3.1.2 \(\int x^3 \sinh ^{-1}(a x) \, dx\) [2]

Optimal. Leaf size=67 \[ \frac {3 x \sqrt {1+a^2 x^2}}{32 a^3}-\frac {x^3 \sqrt {1+a^2 x^2}}{16 a}-\frac {3 \sinh ^{-1}(a x)}{32 a^4}+\frac {1}{4} x^4 \sinh ^{-1}(a x) \]

[Out]

-3/32*arcsinh(a*x)/a^4+1/4*x^4*arcsinh(a*x)+3/32*x*(a^2*x^2+1)^(1/2)/a^3-1/16*x^3*(a^2*x^2+1)^(1/2)/a

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Rubi [A]
time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5776, 327, 221} \begin {gather*} -\frac {3 \sinh ^{-1}(a x)}{32 a^4}-\frac {x^3 \sqrt {a^2 x^2+1}}{16 a}+\frac {3 x \sqrt {a^2 x^2+1}}{32 a^3}+\frac {1}{4} x^4 \sinh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x*Sqrt[1 + a^2*x^2])/(32*a^3) - (x^3*Sqrt[1 + a^2*x^2])/(16*a) - (3*ArcSinh[a*x])/(32*a^4) + (x^4*ArcSinh[a
*x])/4

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 49, normalized size = 0.73 \begin {gather*} \frac {a x \left (3-2 a^2 x^2\right ) \sqrt {1+a^2 x^2}+\left (-3+8 a^4 x^4\right ) \sinh ^{-1}(a x)}{32 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(3 - 2*a^2*x^2)*Sqrt[1 + a^2*x^2] + (-3 + 8*a^4*x^4)*ArcSinh[a*x])/(32*a^4)

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Maple [A]
time = 0.19, size = 58, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {\frac {a^{4} x^{4} \arcsinh \left (a x \right )}{4}-\frac {a^{3} x^{3} \sqrt {a^{2} x^{2}+1}}{16}+\frac {3 a x \sqrt {a^{2} x^{2}+1}}{32}-\frac {3 \arcsinh \left (a x \right )}{32}}{a^{4}}\)	\(58\)
default	\(\frac {\frac {a^{4} x^{4} \arcsinh \left (a x \right )}{4}-\frac {a^{3} x^{3} \sqrt {a^{2} x^{2}+1}}{16}+\frac {3 a x \sqrt {a^{2} x^{2}+1}}{32}-\frac {3 \arcsinh \left (a x \right )}{32}}{a^{4}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(a*x),x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4*a^4*x^4*arcsinh(a*x)-1/16*a^3*x^3*(a^2*x^2+1)^(1/2)+3/32*a*x*(a^2*x^2+1)^(1/2)-3/32*arcsinh(a*x))

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Maxima [A]
time = 0.25, size = 59, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {arsinh}\left (a x\right ) - \frac {1}{32} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} x^{3}}{a^{2}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arcsinh(a*x) - 1/32*(2*sqrt(a^2*x^2 + 1)*x^3/a^2 - 3*sqrt(a^2*x^2 + 1)*x/a^4 + 3*arcsinh(a*x)/a^5)*a

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Fricas [A]
time = 0.35, size = 59, normalized size = 0.88 \begin {gather*} \frac {{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1}}{32 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*((8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 + 1)) - (2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^2 + 1))/a^4

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Sympy [A]
time = 0.28, size = 61, normalized size = 0.91 \begin {gather*} \begin {cases} \frac {x^{4} \operatorname {asinh}{\left (a x \right )}}{4} - \frac {x^{3} \sqrt {a^{2} x^{2} + 1}}{16 a} + \frac {3 x \sqrt {a^{2} x^{2} + 1}}{32 a^{3}} - \frac {3 \operatorname {asinh}{\left (a x \right )}}{32 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),x)

[Out]

Piecewise((x**4*asinh(a*x)/4 - x**3*sqrt(a**2*x**2 + 1)/(16*a) + 3*x*sqrt(a**2*x**2 + 1)/(32*a**3) - 3*asinh(a
*x)/(32*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),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 ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*asinh(a*x), x)

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