3.34 \(\int x^3 \sinh ^{-1}(a x)^4 \, dx\)

Optimal. Leaf size=194 \[ -\frac{45 x^2}{128 a^2}-\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{4 a}-\frac{3 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{32 a}-\frac{9 x^2 \sinh ^{-1}(a x)^2}{16 a^2}+\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{8 a^3}+\frac{45 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{64 a^3}-\frac{3 \sinh ^{-1}(a x)^4}{32 a^4}-\frac{45 \sinh ^{-1}(a x)^2}{128 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4+\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x^4}{128} \]

[Out]

(-45*x^2)/(128*a^2) + (3*x^4)/128 + (45*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(64*a^3) - (3*x^3*Sqrt[1 + a^2*x^2]*
ArcSinh[a*x])/(32*a) - (45*ArcSinh[a*x]^2)/(128*a^4) - (9*x^2*ArcSinh[a*x]^2)/(16*a^2) + (3*x^4*ArcSinh[a*x]^2
)/16 + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(8*a^3) - (x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(4*a) - (3*ArcS
inh[a*x]^4)/(32*a^4) + (x^4*ArcSinh[a*x]^4)/4

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Rubi [A]  time = 0.504404, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5661, 5758, 5675, 30} \[ -\frac{45 x^2}{128 a^2}-\frac{x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{4 a}-\frac{3 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{32 a}-\frac{9 x^2 \sinh ^{-1}(a x)^2}{16 a^2}+\frac{3 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^3}{8 a^3}+\frac{45 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{64 a^3}-\frac{3 \sinh ^{-1}(a x)^4}{32 a^4}-\frac{45 \sinh ^{-1}(a x)^2}{128 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4+\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x^4}{128} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a*x]^4,x]

[Out]

(-45*x^2)/(128*a^2) + (3*x^4)/128 + (45*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(64*a^3) - (3*x^3*Sqrt[1 + a^2*x^2]*
ArcSinh[a*x])/(32*a) - (45*ArcSinh[a*x]^2)/(128*a^4) - (9*x^2*ArcSinh[a*x]^2)/(16*a^2) + (3*x^4*ArcSinh[a*x]^2
)/16 + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(8*a^3) - (x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(4*a) - (3*ArcS
inh[a*x]^4)/(32*a^4) + (x^4*ArcSinh[a*x]^4)/4

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 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]

Rule 5675

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int x^3 \sinh ^{-1}(a x)^4 \, dx &=\frac{1}{4} x^4 \sinh ^{-1}(a x)^4-a \int \frac{x^4 \sinh ^{-1}(a x)^3}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{4 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4+\frac{3}{4} \int x^3 \sinh ^{-1}(a x)^2 \, dx+\frac{3 \int \frac{x^2 \sinh ^{-1}(a x)^3}{\sqrt{1+a^2 x^2}} \, dx}{4 a}\\ &=\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{8 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{4 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4-\frac{3 \int \frac{\sinh ^{-1}(a x)^3}{\sqrt{1+a^2 x^2}} \, dx}{8 a^3}-\frac{9 \int x \sinh ^{-1}(a x)^2 \, dx}{8 a^2}-\frac{1}{8} (3 a) \int \frac{x^4 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{32 a}-\frac{9 x^2 \sinh ^{-1}(a x)^2}{16 a^2}+\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{8 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{4 a}-\frac{3 \sinh ^{-1}(a x)^4}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4+\frac{3 \int x^3 \, dx}{32}+\frac{9 \int \frac{x^2 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{32 a}+\frac{9 \int \frac{x^2 \sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{8 a}\\ &=\frac{3 x^4}{128}+\frac{45 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{64 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{32 a}-\frac{9 x^2 \sinh ^{-1}(a x)^2}{16 a^2}+\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{8 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{4 a}-\frac{3 \sinh ^{-1}(a x)^4}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4-\frac{9 \int \frac{\sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{64 a^3}-\frac{9 \int \frac{\sinh ^{-1}(a x)}{\sqrt{1+a^2 x^2}} \, dx}{16 a^3}-\frac{9 \int x \, dx}{64 a^2}-\frac{9 \int x \, dx}{16 a^2}\\ &=-\frac{45 x^2}{128 a^2}+\frac{3 x^4}{128}+\frac{45 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{64 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)}{32 a}-\frac{45 \sinh ^{-1}(a x)^2}{128 a^4}-\frac{9 x^2 \sinh ^{-1}(a x)^2}{16 a^2}+\frac{3}{16} x^4 \sinh ^{-1}(a x)^2+\frac{3 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{8 a^3}-\frac{x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3}{4 a}-\frac{3 \sinh ^{-1}(a x)^4}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^4\\ \end{align*}

Mathematica [A]  time = 0.070216, size = 133, normalized size = 0.69 \[ \frac{3 a^2 x^2 \left (a^2 x^2-15\right )+4 \left (8 a^4 x^4-3\right ) \sinh ^{-1}(a x)^4-16 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-3\right ) \sinh ^{-1}(a x)^3+3 \left (8 a^4 x^4-24 a^2 x^2-15\right ) \sinh ^{-1}(a x)^2-6 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-15\right ) \sinh ^{-1}(a x)}{128 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a*x]^4,x]

[Out]

