3.15 \(\int x \cosh ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=64 \[ -\frac{\cosh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cosh ^{-1}(a x)^2-\frac{x \sqrt{a x-1} \sqrt{a x+1} \cosh ^{-1}(a x)}{2 a}+\frac{x^2}{4} \]

[Out]

x^2/4 - (x*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x])/(2*a) - ArcCosh[a*x]^2/(4*a^2) + (x^2*ArcCosh[a*x]^2)/2

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Rubi [A]  time = 0.250665, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5662, 5759, 5676, 30} \[ -\frac{\cosh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cosh ^{-1}(a x)^2-\frac{x \sqrt{a x-1} \sqrt{a x+1} \cosh ^{-1}(a x)}{2 a}+\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCosh[a*x]^2,x]

[Out]

x^2/4 - (x*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x])/(2*a) - ArcCosh[a*x]^2/(4*a^2) + (x^2*ArcCosh[a*x]^2)/2

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5759

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[(f*(f*x)^(m - 1)*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x]*(a + b*ArcCosh[c*x])^n)/(e1*e2*m
), x] + (Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)^(m - 2)*(a + b*ArcCosh[c*x])^n)/(Sqrt[d1 + e1*x]*Sqrt[d2 + e2*
x]), x], x] + Dist[(b*f*n*Sqrt[d1 + e1*x]*Sqrt[d2 + e2*x])/(c*d1*d2*m*Sqrt[1 + c*x]*Sqrt[-1 + c*x]), Int[(f*x)
^(m - 1)*(a + b*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2, f}, x] && EqQ[e1 - c*d1, 0]
&& EqQ[e2 + c*d2, 0] && GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

Rule 5676

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_.)*(x_)]), x_Symbol]
:> Simp[(a + b*ArcCosh[c*x])^(n + 1)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n},
x] && EqQ[e1, c*d1] && EqQ[e2, -(c*d2)] && GtQ[d1, 0] && LtQ[d2, 0] && NeQ[n, -1]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x \cosh ^{-1}(a x)^2 \, dx &=\frac{1}{2} x^2 \cosh ^{-1}(a x)^2-a \int \frac{x^2 \cosh ^{-1}(a x)}{\sqrt{-1+a x} \sqrt{1+a x}} \, dx\\ &=-\frac{x \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)}{2 a}+\frac{1}{2} x^2 \cosh ^{-1}(a x)^2+\frac{\int x \, dx}{2}-\frac{\int \frac{\cosh ^{-1}(a x)}{\sqrt{-1+a x} \sqrt{1+a x}} \, dx}{2 a}\\ &=\frac{x^2}{4}-\frac{x \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)}{2 a}-\frac{\cosh ^{-1}(a x)^2}{4 a^2}+\frac{1}{2} x^2 \cosh ^{-1}(a x)^2\\ \end{align*}

Mathematica [A]  time = 0.0498287, size = 58, normalized size = 0.91 \[ \frac{a^2 x^2+\left (2 a^2 x^2-1\right ) \cosh ^{-1}(a x)^2-2 a x \sqrt{a x-1} \sqrt{a x+1} \cosh ^{-1}(a x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCosh[a*x]^2,x]

[Out]

(a^2*x^2 - 2*a*x*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x] + (-1 + 2*a^2*x^2)*ArcCosh[a*x]^2)/(4*a^2)

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Maple [A]  time = 0.029, size = 58, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}{a}^{2}{x}^{2}}{2}}-{\frac{ax{\rm arccosh} \left (ax\right )}{2}\sqrt{ax-1}\sqrt{ax+1}}-{\frac{ \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}}{4}}+{\frac{{a}^{2}{x}^{2}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccosh(a*x)^2,x)

[Out]

1/a^2*(1/2*arccosh(a*x)^2*a^2*x^2-1/2*arccosh(a*x)*a*x*(a*x-1)^(1/2)*(a*x+1)^(1/2)-1/4*arccosh(a*x)^2+1/4*a^2*
x^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, x^{2} \log \left (a x + \sqrt{a x + 1} \sqrt{a x - 1}\right )^{2} - \int \frac{{\left (a^{3} x^{4} + \sqrt{a x + 1} \sqrt{a x - 1} a^{2} x^{3} - a x^{2}\right )} \log \left (a x + \sqrt{a x + 1} \sqrt{a x - 1}\right )}{a^{3} x^{3} +{\left (a^{2} x^{2} - 1\right )} \sqrt{a x + 1} \sqrt{a x - 1} - a x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccosh(a*x)^2,x, algorithm="maxima")

[Out]

1/2*x^2*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))^2 - integrate((a^3*x^4 + sqrt(a*x + 1)*sqrt(a*x - 1)*a^2*x^3 -
a*x^2)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))/(a^3*x^3 + (a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(a*x - 1) - a*x), x)

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Fricas [A]  time = 2.19339, size = 166, normalized size = 2.59 \begin{align*} \frac{a^{2} x^{2} - 2 \, \sqrt{a^{2} x^{2} - 1} a x \log \left (a x + \sqrt{a^{2} x^{2} - 1}\right ) +{\left (2 \, a^{2} x^{2} - 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} - 1}\right )^{2}}{4 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccosh(a*x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.639111, size = 60, normalized size = 0.94 \begin{align*} \begin{cases} \frac{x^{2} \operatorname{acosh}^{2}{\left (a x \right )}}{2} + \frac{x^{2}}{4} - \frac{x \sqrt{a^{2} x^{2} - 1} \operatorname{acosh}{\left (a x \right )}}{2 a} - \frac{\operatorname{acosh}^{2}{\left (a x \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\- \frac{\pi ^{2} x^{2}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acosh(a*x)**2,x)

[Out]

Piecewise((x**2*acosh(a*x)**2/2 + x**2/4 - x*sqrt(a**2*x**2 - 1)*acosh(a*x)/(2*a) - acosh(a*x)**2/(4*a**2), Ne
(a, 0)), (-pi**2*x**2/8, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{arcosh}\left (a x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccosh(a*x)^2,x, algorithm="giac")

[Out]

integrate(x*arccosh(a*x)^2, x)