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]

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

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]]

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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [B]  time = 0.60, size = 73, normalized size = 2.35 \[ \frac {b^{2} x^{2} + 2 \, a b x + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (b x + a\right )} - \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{2 \, b} \]

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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giac [B]  time = 0.29, size = 80, normalized size = 2.58 \[ \frac {1}{2} \, b x^{2} + a x + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (x + \frac {a}{b}\right )} - \frac {\log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{2 \, {\left | b \right |}} \]

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)

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maple [B]  time = 0.00, size = 89, normalized size = 2.87 \[ a x +\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 \sqrt {b^{2}}}+\frac {b \,x^{2}}{2} \]

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 [B]  time = 0.46, size = 141, normalized size = 4.55 \[ \frac {1}{2} \, b x^{2} + a x - \frac {a^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b} + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} x + \frac {{\left (a^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a}{2 \, b} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int a+\sqrt {{\left (a+b\,x\right )}^2+1}+b\,x \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x + \sqrt {\left (a + b x\right )^{2} + 1}\right )\, dx \]

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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