3.35 \(\int x^2 \sinh ^{-1}(a x)^4 \, dx\)

Optimal. Leaf size=162 \[ -\frac {4 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{9 a}-\frac {8 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{27 a}-\frac {160 x}{27 a^2}-\frac {8 x \sinh ^{-1}(a x)^2}{3 a^2}+\frac {8 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{9 a^3}+\frac {160 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{27 a^3}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4+\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 x^3}{81} \]

[Out]

-160/27*x/a^2+8/81*x^3-8/3*x*arcsinh(a*x)^2/a^2+4/9*x^3*arcsinh(a*x)^2+1/3*x^3*arcsinh(a*x)^4+160/27*arcsinh(a
*x)*(a^2*x^2+1)^(1/2)/a^3-8/27*x^2*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a+8/9*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a^3-4
/9*x^2*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a

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Rubi [A]  time = 0.36, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5661, 5758, 5717, 5653, 8, 30} \[ -\frac {4 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{9 a}+\frac {8 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{9 a^3}-\frac {8 x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{27 a}+\frac {160 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{27 a^3}-\frac {160 x}{27 a^2}-\frac {8 x \sinh ^{-1}(a x)^2}{3 a^2}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4+\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 x^3}{81} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-160*x)/(27*a^2) + (8*x^3)/81 + (160*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(27*a^3) - (8*x^2*Sqrt[1 + a^2*x^2]*ArcS
inh[a*x])/(27*a) - (8*x*ArcSinh[a*x]^2)/(3*a^2) + (4*x^3*ArcSinh[a*x]^2)/9 + (8*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]
^3)/(9*a^3) - (4*x^2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(9*a) + (x^3*ArcSinh[a*x]^4)/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 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

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)^4 \, dx &=\frac {1}{3} x^3 \sinh ^{-1}(a x)^4-\frac {1}{3} (4 a) \int \frac {x^3 \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4+\frac {4}{3} \int x^2 \sinh ^{-1}(a x)^2 \, dx+\frac {8 \int \frac {x \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx}{9 a}\\ &=\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a^3}-\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4-\frac {8 \int \sinh ^{-1}(a x)^2 \, dx}{3 a^2}-\frac {1}{9} (8 a) \int \frac {x^3 \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{27 a}-\frac {8 x \sinh ^{-1}(a x)^2}{3 a^2}+\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a^3}-\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4+\frac {8 \int x^2 \, dx}{27}+\frac {16 \int \frac {x \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{27 a}+\frac {16 \int \frac {x \sinh ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{3 a}\\ &=\frac {8 x^3}{81}+\frac {160 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{27 a^3}-\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{27 a}-\frac {8 x \sinh ^{-1}(a x)^2}{3 a^2}+\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a^3}-\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4-\frac {16 \int 1 \, dx}{27 a^2}-\frac {16 \int 1 \, dx}{3 a^2}\\ &=-\frac {160 x}{27 a^2}+\frac {8 x^3}{81}+\frac {160 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{27 a^3}-\frac {8 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{27 a}-\frac {8 x \sinh ^{-1}(a x)^2}{3 a^2}+\frac {4}{9} x^3 \sinh ^{-1}(a x)^2+\frac {8 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a^3}-\frac {4 x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{9 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^4\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 112, normalized size = 0.69 \[ \frac {27 a^3 x^3 \sinh ^{-1}(a x)^4+8 a x \left (a^2 x^2-60\right )-36 \left (a^2 x^2-2\right ) \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3+36 a x \left (a^2 x^2-6\right ) \sinh ^{-1}(a x)^2-24 \left (a^2 x^2-20\right ) \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{81 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*x*(-60 + a^2*x^2) - 24*(-20 + a^2*x^2)*Sqrt[1 + a^2*x^2]*ArcSinh[a*x] + 36*a*x*(-6 + a^2*x^2)*ArcSinh[a*x
]^2 - 36*(-2 + a^2*x^2)*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3 + 27*a^3*x^3*ArcSinh[a*x]^4)/(81*a^3)

