3.4.94 \(\int x^2 (a+b \text {ArcSin}(c+d x^2)) \, dx\) [394]

Optimal. Leaf size=287 \[ \frac {2 b x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{9 d}+\frac {1}{3} x^3 \left (a+b \text {ArcSin}\left (c+d x^2\right )\right )+\frac {8 b \sqrt {1-c} c (1+c) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} E\left (\text {ArcSin}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{9 d^{3/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}-\frac {2 b \sqrt {1-c} (1+c) (1+3 c) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} F\left (\text {ArcSin}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{9 d^{3/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \]

[Out]

1/3*x^3*(a+b*arcsin(d*x^2+c))+8/9*b*c*(1+c)*EllipticE(x*d^(1/2)/(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-c)^(1/2)*
(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/d^(3/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)-2/9*b*(1+c)*(1+3*c)*Ellip
ticF(x*d^(1/2)/(1-c)^(1/2),((-1+c)/(1+c))^(1/2))*(1-c)^(1/2)*(1-d*x^2/(1-c))^(1/2)*(1+d*x^2/(1+c))^(1/2)/d^(3/
2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/9*b*x*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d

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Rubi [A]
time = 0.23, antiderivative size = 287, 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 = {4926, 12, 1136, 1216, 538, 435, 430} \begin {gather*} \frac {1}{3} x^3 \left (a+b \text {ArcSin}\left (c+d x^2\right )\right )-\frac {2 b \sqrt {1-c} (c+1) (3 c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\text {ArcSin}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{9 d^{3/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}+\frac {8 b \sqrt {1-c} c (c+1) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\text {ArcSin}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{9 d^{3/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}+\frac {2 b x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(2*b*x*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(9*d) + (x^3*(a + b*ArcSin[c + d*x^2]))/3 + (8*b*Sqrt[1 - c]*c*(1
+ c)*Sqrt[1 - (d*x^2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]*EllipticE[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/
(1 + c))])/(9*d^(3/2)*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]) - (2*b*Sqrt[1 - c]*(1 + c)*(1 + 3*c)*Sqrt[1 - (d*x^
2)/(1 - c)]*Sqrt[1 + (d*x^2)/(1 + c)]*EllipticF[ArcSin[(Sqrt[d]*x)/Sqrt[1 - c]], -((1 - c)/(1 + c))])/(9*d^(3/
2)*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^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 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 435

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

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 1136

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[d^3*(d*x)^(m - 3)*((a + b*
x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1216

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[Sqrt[1 + 2*c*(x^2/(b - q))]*(Sqrt[1 + 2*c*(x^2/(b + q))]/Sqrt[a + b*x^2 + c*x^4]), Int[(d + e*x^2)/(Sqr
t[1 + 2*c*(x^2/(b - q))]*Sqrt[1 + 2*c*(x^2/(b + q))]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c
, 0] && NegQ[c/a]

Rule 4926

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^2 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {1}{3} b \int \frac {2 d x^4}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {1}{3} (2 b d) \int \frac {x^4}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac {2 b x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{9 d}+\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {(2 b) \int \frac {1-c^2-4 c d x^2}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx}{9 d}\\ &=\frac {2 b x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{9 d}+\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {\left (2 b \sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}} \sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac {1-c^2-4 c d x^2}{\sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}} \sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}} \, dx}{9 d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ &=\frac {2 b x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{9 d}+\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac {\left (8 b c (1+c) \sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}} \sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac {\sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}}}{\sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}} \, dx}{9 d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}-\frac {\left (2 b (1+c) (1+3 c) \sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}} \sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}\right ) \int \frac {1}{\sqrt {1-\frac {2 d^2 x^2}{-2 d-2 c d}} \sqrt {1-\frac {2 d^2 x^2}{2 d-2 c d}}} \, dx}{9 d \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ &=\frac {2 b x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{9 d}+\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac {8 b \sqrt {1-c} c (1+c) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{9 d^{3/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}-\frac {2 b \sqrt {1-c} (1+c) (1+3 c) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {1+\frac {d x^2}{1+c}} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{1+c}\right )}{9 d^{3/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.36, size = 307, normalized size = 1.07 \begin {gather*} \frac {\sqrt {\frac {d}{1+c}} x \left (3 a d x^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}-2 b \left (-1+c^2+2 c d x^2+d^2 x^4\right )+3 b d x^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4} \text {ArcSin}\left (c+d x^2\right )\right )-8 i b (-1+c) c \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )+2 i b \left (1-4 c+3 c^2\right ) \sqrt {\frac {-1+c+d x^2}{-1+c}} \sqrt {\frac {1+c+d x^2}{1+c}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{1+c}} x\right )|\frac {1+c}{-1+c}\right )}{9 d \sqrt {\frac {d}{1+c}} \sqrt {1-c^2-2 c d x^2-d^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d/(1 + c)]*x*(3*a*d*x^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] - 2*b*(-1 + c^2 + 2*c*d*x^2 + d^2*x^4) + 3*b
*d*x^2*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]*ArcSin[c + d*x^2]) - (8*I)*b*(-1 + c)*c*Sqrt[(-1 + c + d*x^2)/(-1 +
c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticE[I*ArcSinh[Sqrt[d/(1 + c)]*x], (1 + c)/(-1 + c)] + (2*I)*b*(1 - 4*
c + 3*c^2)*Sqrt[(-1 + c + d*x^2)/(-1 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticF[I*ArcSinh[Sqrt[d/(1 + c)]*x
], (1 + c)/(-1 + c)])/(9*d*Sqrt[d/(1 + c)]*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])

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Maple [A]
time = 0.01, size = 295, normalized size = 1.03

method	result	size
default	\(\frac {x^{3} a}{3}+b \left (\frac {x^{3} \arcsin \left (d \,x^{2}+c \right )}{3}-\frac {2 d \left (-\frac {x \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{3 d^{2}}+\frac {\left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{3 d^{2} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}+\frac {8 c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{3 d \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right )}{3}\right )\)	\(295\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(d*x^2+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is

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Fricas [A]
time = 0.52, size = 72, normalized size = 0.25 \begin {gather*} \frac {3 \, b d^{2} x^{4} \arcsin \left (d x^{2} + c\right ) + 3 \, a d^{2} x^{4} + 2 \, \sqrt {-d^{2} x^{4} - 2 \, c d x^{2} - c^{2} + 1} {\left (b d x^{2} - 4 \, b c\right )}}{9 \, d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*b*d^2*x^4*arcsin(d*x^2 + c) + 3*a*d^2*x^4 + 2*sqrt(-d^2*x^4 - 2*c*d*x^2 - c^2 + 1)*(b*d*x^2 - 4*b*c))/(
d^2*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {asin}{\left (c + d x^{2} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\left (a+b\,\mathrm {asin}\left (d\,x^2+c\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asin(c + d*x^2)),x)

[Out]

int(x^2*(a + b*asin(c + d*x^2)), x)

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