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

Optimal. Leaf size=90 \[ -\frac{\left (2 a^2+1\right ) \cosh ^{-1}(a+b x)}{4 b^2}+\frac{3 a \sqrt{a+b x-1} \sqrt{a+b x+1}}{4 b^2}+\frac{1}{2} x^2 \cosh ^{-1}(a+b x)-\frac{x \sqrt{a+b x-1} \sqrt{a+b x+1}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0638357, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5866, 5802, 90, 80, 52} \[ -\frac{\left (2 a^2+1\right ) \cosh ^{-1}(a+b x)}{4 b^2}+\frac{3 a \sqrt{a+b x-1} \sqrt{a+b x+1}}{4 b^2}+\frac{1}{2} x^2 \cosh ^{-1}(a+b x)-\frac{x \sqrt{a+b x-1} \sqrt{a+b x+1}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcCosh[a + b*x],x]

[Out]

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

Rule 5866

Int[((a_.) + ArcCosh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcCosh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5802

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

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rubi steps

\begin{align*} \int x \cosh ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right ) \cosh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{2} x^2 \cosh ^{-1}(a+b x)-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=-\frac{x \sqrt{-1+a+b x} \sqrt{1+a+b x}}{4 b}+\frac{1}{2} x^2 \cosh ^{-1}(a+b x)-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\frac{1+2 a^2}{b^2}-\frac{3 a x}{b^2}}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=\frac{3 a \sqrt{-1+a+b x} \sqrt{1+a+b x}}{4 b^2}-\frac{x \sqrt{-1+a+b x} \sqrt{1+a+b x}}{4 b}+\frac{1}{2} x^2 \cosh ^{-1}(a+b x)-\frac{\left (1+2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )}{4 b^2}\\ &=\frac{3 a \sqrt{-1+a+b x} \sqrt{1+a+b x}}{4 b^2}-\frac{x \sqrt{-1+a+b x} \sqrt{1+a+b x}}{4 b}-\frac{\left (1+2 a^2\right ) \cosh ^{-1}(a+b x)}{4 b^2}+\frac{1}{2} x^2 \cosh ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.0730523, size = 87, normalized size = 0.97 \[ \frac{-\left (2 a^2+1\right ) \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )+2 b^2 x^2 \cosh ^{-1}(a+b x)+(3 a-b x) \sqrt{a+b x-1} \sqrt{a+b x+1}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcCosh[a + b*x],x]

[Out]

((3*a - b*x)*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x] + 2*b^2*x^2*ArcCosh[a + b*x] - (1 + 2*a^2)*Log[a + b*x + Sqr
t[-1 + a + b*x]*Sqrt[1 + a + b*x]])/(4*b^2)

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Maple [A]  time = 0.01, size = 120, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}{\rm arccosh} \left (bx+a\right )}{2}}-{\frac{{\rm arccosh} \left (bx+a\right ){a}^{2}}{2\,{b}^{2}}}-{\frac{x}{4\,b}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{3\,a}{4\,{b}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{1}{4\,{b}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1}\ln \left ( bx+a+\sqrt{ \left ( bx+a \right ) ^{2}-1} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccosh(b*x+a),x)

[Out]

1/2*x^2*arccosh(b*x+a)-1/2/b^2*arccosh(b*x+a)*a^2-1/4*x*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/b+3/4*a*(b*x+a-1)^(1/2
)*(b*x+a+1)^(1/2)/b^2-1/4/b^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/((b*x+a)^2-1)^(1/2)*ln(b*x+a+((b*x+a)^2-1)^(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(x*arccosh(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.26021, size = 178, normalized size = 1.98 \begin{align*} \frac{{\left (2 \, b^{2} x^{2} - 2 \, a^{2} - 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (b x - 3 \, a\right )}}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccosh(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.384441, size = 104, normalized size = 1.16 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{acosh}{\left (a + b x \right )}}{2 b^{2}} + \frac{3 a \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{4 b^{2}} + \frac{x^{2} \operatorname{acosh}{\left (a + b x \right )}}{2} - \frac{x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{4 b} - \frac{\operatorname{acosh}{\left (a + b x \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{acosh}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acosh(b*x+a),x)

[Out]

Piecewise((-a**2*acosh(a + b*x)/(2*b**2) + 3*a*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(4*b**2) + x**2*acosh(a +
b*x)/2 - x*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(4*b) - acosh(a + b*x)/(4*b**2), Ne(b, 0)), (x**2*acosh(a)/2,
True))

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Giac [A]  time = 1.20757, size = 151, normalized size = 1.68 \begin{align*} \frac{1}{2} \, x^{2} \log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} - 1}\right ) - \frac{1}{4} \,{\left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (\frac{x}{b^{2}} - \frac{3 \, a}{b^{3}}\right )} - \frac{{\left (2 \, a^{2} + 1\right )} \log \left ({\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 |}\right )}{b^{2}{\left | b \right |}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccosh(b*x+a),x, algorithm="giac")

[Out]

1/2*x^2*log(b*x + a + sqrt((b*x + a)^2 - 1)) - 1/4*(sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(x/b^2 - 3*a/b^3) - (2*a
^2 + 1)*log(abs(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*abs(b)))/(b^2*abs(b)))*b