3.1.12 \(\int x^4 \sinh ^{-1}(a x)^2 \, dx\) [12]

Optimal. Leaf size=117 \[ \frac {16 x}{75 a^4}-\frac {8 x^3}{225 a^2}+\frac {2 x^5}{125}-\frac {16 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^5}+\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac {2 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2 \]

[Out]

16/75*x/a^4-8/225*x^3/a^2+2/125*x^5+1/5*x^5*arcsinh(a*x)^2-16/75*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^5+8/75*x^2*a
rcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^3-2/25*x^4*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a

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Rubi [A]
time = 0.12, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5776, 5812, 5798, 8, 30} \begin {gather*} \frac {16 x}{75 a^4}-\frac {8 x^3}{225 a^2}-\frac {2 x^4 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{25 a}-\frac {16 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^5}+\frac {8 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{75 a^3}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2+\frac {2 x^5}{125} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSinh[a*x]^2,x]

[Out]

(16*x)/(75*a^4) - (8*x^3)/(225*a^2) + (2*x^5)/125 - (16*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^5) + (8*x^2*Sqrt
[1 + a^2*x^2]*ArcSinh[a*x])/(75*a^3) - (2*x^4*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(25*a) + (x^5*ArcSinh[a*x]^2)/5

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 5776

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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5812

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

Rubi steps

\begin {align*} \int x^4 \sinh ^{-1}(a x)^2 \, dx &=\frac {1}{5} x^5 \sinh ^{-1}(a x)^2-\frac {1}{5} (2 a) \int \frac {x^5 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {2 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2+\frac {2 \int x^4 \, dx}{25}+\frac {8 \int \frac {x^3 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{25 a}\\ &=\frac {2 x^5}{125}+\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac {2 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2-\frac {16 \int \frac {x \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{75 a^3}-\frac {8 \int x^2 \, dx}{75 a^2}\\ &=-\frac {8 x^3}{225 a^2}+\frac {2 x^5}{125}-\frac {16 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^5}+\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac {2 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2+\frac {16 \int 1 \, dx}{75 a^4}\\ &=\frac {16 x}{75 a^4}-\frac {8 x^3}{225 a^2}+\frac {2 x^5}{125}-\frac {16 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^5}+\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{75 a^3}-\frac {2 x^4 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{25 a}+\frac {1}{5} x^5 \sinh ^{-1}(a x)^2\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 75, normalized size = 0.64 \begin {gather*} \frac {\frac {240 x}{a^4}-\frac {40 x^3}{a^2}+18 x^5-\frac {30 \sqrt {1+a^2 x^2} \left (8-4 a^2 x^2+3 a^4 x^4\right ) \sinh ^{-1}(a x)}{a^5}+225 x^5 \sinh ^{-1}(a x)^2}{1125} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSinh[a*x]^2,x]

[Out]

((240*x)/a^4 - (40*x^3)/a^2 + 18*x^5 - (30*Sqrt[1 + a^2*x^2]*(8 - 4*a^2*x^2 + 3*a^4*x^4)*ArcSinh[a*x])/a^5 + 2
25*x^5*ArcSinh[a*x]^2)/1125

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{4} \arcsinh \left (a x \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arcsinh(a*x)^2,x)

[Out]

int(x^4*arcsinh(a*x)^2,x)

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Maxima [A]
time = 0.28, size = 99, normalized size = 0.85 \begin {gather*} \frac {1}{5} \, x^{5} \operatorname {arsinh}\left (a x\right )^{2} - \frac {2}{75} \, {\left (\frac {3 \, \sqrt {a^{2} x^{2} + 1} x^{4}}{a^{2}} - \frac {4 \, \sqrt {a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac {8 \, \sqrt {a^{2} x^{2} + 1}}{a^{6}}\right )} a \operatorname {arsinh}\left (a x\right ) + \frac {2 \, {\left (9 \, a^{4} x^{5} - 20 \, a^{2} x^{3} + 120 \, x\right )}}{1125 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2,x, algorithm="maxima")

[Out]

1/5*x^5*arcsinh(a*x)^2 - 2/75*(3*sqrt(a^2*x^2 + 1)*x^4/a^2 - 4*sqrt(a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(a^2*x^2 + 1)
/a^6)*a*arcsinh(a*x) + 2/1125*(9*a^4*x^5 - 20*a^2*x^3 + 120*x)/a^4

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Fricas [A]
time = 0.35, size = 99, normalized size = 0.85 \begin {gather*} \frac {225 \, a^{5} x^{5} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 18 \, a^{5} x^{5} - 40 \, a^{3} x^{3} - 30 \, {\left (3 \, a^{4} x^{4} - 4 \, a^{2} x^{2} + 8\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + 240 \, a x}{1125 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arcsinh(a*x)^2,x, algorithm="fricas")

[Out]

1/1125*(225*a^5*x^5*log(a*x + sqrt(a^2*x^2 + 1))^2 + 18*a^5*x^5 - 40*a^3*x^3 - 30*(3*a^4*x^4 - 4*a^2*x^2 + 8)*
sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)) + 240*a*x)/a^5

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Sympy [A]
time = 0.45, size = 114, normalized size = 0.97 \begin {gather*} \begin {cases} \frac {x^{5} \operatorname {asinh}^{2}{\left (a x \right )}}{5} + \frac {2 x^{5}}{125} - \frac {2 x^{4} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{25 a} - \frac {8 x^{3}}{225 a^{2}} + \frac {8 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{75 a^{3}} + \frac {16 x}{75 a^{4}} - \frac {16 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{75 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*asinh(a*x)**2,x)

[Out]

Piecewise((x**5*asinh(a*x)**2/5 + 2*x**5/125 - 2*x**4*sqrt(a**2*x**2 + 1)*asinh(a*x)/(25*a) - 8*x**3/(225*a**2
) + 8*x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)/(75*a**3) + 16*x/(75*a**4) - 16*sqrt(a**2*x**2 + 1)*asinh(a*x)/(75*a
**5), Ne(a, 0)), (0, 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^4*arcsinh(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^4\,{\mathrm {asinh}\left (a\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*asinh(a*x)^2,x)

[Out]

int(x^4*asinh(a*x)^2, x)

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