3.393 \(\int x^4 (a+b \sin ^{-1}(c+d x^2)) \, dx\)

Optimal. Leaf size=336 \[ \frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {16 b c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{75 d^2}+\frac {2 b x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{25 d}+\frac {2 b \sqrt {1-c} (c+1) \left (15 c^2+8 c+9\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{75 d^{5/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b \sqrt {1-c} (c+1) \left (23 c^2+9\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{75 d^{5/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

[Out]

1/5*x^5*(a+b*arcsin(d*x^2+c))-2/75*b*(1+c)*(23*c^2+9)*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^(5/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+2/75*b*(1+c)*(1
5*c^2+8*c+9)*EllipticF(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^(5/2)/(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)-16/75*b*c*x*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d^2+2/25*b*
x^3*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)/d

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Rubi [A]  time = 0.42, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4842, 12, 1122, 1279, 1202, 524, 424, 419} \[ \frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac {2 b x^3 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{25 d}-\frac {16 b c x \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}{75 d^2}+\frac {2 b \sqrt {1-c} (c+1) \left (15 c^2+8 c+9\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} F\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{75 d^{5/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}}-\frac {2 b \sqrt {1-c} (c+1) \left (23 c^2+9\right ) \sqrt {1-\frac {d x^2}{1-c}} \sqrt {\frac {d x^2}{c+1}+1} E\left (\sin ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {1-c}}\right )|-\frac {1-c}{c+1}\right )}{75 d^{5/2} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(-16*b*c*x*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(75*d^2) + (2*b*x^3*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])/(25*d
) + (x^5*(a + b*ArcSin[c + d*x^2]))/5 - (2*b*Sqrt[1 - c]*(1 + c)*(9 + 23*c^2)*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))])/(75*d^(5/2)*Sqrt[1 - c^2 -
2*c*d*x^2 - d^2*x^4]) + (2*b*Sqrt[1 - c]*(1 + c)*(9 + 8*c + 15*c^2)*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))])/(75*d^(5/2)*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 419

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

Rule 424

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

Rule 524

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 1122

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 1202

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 1279

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

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^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right ) \, dx &=\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {1}{5} b \int \frac {2 d x^6}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {1}{5} (2 b d) \int \frac {x^6}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx\\ &=\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {(2 b) \int \frac {x^2 \left (3 \left (1-c^2\right )-8 c d x^2\right )}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx}{25 d}\\ &=-\frac {16 b c x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{75 d^2}+\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {(2 b) \int \frac {-8 c \left (1-c^2\right ) d+\left (9+23 c^2\right ) d^2 x^2}{\sqrt {1-c^2-2 c d x^2-d^2 x^4}} \, dx}{75 d^3}\\ &=-\frac {16 b c x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{75 d^2}+\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \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 {-8 c \left (1-c^2\right ) d+\left (9+23 c^2\right ) d^2 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}{75 d^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac {16 b c x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{75 d^2}+\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )+\frac {\left (2 b (1+c) \left (9+8 c+15 c^2\right ) \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}{75 d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}-\frac {\left (2 b (1+c) \left (9+23 c^2\right ) \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}{75 d^2 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ &=-\frac {16 b c x \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{75 d^2}+\frac {2 b x^3 \sqrt {1-c^2-2 c d x^2-d^2 x^4}}{25 d}+\frac {1}{5} x^5 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right )-\frac {2 b \sqrt {1-c} (1+c) \left (9+23 c^2\right ) \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 )}{75 d^{5/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}+\frac {2 b \sqrt {1-c} (1+c) \left (9+8 c+15 c^2\right ) \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 )}{75 d^{5/2} \sqrt {1-c^2-2 c d x^2-d^2 x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.66, size = 349, normalized size = 1.04 \[ \frac {x \sqrt {\frac {d}{c+1}} \left (15 a d^2 x^4 \sqrt {-c^2-2 c d x^2-d^2 x^4+1}+15 b d^2 x^4 \sqrt {-c^2-2 c d x^2-d^2 x^4+1} \sin ^{-1}\left (c+d x^2\right )+2 b \left (8 c^3+13 c^2 d x^2+2 c d^2 x^4-8 c-3 d^3 x^6+3 d x^2\right )\right )-2 i b \left (15 c^3-23 c^2+17 c-9\right ) \sqrt {\frac {c+d x^2-1}{c-1}} \sqrt {\frac {c+d x^2+1}{c+1}} F\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )+2 i b \left (23 c^3-23 c^2+9 c-9\right ) \sqrt {\frac {c+d x^2-1}{c-1}} \sqrt {\frac {c+d x^2+1}{c+1}} E\left (i \sinh ^{-1}\left (\sqrt {\frac {d}{c+1}} x\right )|\frac {c+1}{c-1}\right )}{75 d^2 \sqrt {\frac {d}{c+1}} \sqrt {-c^2-2 c d x^2-d^2 x^4+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcSin[c + d*x^2]),x]

