3.1.14 \(\int x^2 \cosh ^{-1}(a x)^2 \, dx\) [14]

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

[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 = 0.20, 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 = {5883, 5939, 5915, 8, 30} \begin {gather*} -\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} \end {gather*}

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 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]

Rule 5915

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_))^(p_), x_Sy
mbol] :> 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/(2*c*(p + 1)))*Simp[(d1 + e1*x)^p/(1 + c*x)^p]*Simp[(d2 + e2*x)^p/(-1 + c*x)^p], Int[(1 + c*x)^(p + 1/2)*(-
1 + c*x)^(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] && EqQ[e2, (-c)*d2] && GtQ[n, 0] && NeQ[p, -1]

Rule 5939

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

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 = 0.07, size = 64, normalized size = 0.71 \begin {gather*} \frac {1}{27} \left (2 x \left (\frac {6}{a^2}+x^2\right )-\frac {6 \sqrt {-1+a x} \sqrt {1+a x} \left (2+a^2 x^2\right ) \cosh ^{-1}(a x)}{a^3}+9 x^3 \cosh ^{-1}(a x)^2\right ) \end {gather*}

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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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \mathrm {arccosh}\left (a x \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 70, normalized size = 0.78 \begin {gather*} \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}} \end {gather*}

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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Fricas [A]
time = 0.37, size = 82, normalized size = 0.91 \begin {gather*} \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}} \end {gather*}

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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Sympy [A]
time = 0.19, size = 85, normalized size = 0.94 \begin {gather*} \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} \end {gather*}

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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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)^2,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 = -0.01 \begin {gather*} \int x^2\,{\mathrm {acosh}\left (a\,x\right )}^2 \,d x \end {gather*}

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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