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

Optimal. Leaf size=31 \[ \frac{\sinh ^{-1}(a+b x)}{2 b}+\frac{e^{2 \sinh ^{-1}(a+b x)}}{4 b} \]

[Out]

E^(2*ArcSinh[a + b*x])/(4*b) + ArcSinh[a + b*x]/(2*b)

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Rubi [A]  time = 0.0174833, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5896, 2282, 12, 14} \[ \frac{\sinh ^{-1}(a+b x)}{2 b}+\frac{e^{2 \sinh ^{-1}(a+b x)}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x],x]

[Out]

E^(2*ArcSinh[a + b*x])/(4*b) + ArcSinh[a + b*x]/(2*b)

Rule 5896

Int[(f_)^(ArcSinh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[f^(c*x^n)*Cosh[x], x], x,
ArcSinh[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0323047, size = 46, normalized size = 1.48 \[ \frac{(a+b x) \left (\sqrt{a^2+2 a b x+b^2 x^2+1}+a+b x\right )+\sinh ^{-1}(a+b x)}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x],x]

[Out]

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

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Maple [B]  time = 0.003, size = 89, normalized size = 2.9 \begin{align*} ax+{\frac{2\,{b}^{2}x+2\,ab}{4\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{1}{2}\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}^{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)

[Out]

a*x+1/4*(2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*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)+1/2*b*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, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.5868, size = 169, normalized size = 5.45 \begin{align*} \frac{b^{2} x^{2} + 2 \, a b x + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (b x + a\right )} - \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \, b} \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, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.48247, size = 108, normalized size = 3.48 \begin{align*} \frac{1}{2} \, b x^{2} + a x + \frac{1}{2} \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (x + \frac{a}{b}\right )} - \frac{\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 \,{\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, algorithm="giac")

[Out]

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