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

Optimal. Leaf size=50 \[ -\frac{x^2 \sqrt{a^2 x^4+1}}{8 a}+\frac{\sinh ^{-1}\left (a x^2\right )}{8 a^2}+\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0365349, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5902, 12, 275, 321, 215} \[ -\frac{x^2 \sqrt{a^2 x^4+1}}{8 a}+\frac{\sinh ^{-1}\left (a x^2\right )}{8 a^2}+\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSinh[a*x^2],x]

[Out]

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

Rule 5902

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 + u^2],
x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)
^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 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]

Rubi steps

\begin{align*} \int x^3 \sinh ^{-1}\left (a x^2\right ) \, dx &=\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right )-\frac{1}{4} \int \frac{2 a x^5}{\sqrt{1+a^2 x^4}} \, dx\\ &=\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right )-\frac{1}{2} a \int \frac{x^5}{\sqrt{1+a^2 x^4}} \, dx\\ &=\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right )-\frac{1}{4} a \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+a^2 x^2}} \, dx,x,x^2\right )\\ &=-\frac{x^2 \sqrt{1+a^2 x^4}}{8 a}+\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+a^2 x^2}} \, dx,x,x^2\right )}{8 a}\\ &=-\frac{x^2 \sqrt{1+a^2 x^4}}{8 a}+\frac{\sinh ^{-1}\left (a x^2\right )}{8 a^2}+\frac{1}{4} x^4 \sinh ^{-1}\left (a x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0183354, size = 44, normalized size = 0.88 \[ \frac{\left (2 a^2 x^4+1\right ) \sinh ^{-1}\left (a x^2\right )-a x^2 \sqrt{a^2 x^4+1}}{8 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSinh[a*x^2],x]

[Out]

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

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Maple [A]  time = 0.03, size = 67, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}{\it Arcsinh} \left ( a{x}^{2} \right ) }{4}}-{\frac{{x}^{2}}{8\,a}\sqrt{{a}^{2}{x}^{4}+1}}+{\frac{1}{8\,a}\ln \left ({{a}^{2}{x}^{2}{\frac{1}{\sqrt{{a}^{2}}}}}+\sqrt{{a}^{2}{x}^{4}+1} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsinh(a*x^2),x)

[Out]

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

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Maxima [B]  time = 1.16603, size = 138, normalized size = 2.76 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{arsinh}\left (a x^{2}\right ) + \frac{1}{16} \, a{\left (\frac{\log \left (a + \frac{\sqrt{a^{2} x^{4} + 1}}{x^{2}}\right )}{a^{3}} - \frac{\log \left (-a + \frac{\sqrt{a^{2} x^{4} + 1}}{x^{2}}\right )}{a^{3}} + \frac{2 \, \sqrt{a^{2} x^{4} + 1}}{{\left (a^{4} - \frac{{\left (a^{2} x^{4} + 1\right )} a^{2}}{x^{4}}\right )} x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsinh(a*x^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.5887, size = 115, normalized size = 2.3 \begin{align*} -\frac{\sqrt{a^{2} x^{4} + 1} a x^{2} -{\left (2 \, a^{2} x^{4} + 1\right )} \log \left (a x^{2} + \sqrt{a^{2} x^{4} + 1}\right )}{8 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.24267, size = 42, normalized size = 0.84 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asinh}{\left (a x^{2} \right )}}{4} - \frac{x^{2} \sqrt{a^{2} x^{4} + 1}}{8 a} + \frac{\operatorname{asinh}{\left (a x^{2} \right )}}{8 a^{2}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asinh(a*x**2),x)

[Out]

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

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Giac [A]  time = 1.32506, size = 100, normalized size = 2. \begin{align*} \frac{1}{4} \, x^{4} \log \left (a x^{2} + \sqrt{a^{2} x^{4} + 1}\right ) - \frac{1}{8} \, a{\left (\frac{\sqrt{a^{2} x^{4} + 1} x^{2}}{a^{2}} + \frac{\log \left (-x^{2}{\left | a \right |} + \sqrt{a^{2} x^{4} + 1}\right )}{a^{2}{\left | a \right |}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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