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

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

[Out]

1/5*x^5*(a+b*arcsin(c*x^2))-6/25*b*EllipticE(x*c^(1/2),I)/c^(5/2)+6/25*b*EllipticF(x*c^(1/2),I)/c^(5/2)+2/25*b
*x^3*(-c^2*x^4+1)^(1/2)/c

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 12

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

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 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 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 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 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 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 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*}

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Mathematica [C]  time = 0.29, size = 93, normalized size = 1.12 \[ \frac {1}{25} \left (5 a x^5+\frac {2 b x^3 \sqrt {1-c^2 x^4}}{c}+5 b x^5 \sin ^{-1}\left (c x^2\right )+\frac {6 i b \left (E\left (\left .i \sinh ^{-1}\left (\sqrt {-c} x\right )\right |-1\right )-F\left (\left .i \sinh ^{-1}\left (\sqrt {-c} x\right )\right |-1\right )\right )}{(-c)^{5/2}}\right ) \]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{4} \arcsin \left (c x^{2}\right ) + a x^{4}, x\right ) \]

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)

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

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)

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maple [A]  time = 0.01, size = 101, normalized size = 1.22 \[ \frac {a \,x^{5}}{5}+b \left (\frac {x^{5} \arcsin \left (c \,x^{2}\right )}{5}-\frac {2 c \left (-\frac {x^{3} \sqrt {-c^{2} x^{4}+1}}{5 c^{2}}-\frac {3 \sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \left (\EllipticF \left (x \sqrt {c}, i\right )-\EllipticE \left (x \sqrt {c}, i\right )\right )}{5 c^{\frac {7}{2}} \sqrt {-c^{2} x^{4}+1}}\right )}{5}\right ) \]

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))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{5} \, a x^{5} + \frac {1}{5} \, {\left (x^{5} \arctan \left (c x^{2}, \sqrt {c x^{2} + 1} \sqrt {-c x^{2} + 1}\right ) + 10 \, c \int \frac {x^{6} e^{\left (\frac {1}{2} \, \log \left (c x^{2} + 1\right ) + \frac {1}{2} \, \log \left (-c x^{2} + 1\right )\right )}}{5 \, {\left (c^{4} x^{8} - c^{2} x^{4} + {\left (c^{2} x^{4} - 1\right )} e^{\left (\log \left (c x^{2} + 1\right ) + \log \left (-c x^{2} + 1\right )\right )}\right )}}\,{d x}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5 + 1/5*(x^5*arctan2(c*x^2, sqrt(c*x^2 + 1)*sqrt(-c*x^2 + 1)) + 10*c*integrate(1/5*x^6*e^(1/2*log(c*x^
2 + 1) + 1/2*log(-c*x^2 + 1))/(c^4*x^8 - c^2*x^4 + (c^2*x^4 - 1)*e^(log(c*x^2 + 1) + log(-c*x^2 + 1))), x))*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\left (a+b\,\mathrm {asin}\left (c\,x^2\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*asin(c*x^2)),x)

[Out]

int(x^4*(a + b*asin(c*x^2)), x)

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sympy [A]  time = 2.09, size = 58, normalized size = 0.70 \[ \frac {a x^{5}}{5} - \frac {b c x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {c^{2} x^{4} e^{2 i \pi }} \right )}}{10 \Gamma \left (\frac {11}{4}\right )} + \frac {b x^{5} \operatorname {asin}{\left (c x^{2} \right )}}{5} \]

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

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