\(\int e^{\text {arcsinh}(a+b x)} x \, dx\) [351]

Optimal result

Integrand size = 10, antiderivative size = 67 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {e^{-\text {arcsinh}(a+b x)}}{4 b^2}-\frac {a e^{2 \text {arcsinh}(a+b x)}}{4 b^2}+\frac {e^{3 \text {arcsinh}(a+b x)}}{12 b^2}-\frac {a \text {arcsinh}(a+b x)}{2 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5873, 2320, 12, 1642} \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=-\frac {a e^{2 \text {arcsinh}(a+b x)}}{4 b^2}-\frac {a \text {arcsinh}(a+b x)}{2 b^2}+\frac {e^{-\text {arcsinh}(a+b x)}}{4 b^2}+\frac {e^{3 \text {arcsinh}(a+b x)}}{12 b^2} \]

[In]

Int[E^ArcSinh[a + b*x]*x,x]

[Out]

1/(4*b^2*E^ArcSinh[a + b*x]) - (a*E^(2*ArcSinh[a + b*x]))/(4*b^2) + E^(3*ArcSinh[a + b*x])/(12*b^2) - (a*ArcSi
nh[a + b*x])/(2*b^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 5873

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^x \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right ) \, dx,x,\text {arcsinh}(a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (-1-x^2\right ) \left (1+2 a x-x^2\right )}{4 b x^2} \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \frac {\left (-1-x^2\right ) \left (1+2 a x-x^2\right )}{x^2} \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{4 b^2} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{x^2}-\frac {2 a}{x}-2 a x+x^2\right ) \, dx,x,e^{\text {arcsinh}(a+b x)}\right )}{4 b^2} \\ & = \frac {e^{-\text {arcsinh}(a+b x)}}{4 b^2}-\frac {a e^{2 \text {arcsinh}(a+b x)}}{4 b^2}+\frac {e^{3 \text {arcsinh}(a+b x)}}{12 b^2}-\frac {a \text {arcsinh}(a+b x)}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {1}{6} \left (3 a x^2+2 b x^3+\frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (2-a^2+a b x+2 b^2 x^2\right )}{b^2}-\frac {3 a \text {arcsinh}(a+b x)}{b^2}\right ) \]

[In]

Integrate[E^ArcSinh[a + b*x]*x,x]

[Out]

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

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 145, normalized size of antiderivative = 2.16

method	result	size
default	\(\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}+\frac {b \,x^{3}}{3}+\frac {a \,x^{2}}{2}\)	\(145\)

[In]

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

[Out]

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))+1/3*b*x^3+1/2*a*x^
2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.39 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {2 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (2 \, b^{2} x^{2} + a b x - a^{2} + 2\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{6 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.94 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {a x^{2}}{2} + \frac {b x^{3}}{3} + \begin {cases} \frac {\left (- \frac {a \left (\frac {1}{3} - \frac {a^{2}}{6}\right )}{b} - \frac {a \left (a^{2} + 1\right )}{6 b}\right ) \log {\left (2 a b + 2 b^{2} x + 2 \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} \sqrt {b^{2}} \right )}}{\sqrt {b^{2}}} + \left (\frac {a x}{6 b} + \frac {x^{2}}{3} + \frac {\frac {1}{3} - \frac {a^{2}}{6}}{b^{2}}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1} & \text {for}\: b^{2} \neq 0 \\\frac {\frac {\left (- a^{2} - 1\right ) \left (a^{2} + 2 a b x + 1\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x + 1\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2} + 1}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (83) = 166\).

Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.61 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} + \frac {a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{2 \, b} - \frac {{\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.58 \[ \int e^{\text {arcsinh}(a+b x)} x \, dx=\frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} + \frac {1}{6} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, x + \frac {a}{b}\right )} x - \frac {a^{2} b - 2 \, b}{b^{3}}\right )} + \frac {a \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, b {\left | b \right |}} \]

[In]

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

[Out]

1/3*b*x^3 + 1/2*a*x^2 + 1/6*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*((2*x + a/b)*x - (a^2*b - 2*b)/b^3) + 1/2*a*log(
-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*abs(b))/(b*abs(b))

Mupad [F(-1)]

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

[In]

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

[Out]

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