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

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

[Out]

1/(4*b^2*E^ArcCosh[a + b*x]) - (a*E^(2*ArcCosh[a + b*x]))/(4*b^2) + E^(3*ArcCosh[a + b*x])/(12*b^2) + (a*ArcCo
sh[a + b*x])/(2*b^2)

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Rubi [A]  time = 0.0702901, 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 = {5899, 2282, 12, 1628} \[ -\frac{a e^{2 \cosh ^{-1}(a+b x)}}{4 b^2}+\frac{a \cosh ^{-1}(a+b x)}{2 b^2}+\frac{e^{-\cosh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \cosh ^{-1}(a+b x)}}{12 b^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]*x,x]

[Out]

1/(4*b^2*E^ArcCosh[a + b*x]) - (a*E^(2*ArcCosh[a + b*x]))/(4*b^2) + E^(3*ArcCosh[a + b*x])/(12*b^2) + (a*ArcCo
sh[a + b*x])/(2*b^2)

Rule 5899

Int[(f_)^(ArcCosh[(a_.) + (b_.)*(x_)]^(n_.)*(c_.))*(x_)^(m_.), x_Symbol] :> Dist[1/b, Subst[Int[(-(a/b) + Cosh
[x]/b)^m*f^(c*x^n)*Sinh[x], x], x, ArcCosh[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^{\cosh ^{-1}(a+b x)} x \, dx &=\frac{\operatorname{Subst}\left (\int e^x \left (-\frac{a}{b}+\frac{\cosh (x)}{b}\right ) \sinh (x) \, dx,x,\cosh ^{-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^{\cosh ^{-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^{\cosh ^{-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^{\cosh ^{-1}(a+b x)}\right )}{4 b^2}\\ &=\frac{e^{-\cosh ^{-1}(a+b x)}}{4 b^2}-\frac{a e^{2 \cosh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \cosh ^{-1}(a+b x)}}{12 b^2}+\frac{a \cosh ^{-1}(a+b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.152642, size = 93, normalized size = 1.39 \[ \frac{1}{6} \left (\frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (-a^2+a b x+2 b^2 x^2-2\right )}{b^2}+\frac{3 a \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )}{b^2}+3 a x^2+2 b x^3\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]*x,x]

[Out]

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

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Maple [C]  time = 0.012, size = 194, normalized size = 2.9 \begin{align*}{\frac{{\it csgn} \left ( b \right ) }{6\,{b}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1} \left ( 2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ){x}^{2}{b}^{2}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) xab-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ){a}^{2}-2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +3\,\ln \left ( \left ( \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +bx+a \right ){\it csgn} \left ( b \right ) \right ) a \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}+{\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+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x,x)

[Out]

1/6*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*(2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*x^2*b^2+(b^2*x^2+2*a*b*x+a^2-1)^(
1/2)*csgn(b)*x*a*b-(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)*a^2-2*(b^2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)+3*ln(((b^
2*x^2+2*a*b*x+a^2-1)^(1/2)*csgn(b)+b*x+a)*csgn(b))*a)*csgn(b)/b^2/(b^2*x^2+2*a*b*x+a^2-1)^(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+(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2))*x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99024, size = 217, normalized size = 3.24 \begin{align*} \frac{2 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} +{\left (2 \, b^{2} x^{2} + a b x - a^{2} - 2\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} - 3 \, a \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right )}{6 \, b^{2}} \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,x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [A]  time = 1.27061, size = 176, normalized size = 2.63 \begin{align*} \frac{2 \,{\left (b x^{3} + \frac{a^{3} + 3 \, a^{2} + 3 \, a + 1}{b^{2}}\right )} b + \frac{3 \,{\left ({\left (b x + a + 1\right )}^{2} - 2 \,{\left (b x + a + 1\right )} a - 2 \, b x - 2 \, a - 2\right )} a}{b} + \frac{{\left ({\left (2 \, b x - a - 2\right )}{\left (b x + a + 1\right )} + 3 \, a\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} - 6 \, a \log \left ({\left | -\sqrt{b x + a + 1} + \sqrt{b x + a - 1} \right |}\right )}{b}}{6 \, 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,x, algorithm="giac")

[Out]

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