3.84 \(\int x^2 \cosh ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=104 \[ -\frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (11 a^2-5 a b x+4\right )}{18 b^3}+\frac{a \left (2 a^2+3\right ) \cosh ^{-1}(a+b x)}{6 b^3}-\frac{x^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{9 b}+\frac{1}{3} x^3 \cosh ^{-1}(a+b x) \]

[Out]

-(x^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(9*b) - (Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(4 + 11*a^2 - 5*a*b*
x))/(18*b^3) + (a*(3 + 2*a^2)*ArcCosh[a + b*x])/(6*b^3) + (x^3*ArcCosh[a + b*x])/3

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Rubi [A]  time = 0.119097, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5866, 5802, 100, 147, 52} \[ -\frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (11 a^2-5 a b x+4\right )}{18 b^3}+\frac{a \left (2 a^2+3\right ) \cosh ^{-1}(a+b x)}{6 b^3}-\frac{x^2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{9 b}+\frac{1}{3} x^3 \cosh ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCosh[a + b*x],x]

[Out]

-(x^2*Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x])/(9*b) - (Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(4 + 11*a^2 - 5*a*b*
x))/(18*b^3) + (a*(3 + 2*a^2)*ArcCosh[a + b*x])/(6*b^3) + (x^3*ArcCosh[a + b*x])/3

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 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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^2 \cosh ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{b}+\frac{x}{b}\right )^2 \cosh ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac{1}{3} x^3 \cosh ^{-1}(a+b x)-\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (-\frac{a}{b}+\frac{x}{b}\right )^3}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=-\frac{x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{9 b}+\frac{1}{3} x^3 \cosh ^{-1}(a+b x)-\frac{1}{9} \operatorname{Subst}\left (\int \frac{\left (\frac{2+3 a^2}{b^2}-\frac{5 a x}{b^2}\right ) \left (-\frac{a}{b}+\frac{x}{b}\right )}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )\\ &=-\frac{x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{9 b}-\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x} \left (4+11 a^2-5 a b x\right )}{18 b^3}+\frac{1}{3} x^3 \cosh ^{-1}(a+b x)+\frac{\left (a \left (3+2 a^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,a+b x\right )}{6 b^3}\\ &=-\frac{x^2 \sqrt{-1+a+b x} \sqrt{1+a+b x}}{9 b}-\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x} \left (4+11 a^2-5 a b x\right )}{18 b^3}+\frac{a \left (3+2 a^2\right ) \cosh ^{-1}(a+b x)}{6 b^3}+\frac{1}{3} x^3 \cosh ^{-1}(a+b x)\\ \end{align*}

Mathematica [A]  time = 0.114483, size = 101, normalized size = 0.97 \[ \frac{-\sqrt{a+b x-1} \sqrt{a+b x+1} \left (11 a^2-5 a b x+2 b^2 x^2+4\right )+\left (6 a^3+9 a\right ) \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )+6 b^3 x^3 \cosh ^{-1}(a+b x)}{18 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCosh[a + b*x],x]

[Out]

(-(Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]*(4 + 11*a^2 - 5*a*b*x + 2*b^2*x^2)) + 6*b^3*x^3*ArcCosh[a + b*x] + (9*
a + 6*a^3)*Log[a + b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x]])/(18*b^3)

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Maple [B]  time = 0.01, size = 207, normalized size = 2. \begin{align*}{\frac{{x}^{3}{\rm arccosh} \left (bx+a\right )}{3}}-{\frac{{x}^{2}}{9\,b}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{{a}^{3}}{3\,{b}^{3}}\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}}}}+{\frac{5\,ax}{18\,{b}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1}}-{\frac{11\,{a}^{2}}{18\,{b}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1}}+{\frac{a}{2\,{b}^{3}}\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}}}}-{\frac{2}{9\,{b}^{3}}\sqrt{bx+a-1}\sqrt{bx+a+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccosh(b*x+a)-1/9*x^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/b+1/3/b^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/((b*x
+a)^2-1)^(1/2)*a^3*ln(b*x+a+((b*x+a)^2-1)^(1/2))+5/18/b^2*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)*x*a-11/18/b^3*(b*x+a
-1)^(1/2)*(b*x+a+1)^(1/2)*a^2+1/2/b^3*(b*x+a-1)^(1/2)*(b*x+a+1)^(1/2)/((b*x+a)^2-1)^(1/2)*a*ln(b*x+a+((b*x+a)^
2-1)^(1/2))-2/9/b^3*(b*x+a-1)^(1/2)*(b*x+a+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^2*arccosh(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.34271, size = 216, normalized size = 2.08 \begin{align*} \frac{3 \,{\left (2 \, b^{3} x^{3} + 2 \, a^{3} + 3 \, a\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (2 \, b^{2} x^{2} - 5 \, a b x + 11 \, a^{2} + 4\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{18 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(3*(2*b^3*x^3 + 2*a^3 + 3*a)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - (2*b^2*x^2 - 5*a*b*x + 11
*a^2 + 4)*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/b^3

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Sympy [A]  time = 0.887139, size = 170, normalized size = 1.63 \begin{align*} \begin{cases} \frac{a^{3} \operatorname{acosh}{\left (a + b x \right )}}{3 b^{3}} - \frac{11 a^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{18 b^{3}} + \frac{5 a x \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{18 b^{2}} + \frac{a \operatorname{acosh}{\left (a + b x \right )}}{2 b^{3}} + \frac{x^{3} \operatorname{acosh}{\left (a + b x \right )}}{3} - \frac{x^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{9 b} - \frac{2 \sqrt{a^{2} + 2 a b x + b^{2} x^{2} - 1}}{9 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \operatorname{acosh}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**3*acosh(a + b*x)/(3*b**3) - 11*a**2*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(18*b**3) + 5*a*x*sqrt(
a**2 + 2*a*b*x + b**2*x**2 - 1)/(18*b**2) + a*acosh(a + b*x)/(2*b**3) + x**3*acosh(a + b*x)/3 - x**2*sqrt(a**2
+ 2*a*b*x + b**2*x**2 - 1)/(9*b) - 2*sqrt(a**2 + 2*a*b*x + b**2*x**2 - 1)/(9*b**3), Ne(b, 0)), (x**3*acosh(a)
/3, True))

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Giac [A]  time = 1.20715, size = 178, normalized size = 1.71 \begin{align*} \frac{1}{3} \, x^{3} \log \left (b x + a + \sqrt{{\left (b x + a\right )}^{2} - 1}\right ) - \frac{1}{18} \,{\left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (x{\left (\frac{2 \, x}{b^{2}} - \frac{5 \, a}{b^{3}}\right )} + \frac{11 \, a^{2} b + 4 \, b}{b^{5}}\right )} + \frac{3 \,{\left (2 \, a^{3} + 3 \, a\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^{3}{\left | b \right |}}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log(b*x + a + sqrt((b*x + a)^2 - 1)) - 1/18*(sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(x*(2*x/b^2 - 5*a/b^3)
+ (11*a^2*b + 4*b)/b^5) + 3*(2*a^3 + 3*a)*log(abs(-a*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*abs(b)
))/(b^3*abs(b)))*b