3.1.3 \(\int x^2 \cosh ^{-1}(a x) \, dx\) [3]

Optimal. Leaf size=65 \[ -\frac {2 \sqrt {-1+a x} \sqrt {1+a x}}{9 a^3}-\frac {x^2 \sqrt {-1+a x} \sqrt {1+a x}}{9 a}+\frac {1}{3} x^3 \cosh ^{-1}(a x) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5883, 102, 12, 75} \begin {gather*} -\frac {2 \sqrt {a x-1} \sqrt {a x+1}}{9 a^3}+\frac {1}{3} x^3 \cosh ^{-1}(a x)-\frac {x^2 \sqrt {a x-1} \sqrt {a x+1}}{9 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 75

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] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 102

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 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 46, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {-1+a x} \sqrt {1+a x} \left (2+a^2 x^2\right )}{9 a^3}+\frac {1}{3} x^3 \cosh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.47, size = 43, normalized size = 0.66

method	result	size
derivativedivides	\(\frac {\frac {a^{3} x^{3} \mathrm {arccosh}\left (a x \right )}{3}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (a^{2} x^{2}+2\right )}{9}}{a^{3}}\)	\(43\)
default	\(\frac {\frac {a^{3} x^{3} \mathrm {arccosh}\left (a x \right )}{3}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (a^{2} x^{2}+2\right )}{9}}{a^{3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccosh(a*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 48, normalized size = 0.74 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {arcosh}\left (a x\right ) - \frac {1}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 52, normalized size = 0.80 \begin {gather*} \frac {3 \, a^{3} x^{3} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) - {\left (a^{2} x^{2} + 2\right )} \sqrt {a^{2} x^{2} - 1}}{9 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 0.13, size = 54, normalized size = 0.83 \begin {gather*} \begin {cases} \frac {x^{3} \operatorname {acosh}{\left (a x \right )}}{3} - \frac {x^{2} \sqrt {a^{2} x^{2} - 1}}{9 a} - \frac {2 \sqrt {a^{2} x^{2} - 1}}{9 a^{3}} & \text {for}\: a \neq 0 \\\frac {i \pi x^{3}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*acosh(a*x)/3 - x**2*sqrt(a**2*x**2 - 1)/(9*a) - 2*sqrt(a**2*x**2 - 1)/(9*a**3), Ne(a, 0)), (I*
pi*x**3/6, True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,\mathrm {acosh}\left (a\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*acosh(a*x), x)

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