3.4.44 \(\int \frac {x^2 \sinh ^{-1}(a x)^3}{\sqrt {1+a^2 x^2}} \, dx\) [344]

Optimal. Leaf size=105 \[ -\frac {3 x^2}{8 a}+\frac {3 x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{4 a^2}-\frac {3 \sinh ^{-1}(a x)^2}{8 a^3}-\frac {3 x^2 \sinh ^{-1}(a x)^2}{4 a}+\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{2 a^2}-\frac {\sinh ^{-1}(a x)^4}{8 a^3} \]

[Out]

-3/8*x^2/a-3/8*arcsinh(a*x)^2/a^3-3/4*x^2*arcsinh(a*x)^2/a-1/8*arcsinh(a*x)^4/a^3+3/4*x*arcsinh(a*x)*(a^2*x^2+
1)^(1/2)/a^2+1/2*x*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)/a^2

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Rubi [A]
time = 0.14, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5812, 5783, 5776, 30} \begin {gather*} -\frac {\sinh ^{-1}(a x)^4}{8 a^3}-\frac {3 \sinh ^{-1}(a x)^2}{8 a^3}+\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{2 a^2}+\frac {3 x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{4 a^2}-\frac {3 x^2}{8 a}-\frac {3 x^2 \sinh ^{-1}(a x)^2}{4 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

(-3*x^2)/(8*a) + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(4*a^2) - (3*ArcSinh[a*x]^2)/(8*a^3) - (3*x^2*ArcSinh[a*
x]^2)/(4*a) + (x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^3)/(2*a^2) - ArcSinh[a*x]^4/(8*a^3)

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

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Mathematica [A]
time = 0.05, size = 83, normalized size = 0.79 \begin {gather*} -\frac {3 a^2 x^2-6 a x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)+\left (3+6 a^2 x^2\right ) \sinh ^{-1}(a x)^2-4 a x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3+\sinh ^{-1}(a x)^4}{8 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2],x]

[Out]

-1/8*(3*a^2*x^2 - 6*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x] + (3 + 6*a^2*x^2)*ArcSinh[a*x]^2 - 4*a*x*Sqrt[1 + a^2*x
^2]*ArcSinh[a*x]^3 + ArcSinh[a*x]^4)/a^3

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Maple [A]
time = 2.59, size = 84, normalized size = 0.80

method	result	size
default	\(-\frac {-4 \arcsinh \left (a x \right )^{3} \sqrt {a^{2} x^{2}+1}\, a x +6 \arcsinh \left (a x \right )^{2} a^{2} x^{2}+\arcsinh \left (a x \right )^{4}-6 \sqrt {a^{2} x^{2}+1}\, \arcsinh \left (a x \right ) a x +3 a^{2} x^{2}+3 \arcsinh \left (a x \right )^{2}+3}{8 a^{3}}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*(-4*arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)*a*x+6*arcsinh(a*x)^2*a^2*x^2+arcsinh(a*x)^4-6*(a^2*x^2+1)^(1/2)*arcs
inh(a*x)*a*x+3*a^2*x^2+3*arcsinh(a*x)^2+3)/a^3

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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^2*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2*arcsinh(a*x)^3/sqrt(a^2*x^2 + 1), x)

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Fricas [A]
time = 0.41, size = 128, normalized size = 1.22 \begin {gather*} \frac {4 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} - 3 \, a^{2} x^{2} - \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{4} + 6 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 3 \, {\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{8 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/8*(4*sqrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1))^3 - 3*a^2*x^2 - log(a*x + sqrt(a^2*x^2 + 1))^4 + 6*s
qrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1)) - 3*(2*a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1))^2)/a^3

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Sympy [A]
time = 0.47, size = 100, normalized size = 0.95 \begin {gather*} \begin {cases} - \frac {3 x^{2} \operatorname {asinh}^{2}{\left (a x \right )}}{4 a} - \frac {3 x^{2}}{8 a} + \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{3}{\left (a x \right )}}{2 a^{2}} + \frac {3 x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{4 a^{2}} - \frac {\operatorname {asinh}^{4}{\left (a x \right )}}{8 a^{3}} - \frac {3 \operatorname {asinh}^{2}{\left (a x \right )}}{8 a^{3}} & \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**2*asinh(a*x)**3/(a**2*x**2+1)**(1/2),x)

[Out]

Piecewise((-3*x**2*asinh(a*x)**2/(4*a) - 3*x**2/(8*a) + x*sqrt(a**2*x**2 + 1)*asinh(a*x)**3/(2*a**2) + 3*x*sqr
t(a**2*x**2 + 1)*asinh(a*x)/(4*a**2) - asinh(a*x)**4/(8*a**3) - 3*asinh(a*x)**2/(8*a**3), Ne(a, 0)), (0, True)
)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(a*x)^3/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^2*arcsinh(a*x)^3/sqrt(a^2*x^2 + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\mathrm {asinh}\left (a\,x\right )}^3}{\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*asinh(a*x)^3)/(a^2*x^2 + 1)^(1/2),x)

[Out]

int((x^2*asinh(a*x)^3)/(a^2*x^2 + 1)^(1/2), x)

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