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

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

[Out]

(-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])

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Rubi [A]  time = 0.417522, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, 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 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 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 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 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 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 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 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)])

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*}

Mathematica [F]  time = 0.753176, size = 0, normalized size = 0. \[ \int x^4 \left (a+b \sin ^{-1}\left (c+d x^2\right )\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 346, normalized size = 1. \begin{align*}{\frac{a{x}^{5}}{5}}+b \left ({\frac{{x}^{5}\arcsin \left ( d{x}^{2}+c \right ) }{5}}-{\frac{2\,d}{5} \left ( -{\frac{{x}^{3}}{5\,{d}^{2}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}+{\frac{8\,cx}{15\,{d}^{3}}\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}-{\frac{8\,c \left ( -{c}^{2}+1 \right ) }{15\,{d}^{3}}\sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}}{\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}{\frac{1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1}}}}-2\,{\frac{-{c}^{2}+1}{\sqrt{-{d}^{2}{x}^{4}-2\,cd{x}^{2}-{c}^{2}+1} \left ( -2\,dc+2\,d \right ) } \left ( 1/5\,{\frac{-3\,{c}^{2}+3}{{d}^{2}}}+{\frac{32\,{c}^{2}}{15\,{d}^{2}}} \right ) \sqrt{1+{\frac{d{x}^{2}}{-1+c}}}\sqrt{1+{\frac{d{x}^{2}}{1+c}}} \left ({\it EllipticF} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) -{\it EllipticE} \left ( x\sqrt{-{\frac{d}{-1+c}}},\sqrt{-1+2\,{\frac{c}{1+c}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{d}{-1+c}}}}}} \right ) } \right ) \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{4} \arcsin \left (d x^{2} + c\right ) + a x^{4}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{asin}{\left (c + d x^{2} \right )}\right )\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (d x^{2} + c\right ) + a\right )} x^{4}\,{d x} \end{align*}

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)