3.4.54 \(\int x^2 (a+b \text {ArcSin}(c x^2)) \, dx\) [354]

Optimal. Leaf size=61 \[ \frac {2 b x \sqrt {1-c^2 x^4}}{9 c}+\frac {1}{3} x^3 \left (a+b \text {ArcSin}\left (c x^2\right )\right )-\frac {2 b F\left (\left .\text {ArcSin}\left (\sqrt {c} x\right )\right |-1\right )}{9 c^{3/2}} \]

[Out]

1/3*x^3*(a+b*arcsin(c*x^2))-2/9*b*EllipticF(x*c^(1/2),I)/c^(3/2)+2/9*b*x*(-c^2*x^4+1)^(1/2)/c

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Rubi [A]
time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4926, 12, 327, 227} \begin {gather*} \frac {1}{3} x^3 \left (a+b \text {ArcSin}\left (c x^2\right )\right )-\frac {2 b F\left (\left .\text {ArcSin}\left (\sqrt {c} x\right )\right |-1\right )}{9 c^{3/2}}+\frac {2 b x \sqrt {1-c^2 x^4}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c*x^2]),x]

[Out]

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

Rule 12

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

Rule 227

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 327

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.10, size = 72, normalized size = 1.18 \begin {gather*} \frac {1}{9} \left (3 a x^3+\frac {2 b x \sqrt {1-c^2 x^4}}{c}+3 b x^3 \text {ArcSin}\left (c x^2\right )-\frac {2 i b F\left (\left .i \sinh ^{-1}\left (\sqrt {-c} x\right )\right |-1\right )}{(-c)^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.01, size = 88, normalized size = 1.44

method	result	size
default	\(\frac {x^{3} a}{3}+b \left (\frac {x^{3} \arcsin \left (c \,x^{2}\right )}{3}-\frac {2 c \left (-\frac {x \sqrt {-c^{2} x^{4}+1}}{3 c^{2}}+\frac {\sqrt {-c \,x^{2}+1}\, \sqrt {c \,x^{2}+1}\, \EllipticF \left (x \sqrt {c}, i\right )}{3 c^{\frac {5}{2}} \sqrt {-c^{2} x^{4}+1}}\right )}{3}\right )\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*a+b*(1/3*x^3*arcsin(c*x^2)-2/3*c*(-1/3/c^2*x*(-c^2*x^4+1)^(1/2)+1/3/c^(5/2)*(-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)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/3*(x^3*arctan2(c*x^2, sqrt(c*x^2 + 1)*sqrt(-c*x^2 + 1)) + 6*c*integrate(1/3*x^4*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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Fricas [A]
time = 0.45, size = 42, normalized size = 0.69 \begin {gather*} \frac {3 \, b c x^{3} \arcsin \left (c x^{2}\right ) + 3 \, a c x^{3} + 2 \, \sqrt {-c^{2} x^{4} + 1} b x}{9 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.09, size = 58, normalized size = 0.95 \begin {gather*} \frac {a x^{3}}{3} - \frac {b c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {c^{2} x^{4} e^{2 i \pi }} \right )}}{6 \Gamma \left (\frac {9}{4}\right )} + \frac {b x^{3} \operatorname {asin}{\left (c x^{2} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 - b*c*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c**2*x**4*exp_polar(2*I*pi))/(6*gamma(9/4)) + b*x**3*
asin(c*x**2)/3

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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(c*x^2)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^2\,\left (a+b\,\mathrm {asin}\left (c\,x^2\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asin(c*x^2)),x)

[Out]

int(x^2*(a + b*asin(c*x^2)), x)

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