3.1.34 \(\int x^3 \sinh ^{-1}(a x)^4 \, dx\) [34]

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

[Out]

-45/128*x^2/a^2+3/128*x^4-45/128*arcsinh(a*x)^2/a^4-9/16*x^2*arcsinh(a*x)^2/a^2+3/16*x^4*arcsinh(a*x)^2-3/32*a
rcsinh(a*x)^4/a^4+1/4*x^4*arcsinh(a*x)^4+45/64*x*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^3-3/32*x^3*arcsinh(a*x)*(a^2
*x^2+1)^(1/2)/a+3/8*x*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a^3-1/4*x^3*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a

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Rubi [A]
time = 0.33, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5776, 5812, 5783, 30} \begin {gather*} -\frac {3 \sinh ^{-1}(a x)^4}{32 a^4}-\frac {45 \sinh ^{-1}(a x)^2}{128 a^4}-\frac {45 x^2}{128 a^2}-\frac {9 x^2 \sinh ^{-1}(a x)^2}{16 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 {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 {1}{4} x^4 \sinh ^{-1}(a x)^4+\frac {3}{16} x^4 \sinh ^{-1}(a x)^2+\frac {3 x^4}{128} \end {gather*}

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 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 5783

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

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Mathematica [A]
time = 0.05, size = 133, normalized size = 0.69 \begin {gather*} \frac {3 a^2 x^2 \left (-15+a^2 x^2\right )-6 a x \sqrt {1+a^2 x^2} \left (-15+2 a^2 x^2\right ) \sinh ^{-1}(a x)+3 \left (-15-24 a^2 x^2+8 a^4 x^4\right ) \sinh ^{-1}(a x)^2-16 a x \sqrt {1+a^2 x^2} \left (-3+2 a^2 x^2\right ) \sinh ^{-1}(a x)^3+4 \left (-3+8 a^4 x^4\right ) \sinh ^{-1}(a x)^4}{128 a^4} \end {gather*}

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 [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{3} \arcsinh \left (a x \right )^{4}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.36, size = 176, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.69, size = 190, normalized size = 0.98 \begin {gather*} \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 {gather*}

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(-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^3*arcsinh(a*x)^4,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^3\,{\mathrm {asinh}\left (a\,x\right )}^4 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*asinh(a*x)^4, x)

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