3.343 \(\int x^3 (a+b \sin ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b x^2 \sqrt{1-c^2 x^4}}{8 c}-\frac{b \sin ^{-1}\left (c x^2\right )}{8 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0435031, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4842, 12, 275, 321, 216} \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b x^2 \sqrt{1-c^2 x^4}}{8 c}-\frac{b \sin ^{-1}\left (c x^2\right )}{8 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSin[c*x^2]),x]

[Out]

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

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
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 216

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

Rubi steps

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

Mathematica [A]  time = 0.0232808, size = 62, normalized size = 1.09 \[ \frac{a x^4}{4}+\frac{b x^2 \sqrt{1-c^2 x^4}}{8 c}-\frac{b \sin ^{-1}\left (c x^2\right )}{8 c^2}+\frac{1}{4} b x^4 \sin ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSin[c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.013, size = 74, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}\arcsin \left ( c{x}^{2} \right ) }{4}}+{\frac{b{x}^{2}}{8\,c}\sqrt{-{c}^{2}{x}^{4}+1}}-{\frac{b}{8\,c}\arctan \left ({{x}^{2}\sqrt{{c}^{2}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ){\frac{1}{\sqrt{{c}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsin(c*x^2)),x)

[Out]

1/4*x^4*a+1/4*b*x^4*arcsin(c*x^2)+1/8*b*x^2*(-c^2*x^4+1)^(1/2)/c-1/8*b/c/(c^2)^(1/2)*arctan((c^2)^(1/2)*x^2/(-
c^2*x^4+1)^(1/2))

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Maxima [A]  time = 1.42222, size = 119, normalized size = 2.09 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{8} \,{\left (2 \, x^{4} \arcsin \left (c x^{2}\right ) + c{\left (\frac{\arctan \left (\frac{\sqrt{-c^{2} x^{4} + 1}}{c x^{2}}\right )}{c^{3}} + \frac{\sqrt{-c^{2} x^{4} + 1}}{{\left (c^{4} - \frac{{\left (c^{2} x^{4} - 1\right )} c^{2}}{x^{4}}\right )} x^{2}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x^2)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/8*(2*x^4*arcsin(c*x^2) + c*(arctan(sqrt(-c^2*x^4 + 1)/(c*x^2))/c^3 + sqrt(-c^2*x^4 + 1)/((c^4 -
(c^2*x^4 - 1)*c^2/x^4)*x^2)))*b

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Fricas [A]  time = 2.52245, size = 116, normalized size = 2.04 \begin{align*} \frac{2 \, a c^{2} x^{4} + \sqrt{-c^{2} x^{4} + 1} b c x^{2} +{\left (2 \, b c^{2} x^{4} - b\right )} \arcsin \left (c x^{2}\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x^2)),x, algorithm="fricas")

[Out]

1/8*(2*a*c^2*x^4 + sqrt(-c^2*x^4 + 1)*b*c*x^2 + (2*b*c^2*x^4 - b)*arcsin(c*x^2))/c^2

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Sympy [A]  time = 1.15005, size = 60, normalized size = 1.05 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{asin}{\left (c x^{2} \right )}}{4} + \frac{b x^{2} \sqrt{- c^{2} x^{4} + 1}}{8 c} - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{8 c^{2}} & \text{for}\: c \neq 0 \\\frac{a x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asin(c*x**2)),x)

[Out]

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

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Giac [A]  time = 1.16703, size = 80, normalized size = 1.4 \begin{align*} \frac{2 \, a c x^{4} + \frac{{\left (\sqrt{-c^{2} x^{4} + 1} c x^{2} + 2 \,{\left (c^{2} x^{4} - 1\right )} \arcsin \left (c x^{2}\right ) + \arcsin \left (c x^{2}\right )\right )} b}{c}}{8 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x^2)),x, algorithm="giac")

[Out]

1/8*(2*a*c*x^4 + (sqrt(-c^2*x^4 + 1)*c*x^2 + 2*(c^2*x^4 - 1)*arcsin(c*x^2) + arcsin(c*x^2))*b/c)/c