3.1.15 \(\int x \sinh ^{-1}(a x)^2 \, dx\) [15]

Optimal. Leaf size=59 \[ \frac {x^2}{4}-\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{2 a}+\frac {\sinh ^{-1}(a x)^2}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^2 \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5776, 5812, 5783, 30} \begin {gather*} -\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac {\sinh ^{-1}(a x)^2}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^2+\frac {x^2}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.02, size = 53, normalized size = 0.90 \begin {gather*} \frac {a^2 x^2-2 a x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)+\left (1+2 a^2 x^2\right ) \sinh ^{-1}(a x)^2}{4 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.08, size = 59, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\left (a^{2} x^{2}+1\right ) \arcsinh \left (a x \right )^{2}}{2}-\frac {a x \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}}{2}-\frac {\arcsinh \left (a x \right )^{2}}{4}+\frac {a^{2} x^{2}}{4}+\frac {1}{4}}{a^{2}}\)	\(59\)
default	\(\frac {\frac {\left (a^{2} x^{2}+1\right ) \arcsinh \left (a x \right )^{2}}{2}-\frac {a x \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}}{2}-\frac {\arcsinh \left (a x \right )^{2}}{4}+\frac {a^{2} x^{2}}{4}+\frac {1}{4}}{a^{2}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arcsinh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 81, normalized size = 1.37 \begin {gather*} \frac {1}{2} \, x^{2} \operatorname {arsinh}\left (a x\right )^{2} + \frac {1}{4} \, a^{2} {\left (\frac {x^{2}}{a^{2}} - \frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{a^{4}}\right )} - \frac {1}{2} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )} \operatorname {arsinh}\left (a x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 73, normalized size = 1.24 \begin {gather*} \frac {a^{2} x^{2} - 2 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{4 \, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.12, size = 51, normalized size = 0.86 \begin {gather*} \begin {cases} \frac {x^{2} \operatorname {asinh}^{2}{\left (a x \right )}}{2} + \frac {x^{2}}{4} - \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{2 a} + \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{4 a^{2}} & \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*asinh(a*x)**2,x)

[Out]

Piecewise((x**2*asinh(a*x)**2/2 + x**2/4 - x*sqrt(a**2*x**2 + 1)*asinh(a*x)/(2*a) + asinh(a*x)**2/(4*a**2), 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*arcsinh(a*x)^2,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 = -0.02 \begin {gather*} \int x\,{\mathrm {asinh}\left (a\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*asinh(a*x)^2, x)

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