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fricas [A]  time = 0.42, size = 154, normalized size = 0.95 \[ \frac {27 \, a^{3} x^{3} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{4} + 8 \, a^{3} x^{3} - 36 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 2\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} + 36 \, {\left (a^{3} x^{3} - 6 \, a x\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - 24 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 20\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 480 \, a x}{81 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/81*(27*a^3*x^3*log(a*x + sqrt(a^2*x^2 + 1))^4 + 8*a^3*x^3 - 36*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 2)*log(a*x + sqr
t(a^2*x^2 + 1))^3 + 36*(a^3*x^3 - 6*a*x)*log(a*x + sqrt(a^2*x^2 + 1))^2 - 24*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 20)*
log(a*x + sqrt(a^2*x^2 + 1)) - 480*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*arcsinh(a*x)^4,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 = 0.38, size = 140, normalized size = 0.86 \[ \frac {\frac {a^{3} x^{3} \arcsinh \left (a x \right )^{4}}{3}+\frac {8 \arcsinh \left (a x \right )^{3} \sqrt {a^{2} x^{2}+1}}{9}-\frac {4 a^{2} x^{2} \arcsinh \left (a x \right )^{3} \sqrt {a^{2} x^{2}+1}}{9}-\frac {8 a x \arcsinh \left (a x \right )^{2}}{3}+\frac {160 \sqrt {a^{2} x^{2}+1}\, \arcsinh \left (a x \right )}{27}-\frac {160 a x}{27}+\frac {4 a^{3} x^{3} \arcsinh \left (a x \right )^{2}}{9}-\frac {8 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}}{27}+\frac {8 a^{3} x^{3}}{81}}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*(1/3*a^3*x^3*arcsinh(a*x)^4+8/9*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)-4/9*a^2*x^2*arcsinh(a*x)^3*(a^2*x^2+1)^
(1/2)-8/3*a*x*arcsinh(a*x)^2+160/27*(a^2*x^2+1)^(1/2)*arcsinh(a*x)-160/27*a*x+4/9*a^3*x^3*arcsinh(a*x)^2-8/27*
arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a^2*x^2+8/81*a^3*x^3)

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maxima [A]  time = 0.33, size = 143, normalized size = 0.88 \[ \frac {1}{3} \, x^{3} \operatorname {arsinh}\left (a x\right )^{4} - \frac {4}{9} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname {arsinh}\left (a x\right )^{3} - \frac {4}{81} \, {\left (2 \, a {\left (\frac {3 \, {\left (\sqrt {a^{2} x^{2} + 1} x^{2} - \frac {20 \, \sqrt {a^{2} x^{2} + 1}}{a^{2}}\right )} \operatorname {arsinh}\left (a x\right )}{a^{3}} - \frac {a^{2} x^{3} - 60 \, x}{a^{4}}\right )} - \frac {9 \, {\left (a^{2} x^{3} - 6 \, x\right )} \operatorname {arsinh}\left (a x\right )^{2}}{a^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arcsinh(a*x)^4 - 4/9*a*(sqrt(a^2*x^2 + 1)*x^2/a^2 - 2*sqrt(a^2*x^2 + 1)/a^4)*arcsinh(a*x)^3 - 4/81*(2*
a*(3*(sqrt(a^2*x^2 + 1)*x^2 - 20*sqrt(a^2*x^2 + 1)/a^2)*arcsinh(a*x)/a^3 - (a^2*x^3 - 60*x)/a^4) - 9*(a^2*x^3
- 6*x)*arcsinh(a*x)^2/a^3)*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {asinh}\left (a\,x\right )}^4 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*asinh(a*x)^4, x)

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sympy [A]  time = 3.08, size = 158, normalized size = 0.98 \[ \begin {cases} \frac {x^{3} \operatorname {asinh}^{4}{\left (a x \right )}}{3} + \frac {4 x^{3} \operatorname {asinh}^{2}{\left (a x \right )}}{9} + \frac {8 x^{3}}{81} - \frac {4 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{9 a} - \frac {8 x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{27 a} - \frac {8 x \operatorname {asinh}^{2}{\left (a x \right )}}{3 a^{2}} - \frac {160 x}{27 a^{2}} + \frac {8 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{9 a^{3}} + \frac {160 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{27 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**3*asinh(a*x)**4/3 + 4*x**3*asinh(a*x)**2/9 + 8*x**3/81 - 4*x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)**
3/(9*a) - 8*x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)/(27*a) - 8*x*asinh(a*x)**2/(3*a**2) - 160*x/(27*a**2) + 8*sqrt
(a**2*x**2 + 1)*asinh(a*x)**3/(9*a**3) + 160*sqrt(a**2*x**2 + 1)*asinh(a*x)/(27*a**3), Ne(a, 0)), (0, True))

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