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

Optimal. Leaf size=115 \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b x^2 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{8 d}-\frac{3 b c \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{8 d^2}-\frac{b \left (2 c^2+1\right ) \sin ^{-1}\left (c+d x^2\right )}{8 d^2} \]

[Out]

(-3*b*c*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(8*d^2) + (b*x^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(8*d) - (b*
(1 + 2*c^2)*ArcSin[c + d*x^2])/(8*d^2) + (x^4*(a + b*ArcSin[c + d*x^2]))/4

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Rubi [A]  time = 0.13019, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {4842, 12, 1114, 742, 640, 619, 216} \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b x^2 \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{8 d}-\frac{3 b c \sqrt{-c^2-2 c d x^2-d^2 x^4+1}}{8 d^2}-\frac{b \left (2 c^2+1\right ) \sin ^{-1}\left (c+d x^2\right )}{8 d^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(-3*b*c*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(8*d^2) + (b*x^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(8*d) - (b*
(1 + 2*c^2)*ArcSin[c + d*x^2])/(8*d^2) + (x^4*(a + b*ArcSin[c + d*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 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

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+d x^2\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{4} b \int \frac{2 d x^5}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{2} (b d) \int \frac{x^5}{\sqrt{1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{1}{4} (b d) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )\\ &=\frac{b x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{b \operatorname{Subst}\left (\int \frac{-1+c^2+3 c d x}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{8 d}\\ &=-\frac{3 b c \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d^2}+\frac{b x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac{\left (b \left (1+2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2-2 c d x-d^2 x^2}} \, dx,x,x^2\right )}{8 d}\\ &=-\frac{3 b c \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d^2}+\frac{b x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac{\left (b \left (1+2 c^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 d^2}}} \, dx,x,-2 d \left (c+d x^2\right )\right )}{16 d^3}\\ &=-\frac{3 b c \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d^2}+\frac{b x^2 \sqrt{1-c^2-2 c d x^2-d^2 x^4}}{8 d}-\frac{b \left (1+2 c^2\right ) \sin ^{-1}\left (c+d x^2\right )}{8 d^2}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0785011, size = 98, normalized size = 0.85 \[ \frac{a x^4}{4}+\frac{1}{2} b \left (\frac{x^2}{4 d}-\frac{3 c}{4 d^2}\right ) \sqrt{-c^2-2 c d x^2-d^2 x^4+1}-\frac{b \left (2 c^2+1\right ) \sin ^{-1}\left (c+d x^2\right )}{8 d^2}+\frac{1}{4} b x^4 \sin ^{-1}\left (c+d x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^4)/4 + (b*((-3*c)/(4*d^2) + x^2/(4*d))*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/2 - (b*(1 + 2*c^2)*ArcSin[c +
d*x^2])/(8*d^2) + (b*x^4*ArcSin[c + d*x^2])/4

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Maple [A]  time = 0.016, size = 191, normalized size = 1.7 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}\arcsin \left ( d{x}^{2}+c \right ) }{4}}+{\frac{b{x}^{2}}{8\,d}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{3\,bc}{8\,{d}^{2}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{b{c}^{2}}{4\,d}\arctan \left ({\sqrt{{d}^{2}} \left ({x}^{2}+{\frac{c}{d}} \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}} \right ){\frac{1}{\sqrt{{d}^{2}}}}}-{\frac{b}{8\,d}\arctan \left ({\sqrt{{d}^{2}} \left ({x}^{2}+{\frac{c}{d}} \right ){\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}} \right ){\frac{1}{\sqrt{{d}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.34112, size = 173, normalized size = 1.5 \begin{align*} \frac{2 \, a d^{2} x^{4} +{\left (2 \, b d^{2} x^{4} - 2 \, b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) + \sqrt{-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1}{\left (b d x^{2} - 3 \, b c\right )}}{8 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.36988, size = 133, normalized size = 1.16 \begin{align*} \begin{cases} \frac{a x^{4}}{4} - \frac{b c^{2} \operatorname{asin}{\left (c + d x^{2} \right )}}{4 d^{2}} - \frac{3 b c \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{8 d^{2}} + \frac{b x^{4} \operatorname{asin}{\left (c + d x^{2} \right )}}{4} + \frac{b x^{2} \sqrt{- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{8 d} - \frac{b \operatorname{asin}{\left (c + d x^{2} \right )}}{8 d^{2}} & \text{for}\: d \neq 0 \\\frac{x^{4} \left (a + b \operatorname{asin}{\left (c \right )}\right )}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**4/4 - b*c**2*asin(c + d*x**2)/(4*d**2) - 3*b*c*sqrt(-c**2 - 2*c*d*x**2 - d**2*x**4 + 1)/(8*d**
2) + b*x**4*asin(c + d*x**2)/4 + b*x**2*sqrt(-c**2 - 2*c*d*x**2 - d**2*x**4 + 1)/(8*d) - b*asin(c + d*x**2)/(8
*d**2), Ne(d, 0)), (x**4*(a + b*asin(c))/4, True))

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Giac [A]  time = 1.13422, size = 176, normalized size = 1.53 \begin{align*} \frac{\frac{2 \,{\left ({\left (d x^{2} + c\right )}^{2} - 2 \,{\left (d x^{2} + c\right )} c\right )} a}{d} - \frac{{\left (4 \,{\left (d x^{2} + c\right )} c \arcsin \left (d x^{2} + c\right ) - 2 \,{\left ({\left (d x^{2} + c\right )}^{2} - 1\right )} \arcsin \left (d x^{2} + c\right ) -{\left (d x^{2} + c\right )} \sqrt{-{\left (d x^{2} + c\right )}^{2} + 1} + 4 \, \sqrt{-{\left (d x^{2} + c\right )}^{2} + 1} c - \arcsin \left (d x^{2} + c\right )\right )} b}{d}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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