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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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 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]

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 = 0.10, size = 98, normalized size = 0.85 \[ \frac {a x^4}{4}-\frac {b \left (2 c^2+1\right ) \sin ^{-1}\left (c+d x^2\right )}{8 d^2}+\frac {1}{2} b \left (\frac {x^2}{4 d}-\frac {3 c}{4 d^2}\right ) \sqrt {-c^2-2 c d x^2-d^2 x^4+1}+\frac {1}{4} b x^4 \sin ^{-1}\left (c+d x^2\right ) \]

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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 79, normalized size = 0.69 \[ \frac {2 \, a d^{2} x^{4} + {\left (2 \, b d^{2} x^{4} - 2 \, b c^{2} - b\right )} \arcsin \left (d x^{2} + c\right ) + \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b d x^{2} - 3 \, b c\right )}}{8 \, d^{2}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.93, size = 130, normalized size = 1.13 \[ \frac {\frac {2 \, {\left ({\left (d x^{2} + c\right )}^{2} - 2 \, {\left (d x^{2} + c\right )} c\right )} a}{d} - \frac {{\left (4 \, {\left (d x^{2} + c\right )} c \arcsin \left (d x^{2} + c\right ) - 2 \, {\left ({\left (d x^{2} + c\right )}^{2} - 1\right )} \arcsin \left (d x^{2} + c\right ) - {\left (d x^{2} + c\right )} \sqrt {-{\left (d x^{2} + c\right )}^{2} + 1} + 4 \, \sqrt {-{\left (d x^{2} + c\right )}^{2} + 1} c - \arcsin \left (d x^{2} + c\right )\right )} b}{d}}{8 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 191, normalized size = 1.66 \[ \frac {x^{4} a}{4}+\frac {b \,x^{4} \arcsin \left (d \,x^{2}+c \right )}{4}+\frac {b \,x^{2} \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{8 d}-\frac {3 b c \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{8 d^{2}}-\frac {b \,c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, \left (x^{2}+\frac {c}{d}\right )}{\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}\right )}{4 d \sqrt {d^{2}}}-\frac {b \arctan \left (\frac {\sqrt {d^{2}}\, \left (x^{2}+\frac {c}{d}\right )}{\sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}\right )}{8 d \sqrt {d^{2}}} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 174, normalized size = 1.51 \[ \frac {1}{4} \, a x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \arcsin \left (d x^{2} + c\right ) + d {\left (\frac {\sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} x^{2}}{d^{2}} + \frac {3 \, c^{2} \arcsin \left (-\frac {d^{2} x^{2} + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {{\left (c^{2} - 1\right )} \arcsin \left (-\frac {d^{2} x^{2} + c d}{\sqrt {c^{2} d^{2} - {\left (c^{2} - 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} c}{d^{3}}\right )}\right )} b \]

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]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\left (a+b\,\mathrm {asin}\left (d\,x^2+c\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*asin(c + d*x^2)),x)

[Out]

int(x^3*(a + b*asin(c + d*x^2)), x)

________________________________________________________________________________________

sympy [A]  time = 0.91, size = 133, normalized size = 1.16 \[ \begin {cases} \frac {a x^{4}}{4} - \frac {b c^{2} \operatorname {asin}{\left (c + d x^{2} \right )}}{4 d^{2}} - \frac {3 b c \sqrt {- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{8 d^{2}} + \frac {b x^{4} \operatorname {asin}{\left (c + d x^{2} \right )}}{4} + \frac {b x^{2} \sqrt {- c^{2} - 2 c d x^{2} - d^{2} x^{4} + 1}}{8 d} - \frac {b \operatorname {asin}{\left (c + d x^{2} \right )}}{8 d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{4} \left (a + b \operatorname {asin}{\relax (c )}\right )}{4} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________