[Out]

(Sqrt[d/(1 + c)]*x*(15*a*d^2*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4] + 2*b*(-8*c + 8*c^3 + 3*d*x^2 + 13*c^2*d*
x^2 + 2*c*d^2*x^4 - 3*d^3*x^6) + 15*b*d^2*x^4*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4]*ArcSin[c + d*x^2]) + (2*I)*b
*(-9 + 9*c - 23*c^2 + 23*c^3)*Sqrt[(-1 + c + d*x^2)/(-1 + c)]*Sqrt[(1 + c + d*x^2)/(1 + c)]*EllipticE[I*ArcSin
h[Sqrt[d/(1 + c)]*x], (1 + c)/(-1 + c)] - (2*I)*b*(-9 + 17*c - 23*c^2 + 15*c^3)*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)])/(75*d^2*Sqrt[d/(1 +
c)]*Sqrt[1 - c^2 - 2*c*d*x^2 - d^2*x^4])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{4} \arcsin \left (d x^{2} + c\right ) + a x^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(d*x^2+c)),x, algorithm="fricas")

[Out]

integral(b*x^4*arcsin(d*x^2 + c) + a*x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arcsin \left (d x^{2} + c\right ) + a\right )} x^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(d*x^2+c)),x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x^2 + c) + a)*x^4, x)

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maple [A]  time = 0.04, size = 346, normalized size = 1.03 \[ \frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (d \,x^{2}+c \right )}{5}-\frac {2 d \left (-\frac {x^{3} \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{5 d^{2}}+\frac {8 c x \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}{15 d^{3}}-\frac {8 c \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )}{15 d^{3} \sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}}-\frac {2 \left (\frac {-3 c^{2}+3}{5 d^{2}}+\frac {32 c^{2}}{15 d^{2}}\right ) \left (-c^{2}+1\right ) \sqrt {1+\frac {d \,x^{2}}{-1+c}}\, \sqrt {1+\frac {d \,x^{2}}{1+c}}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )-\EllipticE \left (x \sqrt {-\frac {d}{-1+c}}, \sqrt {-1+\frac {2 c}{1+c}}\right )\right )}{\sqrt {-\frac {d}{-1+c}}\, \sqrt {-d^{2} x^{4}-2 c d \,x^{2}-c^{2}+1}\, \left (-2 d c +2 d \right )}\right )}{5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsin(d*x^2+c)),x)

[Out]

1/5*a*x^5+b*(1/5*x^5*arcsin(d*x^2+c)-2/5*d*(-1/5/d^2*x^3*(-d^2*x^4-2*c*d*x^2-c^2+1)^(1/2)+8/15*c/d^3*x*(-d^2*x
^4-2*c*d*x^2-c^2+1)^(1/2)-8/15*c/d^3*(-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))-2*(1/5/d^2*(-3*c^2+3)+32/1
5*c^2/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)/(-2*c*d+2*d)*(EllipticF(x*(-d/(-1+c))^(1/2),(-1+2*c/(1+c))^(1/2))-EllipticE(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(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 c-1 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^4\,\left (a+b\,\mathrm {asin}\left (d\,x^2+c\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*asin(c + d*x^2)),x)

[Out]

int(x^4*(a + b*asin(c + d*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{4} \left (a + b \operatorname {asin}{\left (c + d x^{2} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(d*x**2+c)),x)

[Out]

Integral(x**4*(a + b*asin(c + d*x**2)), x)

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