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

Optimal. Leaf size=90 \[ -\frac {4 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{9 a^3}+\frac {4 x}{9 a^2}+\frac {1}{3} x^3 \cosh ^{-1}(a x)^2-\frac {2 x^2 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{9 a}+\frac {2 x^3}{27} \]

[Out]

4/9*x/a^2+2/27*x^3+1/3*x^3*arccosh(a*x)^2-4/9*arccosh(a*x)*(a*x-1)^(1/2)*(a*x+1)^(1/2)/a^3-2/9*x^2*arccosh(a*x
)*(a*x-1)^(1/2)*(a*x+1)^(1/2)/a

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Rubi [A]  time = 0.31, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5662, 5759, 5718, 8, 30} \[ \frac {4 x}{9 a^2}-\frac {4 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{9 a^3}+\frac {1}{3} x^3 \cosh ^{-1}(a x)^2-\frac {2 x^2 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{9 a}+\frac {2 x^3}{27} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x)/(9*a^2) + (2*x^3)/27 - (4*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x])/(9*a^3) - (2*x^2*Sqrt[-1 + a*x]*Sqr
t[1 + a*x]*ArcCosh[a*x])/(9*a) + (x^3*ArcCosh[a*x]^2)/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

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 5718

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_.), x_
Symbol] :> Simp[((d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(2*e1*e2*(p + 1)), x] - Dist[
(b*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(2*c*(p + 1)*(1 + c*x)^FracPart[p]
*(-1 + c*x)^FracPart[p]), Int[(-1 + c^2*x^2)^(p + 1/2)*(a + b*ArcCosh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c,
d1, e1, d2, e2, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && GtQ[n, 0] && NeQ[p, -1] && IntegerQ[p + 1
/2]

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]

Rubi steps

\begin {align*} \int x^2 \cosh ^{-1}(a x)^2 \, dx &=\frac {1}{3} x^3 \cosh ^{-1}(a x)^2-\frac {1}{3} (2 a) \int \frac {x^3 \cosh ^{-1}(a x)}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx\\ &=-\frac {2 x^2 \sqrt {-1+a x} \sqrt {1+a x} \cosh ^{-1}(a x)}{9 a}+\frac {1}{3} x^3 \cosh ^{-1}(a x)^2+\frac {2 \int x^2 \, dx}{9}-\frac {4 \int \frac {x \cosh ^{-1}(a x)}{\sqrt {-1+a x} \sqrt {1+a x}} \, dx}{9 a}\\ &=\frac {2 x^3}{27}-\frac {4 \sqrt {-1+a x} \sqrt {1+a x} \cosh ^{-1}(a x)}{9 a^3}-\frac {2 x^2 \sqrt {-1+a x} \sqrt {1+a x} \cosh ^{-1}(a x)}{9 a}+\frac {1}{3} x^3 \cosh ^{-1}(a x)^2+\frac {4 \int 1 \, dx}{9 a^2}\\ &=\frac {4 x}{9 a^2}+\frac {2 x^3}{27}-\frac {4 \sqrt {-1+a x} \sqrt {1+a x} \cosh ^{-1}(a x)}{9 a^3}-\frac {2 x^2 \sqrt {-1+a x} \sqrt {1+a x} \cosh ^{-1}(a x)}{9 a}+\frac {1}{3} x^3 \cosh ^{-1}(a x)^2\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 64, normalized size = 0.71 \[ \frac {1}{27} \left (2 x \left (\frac {6}{a^2}+x^2\right )-\frac {6 \sqrt {a x-1} \sqrt {a x+1} \left (a^2 x^2+2\right ) \cosh ^{-1}(a x)}{a^3}+9 x^3 \cosh ^{-1}(a x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(6/a^2 + x^2) - (6*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*(2 + a^2*x^2)*ArcCosh[a*x])/a^3 + 9*x^3*ArcCosh[a*x]^2)/2
7

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fricas [A]  time = 0.62, size = 82, normalized size = 0.91 \[ \frac {9 \, a^{3} x^{3} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )^{2} + 2 \, a^{3} x^{3} - 6 \, {\left (a^{2} x^{2} + 2\right )} \sqrt {a^{2} x^{2} - 1} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) + 12 \, a x}{27 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*(9*a^3*x^3*log(a*x + sqrt(a^2*x^2 - 1))^2 + 2*a^3*x^3 - 6*(a^2*x^2 + 2)*sqrt(a^2*x^2 - 1)*log(a*x + sqrt(
a^2*x^2 - 1)) + 12*a*x)/a^3

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 78, normalized size = 0.87 \[ \frac {\frac {a^{3} x^{3} \mathrm {arccosh}\left (a x \right )^{2}}{3}-\frac {4 \sqrt {a x -1}\, \sqrt {a x +1}\, \mathrm {arccosh}\left (a x \right )}{9}-\frac {2 \,\mathrm {arccosh}\left (a x \right ) \sqrt {a x -1}\, \sqrt {a x +1}\, a^{2} x^{2}}{9}+\frac {4 a x}{9}+\frac {2 x^{3} a^{3}}{27}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*(1/3*a^3*x^3*arccosh(a*x)^2-4/9*(a*x-1)^(1/2)*(a*x+1)^(1/2)*arccosh(a*x)-2/9*arccosh(a*x)*(a*x-1)^(1/2)*
(a*x+1)^(1/2)*a^2*x^2+4/9*a*x+2/27*x^3*a^3)

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maxima [A]  time = 0.69, size = 70, normalized size = 0.78 \[ \frac {1}{3} \, x^{3} \operatorname {arcosh}\left (a x\right )^{2} - \frac {2}{9} \, a {\left (\frac {\sqrt {a^{2} x^{2} - 1} x^{2}}{a^{2}} + \frac {2 \, \sqrt {a^{2} x^{2} - 1}}{a^{4}}\right )} \operatorname {arcosh}\left (a x\right ) + \frac {2 \, {\left (a^{2} x^{3} + 6 \, x\right )}}{27 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arccosh(a*x)^2 - 2/9*a*(sqrt(a^2*x^2 - 1)*x^2/a^2 + 2*sqrt(a^2*x^2 - 1)/a^4)*arccosh(a*x) + 2/27*(a^2*
x^3 + 6*x)/a^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {acosh}\left (a\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*acosh(a*x)^2, x)

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sympy [A]  time = 0.97, size = 85, normalized size = 0.94 \[ \begin {cases} \frac {x^{3} \operatorname {acosh}^{2}{\left (a x \right )}}{3} + \frac {2 x^{3}}{27} - \frac {2 x^{2} \sqrt {a^{2} x^{2} - 1} \operatorname {acosh}{\left (a x \right )}}{9 a} + \frac {4 x}{9 a^{2}} - \frac {4 \sqrt {a^{2} x^{2} - 1} \operatorname {acosh}{\left (a x \right )}}{9 a^{3}} & \text {for}\: a \neq 0 \\- \frac {\pi ^{2} x^{3}}{12} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*acosh(a*x)**2/3 + 2*x**3/27 - 2*x**2*sqrt(a**2*x**2 - 1)*acosh(a*x)/(9*a) + 4*x/(9*a**2) - 4*s
qrt(a**2*x**2 - 1)*acosh(a*x)/(9*a**3), Ne(a, 0)), (-pi**2*x**3/12, True))

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