - 3 \, b^{3}}{b^{5}}\right )} x + \frac {2 \, a^{3} b^{2} - 13 \, a b^{2}}{b^{5}}\right )} - \frac {{\left (4 \, a^{2} - 1\right )} \log \left ({\left | -a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |} \right |}\right )}{8 \, b^{2} {\left | b \right |}} \] Input:

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

Output:

1/4*b*x^4 + 1/3*a*x^3 + 1/24*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*(3*x + 
a/b)*x - (2*a^2*b^3 - 3*b^3)/b^5)*x + (2*a^3*b^2 - 13*a*b^2)/b^5) - 1/8*(4 
*a^2 - 1)*log(abs(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*ab 
s(b)))/(b^2*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.93 \[ \int e^{\text {arcsinh}(a+b x)} x^2 \, dx=\frac {2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{3}-2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a^{2} b x +2 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a \,b^{2} x^{2}-13 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, a +6 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b^{3} x^{3}+3 \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x +12 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right ) a^{2}-3 \,\mathrm {log}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a +b x \right )+8 a \,b^{3} x^{3}+6 b^{4} x^{4}}{24 b^{3}} \] Input:

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

Output:

(2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a**3 - 2*sqrt(a**2 + 2*a*b*x + b** 
2*x**2 + 1)*a**2*b*x + 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a*b**2*x**2 
- 13*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*a + 6*sqrt(a**2 + 2*a*b*x + b**2 
*x**2 + 1)*b**3*x**3 + 3*sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)*b*x + 12*log 
(sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1) + a + b*x)*a**2 - 3*log(sqrt(a**2 + 
2*a*b*x + b**2*x**2 + 1) + a + b*x) + 8*a*b**3*x**3 + 6*b**4*x**4)/(24*b** 
3)