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

Optimal. Leaf size=83 \[ \frac{6 b \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{c} x\right ),-1\right )}{25 c^{5/2}}+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}-\frac{6 b E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}} \]

[Out]

(2*b*x^3*Sqrt[1 - c^2*x^4])/(25*c) + (x^5*(a + b*ArcSin[c*x^2]))/5 - (6*b*EllipticE[ArcSin[Sqrt[c]*x], -1])/(2
5*c^(5/2)) + (6*b*EllipticF[ArcSin[Sqrt[c]*x], -1])/(25*c^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0609729, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4842, 12, 321, 307, 221, 1199, 424} \[ \frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}+\frac{6 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}}-\frac{6 b E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcSin[c*x^2]),x]

[Out]

(2*b*x^3*Sqrt[1 - c^2*x^4])/(25*c) + (x^5*(a + b*ArcSin[c*x^2]))/5 - (6*b*EllipticE[ArcSin[Sqrt[c]*x], -1])/(2
5*c^(5/2)) + (6*b*EllipticF[ArcSin[Sqrt[c]*x], -1])/(25*c^(5/2))

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]

Rule 12

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

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]

Rubi steps

\begin{align*} \int x^4 \left (a+b \sin ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{5} b \int \frac{2 c x^6}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{5} (2 b c) \int \frac{x^6}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{(6 b) \int \frac{x^2}{\sqrt{1-c^2 x^4}} \, dx}{25 c}\\ &=\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{(6 b) \int \frac{1}{\sqrt{1-c^2 x^4}} \, dx}{25 c^2}-\frac{(6 b) \int \frac{1+c x^2}{\sqrt{1-c^2 x^4}} \, dx}{25 c^2}\\ &=\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{6 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}}-\frac{(6 b) \int \frac{\sqrt{1+c x^2}}{\sqrt{1-c x^2}} \, dx}{25 c^2}\\ &=\frac{2 b x^3 \sqrt{1-c^2 x^4}}{25 c}+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{6 b E\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}}+\frac{6 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{25 c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.232384, size = 93, normalized size = 1.12 \[ \frac{1}{25} \left (\frac{6 i b \left (E\left (\left .i \sinh ^{-1}\left (\sqrt{-c} x\right )\right |-1\right )-\text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c} x\right ),-1\right )\right )}{(-c)^{5/2}}+5 a x^5+\frac{2 b x^3 \sqrt{1-c^2 x^4}}{c}+5 b x^5 \sin ^{-1}\left (c x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x^5 + (2*b*x^3*Sqrt[1 - c^2*x^4])/c + 5*b*x^5*ArcSin[c*x^2] + ((6*I)*b*(EllipticE[I*ArcSinh[Sqrt[-c]*x],
-1] - EllipticF[I*ArcSinh[Sqrt[-c]*x], -1]))/(-c)^(5/2))/25

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 101, normalized size = 1.2 \begin{align*}{\frac{a{x}^{5}}{5}}+b \left ({\frac{{x}^{5}\arcsin \left ( c{x}^{2} \right ) }{5}}-{\frac{2\,c}{5} \left ( -{\frac{{x}^{3}}{5\,{c}^{2}}\sqrt{-{c}^{2}{x}^{4}+1}}-{\frac{3}{5}\sqrt{-c{x}^{2}+1}\sqrt{c{x}^{2}+1} \left ({\it EllipticF} \left ( x\sqrt{c},i \right ) -{\it EllipticE} \left ( x\sqrt{c},i \right ) \right ){c}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5+b*(1/5*x^5*arcsin(c*x^2)-2/5*c*(-1/5/c^2*x^3*(-c^2*x^4+1)^(1/2)-3/5/c^(7/2)*(-c*x^2+1)^(1/2)*(c*x^2+
1)^(1/2)/(-c^2*x^4+1)^(1/2)*(EllipticF(x*c^(1/2),I)-EllipticE(x*c^(1/2),I))))

________________________________________________________________________________________

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(c*x^2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{4} \arcsin \left (c x^{2}\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(c*x^2)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 2.56983, size = 58, normalized size = 0.7 \begin{align*} \frac{a x^{5}}{5} - \frac{b c x^{7} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{c^{2} x^{4} e^{2 i \pi }} \right )}}{10 \Gamma \left (\frac{11}{4}\right )} + \frac{b x^{5} \operatorname{asin}{\left (c x^{2} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**5/5 - b*c*x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), c**2*x**4*exp_polar(2*I*pi))/(10*gamma(11/4)) + b*x*
*5*asin(c*x**2)/5

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (c x^{2}\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(c*x^2)),x, algorithm="giac")

[Out]

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