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

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

[Out]

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

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Rubi [A]  time = 0.0173356, 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 = {5897, 2282, 12, 14} \[ \frac{e^{2 \cosh ^{-1}(a+b x)}}{4 b}-\frac{\cosh ^{-1}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x],x]

[Out]

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

Rule 5897

Int[(f_)^(ArcCosh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Dist[1/b, Subst[Int[f^(c*x^n)*Sinh[x], x], x,
ArcCosh[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^{\cosh ^{-1}(a+b x)} \, dx &=\frac{\operatorname{Subst}\left (\int e^x \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{2 x} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x}+x\right ) \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )}{2 b}\\ &=\frac{e^{2 \cosh ^{-1}(a+b x)}}{4 b}-\frac{\cosh ^{-1}(a+b x)}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.0274963, size = 69, normalized size = 2.23 \[ \frac{(a+b x) \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )-\log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )}{2 b} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x],x]

[Out]

((a + b*x)*(a + b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]) - Log[a + b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*
x]])/(2*b)

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Maple [B]  time = 0.007, size = 147, normalized size = 4.7 \begin{align*}{\frac{b{x}^{2}}{2}}+ax+{\frac{1}{2\,b}\sqrt{bx+a-1} \left ( bx+a+1 \right ) ^{{\frac{3}{2}}}}-{\frac{1}{2\,b}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{1}{2}\sqrt{ \left ( bx+a+1 \right ) \left ( bx+a-1 \right ) }\ln \left ({ \left ({\frac{b \left ( 1+a \right ) }{2}}+{\frac{ \left ( a-1 \right ) b}{2}}+{b}^{2}x \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+ \left ( b \left ( 1+a \right ) + \left ( a-1 \right ) b \right ) x+ \left ( 1+a \right ) \left ( a-1 \right ) } \right ){\frac{1}{\sqrt{bx+a-1}}}{\frac{1}{\sqrt{bx+a+1}}}{\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.00254, size = 174, normalized size = 5.61 \begin{align*} \frac{b^{2} x^{2} + 2 \, a b x + \sqrt{b x + a + 1}{\left (b x + a\right )} \sqrt{b x + a - 1} + \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32156, size = 82, normalized size = 2.65 \begin{align*} \frac{1}{2} \, b x^{2} + a x + \frac{\sqrt{b x + a + 1}{\left (b x + a\right )} \sqrt{b x + a - 1} + 2 \, \log \left ({\left | -\sqrt{b x + a + 1} + \sqrt{b x + a - 1} \right |}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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