3.284 \(\int \frac {x^2 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx\)

Optimal. Leaf size=87 \[ -\frac {\sinh ^{-1}(a x)^3}{6 a^3}-\frac {\sinh ^{-1}(a x)}{4 a^3}+\frac {x \sqrt {a^2 x^2+1}}{4 a^2}+\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {x^2 \sinh ^{-1}(a x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5758, 5675, 5661, 321, 215} \[ \frac {x \sqrt {a^2 x^2+1}}{4 a^2}+\frac {x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {\sinh ^{-1}(a x)^3}{6 a^3}-\frac {\sinh ^{-1}(a x)}{4 a^3}-\frac {x^2 \sinh ^{-1}(a x)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 215

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 321

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 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 \frac {x^2 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx &=\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {\int \frac {\sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{2 a^2}-\frac {\int x \sinh ^{-1}(a x) \, dx}{a}\\ &=-\frac {x^2 \sinh ^{-1}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {\sinh ^{-1}(a x)^3}{6 a^3}+\frac {1}{2} \int \frac {x^2}{\sqrt {1+a^2 x^2}} \, dx\\ &=\frac {x \sqrt {1+a^2 x^2}}{4 a^2}-\frac {x^2 \sinh ^{-1}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {\sinh ^{-1}(a x)^3}{6 a^3}-\frac {\int \frac {1}{\sqrt {1+a^2 x^2}} \, dx}{4 a^2}\\ &=\frac {x \sqrt {1+a^2 x^2}}{4 a^2}-\frac {\sinh ^{-1}(a x)}{4 a^3}-\frac {x^2 \sinh ^{-1}(a x)}{2 a}+\frac {x \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{2 a^2}-\frac {\sinh ^{-1}(a x)^3}{6 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 72, normalized size = 0.83 \[ \frac {3 a x \sqrt {a^2 x^2+1}+6 a x \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2-3 \left (2 a^2 x^2+1\right ) \sinh ^{-1}(a x)-2 \sinh ^{-1}(a x)^3}{12 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x*Sqrt[1 + a^2*x^2] - 3*(1 + 2*a^2*x^2)*ArcSinh[a*x] + 6*a*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2 - 2*ArcSinh
[a*x]^3)/(12*a^3)

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fricas [A]  time = 0.53, size = 102, normalized size = 1.17 \[ \frac {6 \, \sqrt {a^{2} x^{2} + 1} a x \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - 2 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} + 3 \, \sqrt {a^{2} x^{2} + 1} a x - 3 \, {\left (2 \, a^{2} x^{2} + 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{12 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*sqrt(a^2*x^2 + 1)*a*x*log(a*x + sqrt(a^2*x^2 + 1))^2 - 2*log(a*x + sqrt(a^2*x^2 + 1))^3 + 3*sqrt(a^2*x
^2 + 1)*a*x - 3*(2*a^2*x^2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 69, normalized size = 0.79 \[ -\frac {-6 \arcsinh \left (a x \right )^{2} \sqrt {a^{2} x^{2}+1}\, a x +6 \arcsinh \left (a x \right ) x^{2} a^{2}+2 \arcsinh \left (a x \right )^{3}-3 \sqrt {a^{2} x^{2}+1}\, x a +3 \arcsinh \left (a x \right )}{12 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(-6*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)*a*x+6*arcsinh(a*x)*x^2*a^2+2*arcsinh(a*x)^3-3*(a^2*x^2+1)^(1/2)*x*a
+3*arcsinh(a*x))/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.24, size = 78, normalized size = 0.90 \[ \begin {cases} - \frac {x^{2} \operatorname {asinh}{\left (a x \right )}}{2 a} + \frac {x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{2 a^{2}} + \frac {x \sqrt {a^{2} x^{2} + 1}}{4 a^{2}} - \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{6 a^{3}} - \frac {\operatorname {asinh}{\left (a x \right )}}{4 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asinh(a*x)**2/(a**2*x**2+1)**(1/2),x)

[Out]

Piecewise((-x**2*asinh(a*x)/(2*a) + x*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(2*a**2) + x*sqrt(a**2*x**2 + 1)/(4*a*
*2) - asinh(a*x)**3/(6*a**3) - asinh(a*x)/(4*a**3), Ne(a, 0)), (0, True))

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