3.351 \(\int e^{\sinh ^{-1}(a+b x)} x \, dx\)

Optimal. Leaf size=67 \[ -\frac{a e^{2 \sinh ^{-1}(a+b x)}}{4 b^2}-\frac{a \sinh ^{-1}(a+b x)}{2 b^2}+\frac{e^{-\sinh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \sinh ^{-1}(a+b x)}}{12 b^2} \]

[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)

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Rubi [A]  time = 0.070877, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5898, 2282, 12, 1628} \[ -\frac{a e^{2 \sinh ^{-1}(a+b x)}}{4 b^2}-\frac{a \sinh ^{-1}(a+b x)}{2 b^2}+\frac{e^{-\sinh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \sinh ^{-1}(a+b x)}}{12 b^2} \]

Antiderivative was successfully verified.

[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 5898

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]

Rule 2282

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 12

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

Rule 1628

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]

Rubi steps

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

Mathematica [A]  time = 0.0987441, size = 73, normalized size = 1.09 \[ \frac{1}{6} \left (\frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-a^2+a b x+2 b^2 x^2+2\right )}{b^2}-\frac{3 a \sinh ^{-1}(a+b x)}{b^2}+3 a x^2+2 b x^3\right ) \]

Warning: Unable to verify antiderivative.

[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

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Maple [A]  time = 0.003, size = 138, normalized size = 2.1 \begin{align*}{\frac{1}{3\,{b}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{ax}{2\,b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{2}}{2\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{a}{2\,b}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{b{x}^{3}}{3}}+{\frac{a{x}^{2}}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.60614, size = 212, normalized size = 3.16 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b x + \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1)), x)

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Giac [A]  time = 1.34604, size = 143, normalized size = 2.13 \begin{align*} \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 |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))