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

Optimal. Leaf size=80 \[ -\frac {2 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a}-\frac {4 x}{9 a^2}+\frac {4 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a^3}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^2+\frac {2 x^3}{27} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 80, 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 = {5661, 5758, 5717, 8, 30} \[ -\frac {2 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a}+\frac {4 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{9 a^3}-\frac {4 x}{9 a^2}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^2+\frac {2 x^3}{27} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x)/(9*a^2) + (2*x^3)/27 + (4*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(9*a^3) - (2*x^2*Sqrt[1 + a^2*x^2]*ArcSinh[a*
x])/(9*a) + (x^3*ArcSinh[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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5717

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

Rule 5758

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 59, normalized size = 0.74 \[ \frac {1}{27} \left (2 x \left (x^2-\frac {6}{a^2}\right )-\frac {6 \left (a^2 x^2-2\right ) \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{a^3}+9 x^3 \sinh ^{-1}(a x)^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-6/a^2 + x^2) - (6*(-2 + a^2*x^2)*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/a^3 + 9*x^3*ArcSinh[a*x]^2)/27

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 82, normalized size = 1.02 \[ \frac {9 \, a^{3} x^{3} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 2 \, a^{3} x^{3} - 6 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 2\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 12 \, a x}{27 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(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*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 2)*log(a*x + sqrt(
a^2*x^2 + 1)) - 12*a*x)/a^3

________________________________________________________________________________________

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*arcsinh(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

________________________________________________________________________________________

maple [A]  time = 0.38, size = 72, normalized size = 0.90 \[ \frac {\frac {a^{3} x^{3} \arcsinh \left (a x \right )^{2}}{3}+\frac {4 \sqrt {a^{2} x^{2}+1}\, \arcsinh \left (a x \right )}{9}-\frac {2 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}}{9}-\frac {4 a x}{9}+\frac {2 a^{3} x^{3}}{27}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 70, normalized size = 0.88 \[ \frac {1}{3} \, x^{3} \operatorname {arsinh}\left (a x\right )^{2} - \frac {2}{9} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname {arsinh}\left (a x\right ) + \frac {2 \, {\left (a^{2} x^{3} - 6 \, x\right )}}{27 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsinh(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)*arcsinh(a*x) + 2/27*(a^2*
x^3 - 6*x)/a^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*asinh(a*x)^2, x)

________________________________________________________________________________________

sympy [A]  time = 0.86, size = 76, normalized size = 0.95 \[ \begin {cases} \frac {x^{3} \operatorname {asinh}^{2}{\left (a x \right )}}{3} + \frac {2 x^{3}}{27} - \frac {2 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{9 a} - \frac {4 x}{9 a^{2}} + \frac {4 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{9 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________