(3*a^2*x^2*(-15 + a^2*x^2) - 6*a*x*Sqrt[1 + a^2*x^2]*(-15 + 2*a^2*x^2)*ArcSinh[a*x] + 3*(-15 - 24*a^2*x^2 + 8*
a^4*x^4)*ArcSinh[a*x]^2 - 16*a*x*Sqrt[1 + a^2*x^2]*(-3 + 2*a^2*x^2)*ArcSinh[a*x]^3 + 4*(-3 + 8*a^4*x^4)*ArcSin
h[a*x]^4)/(128*a^4)

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Maple [A]  time = 0.036, size = 208, normalized size = 1.1 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}{a}^{2}{x}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}ax}{4} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{5\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}ax}{8}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{5\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{4}}{32}}+{\frac{3\,{a}^{2}{x}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{16}}-{\frac{3\,{\it Arcsinh} \left ( ax \right ) ax}{32} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{51\,{\it Arcsinh} \left ( ax \right ) ax}{64}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{51\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{128}}+{\frac{3\,{a}^{2}{x}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{128}}-{\frac{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{3\,{a}^{2}{x}^{2}}{8}}-{\frac{3}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(a*x)^4,x)

[Out]

1/a^4*(1/4*arcsinh(a*x)^4*a^2*x^2*(a^2*x^2+1)-1/4*arcsinh(a*x)^4*(a^2*x^2+1)-1/4*arcsinh(a*x)^3*a*x*(a^2*x^2+1
)^(3/2)+5/8*arcsinh(a*x)^3*a*x*(a^2*x^2+1)^(1/2)+5/32*arcsinh(a*x)^4+3/16*a^2*x^2*arcsinh(a*x)^2*(a^2*x^2+1)-3
/32*arcsinh(a*x)*a*x*(a^2*x^2+1)^(3/2)+51/64*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a*x+51/128*arcsinh(a*x)^2+3/128*a^
2*x^2*(a^2*x^2+1)-3/4*arcsinh(a*x)^2*(a^2*x^2+1)-3/8*a^2*x^2-3/8)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, x^{4} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{4} - \int \frac{{\left (a^{3} x^{6} + \sqrt{a^{2} x^{2} + 1} a^{2} x^{5} + a x^{4}\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3}}{a^{3} x^{3} + a x +{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^4,x, algorithm="maxima")

[Out]

1/4*x^4*log(a*x + sqrt(a^2*x^2 + 1))^4 - integrate((a^3*x^6 + sqrt(a^2*x^2 + 1)*a^2*x^5 + a*x^4)*log(a*x + sqr
t(a^2*x^2 + 1))^3/(a^3*x^3 + a*x + (a^2*x^2 + 1)^(3/2)), x)

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Fricas [A]  time = 2.10842, size = 402, normalized size = 2.07 \begin{align*} \frac{3 \, a^{4} x^{4} + 4 \,{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{4} - 16 \,{\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} - 45 \, a^{2} x^{2} + 3 \,{\left (8 \, a^{4} x^{4} - 24 \, a^{2} x^{2} - 15\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} - 6 \,{\left (2 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{128 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^4,x, algorithm="fricas")

[Out]

1/128*(3*a^4*x^4 + 4*(8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 + 1))^4 - 16*(2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^2 + 1)
*log(a*x + sqrt(a^2*x^2 + 1))^3 - 45*a^2*x^2 + 3*(8*a^4*x^4 - 24*a^2*x^2 - 15)*log(a*x + sqrt(a^2*x^2 + 1))^2
- 6*(2*a^3*x^3 - 15*a*x)*sqrt(a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^4

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Sympy [A]  time = 7.5164, size = 190, normalized size = 0.98 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asinh}^{4}{\left (a x \right )}}{4} + \frac{3 x^{4} \operatorname{asinh}^{2}{\left (a x \right )}}{16} + \frac{3 x^{4}}{128} - \frac{x^{3} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{3}{\left (a x \right )}}{4 a} - \frac{3 x^{3} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{32 a} - \frac{9 x^{2} \operatorname{asinh}^{2}{\left (a x \right )}}{16 a^{2}} - \frac{45 x^{2}}{128 a^{2}} + \frac{3 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{3}{\left (a x \right )}}{8 a^{3}} + \frac{45 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}{\left (a x \right )}}{64 a^{3}} - \frac{3 \operatorname{asinh}^{4}{\left (a x \right )}}{32 a^{4}} - \frac{45 \operatorname{asinh}^{2}{\left (a x \right )}}{128 a^{4}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asinh(a*x)**4,x)

[Out]

Piecewise((x**4*asinh(a*x)**4/4 + 3*x**4*asinh(a*x)**2/16 + 3*x**4/128 - x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)**
3/(4*a) - 3*x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)/(32*a) - 9*x**2*asinh(a*x)**2/(16*a**2) - 45*x**2/(128*a**2) +
3*x*sqrt(a**2*x**2 + 1)*asinh(a*x)**3/(8*a**3) + 45*x*sqrt(a**2*x**2 + 1)*asinh(a*x)/(64*a**3) - 3*asinh(a*x)
**4/(32*a**4) - 45*asinh(a*x)**2/(128*a**4), Ne(a, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsinh}\left (a x\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x)^4,x, algorithm="giac")

[Out]

integrate(x^3*arcsinh(a*x)^4, x)