3.15 \(\int x \sinh ^{-1}(a x)^2 \, dx\)

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

[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 = 0.09, 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 = {5661, 5758, 5675, 30} \[ -\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)}{2 a}+\frac {\sinh ^{-1}(a x)^2}{4 a^2}+\frac {1}{2} x^2 \sinh ^{-1}(a x)^2+\frac {x^2}{4} \]

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 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 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 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]

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 = 0.03, size = 53, normalized size = 0.90 \[ \frac {a^2 x^2-2 a x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)+\left (2 a^2 x^2+1\right ) \sinh ^{-1}(a x)^2}{4 a^2} \]

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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fricas [A]  time = 0.41, size = 73, normalized size = 1.24 \[ \frac {a^{2} x^{2} - 2 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{4 \, a^{2}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.08, size = 59, normalized size = 1.00 \[ \frac {\frac {\left (a^{2} x^{2}+1\right ) \arcsinh \left (a x \right )^{2}}{2}-\frac {\arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x}{2}-\frac {\arcsinh \left (a x \right )^{2}}{4}+\frac {a^{2} x^{2}}{4}+\frac {1}{4}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 81, normalized size = 1.37 \[ \frac {1}{2} \, x^{2} \operatorname {arsinh}\left (a x\right )^{2} + \frac {1}{4} \, a^{2} {\left (\frac {x^{2}}{a^{2}} - \frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{a^{4}}\right )} - \frac {1}{2} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )} \operatorname {arsinh}\left (a x\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,{\mathrm {asinh}\left (a\,x\right )}^2 \,d x \]

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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sympy [A]  time = 0.43, size = 51, normalized size = 0.86 \[ \begin {cases} \frac {x^{2} \operatorname {asinh}^{2}{\left (a x \right )}}{2} + \frac {x^{2}}{4} - \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{2 a} + \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{4 a^{2}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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