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

Optimal. Leaf size=68 \[ \frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^n\right )\right )-\frac {b c n x^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2 n};\frac {3 (n+1)}{2 n};c^2 x^{2 n}\right )}{3 (n+3)} \]

[Out]

1/3*x^3*(a+b*arcsin(c*x^n))-1/3*b*c*n*x^(3+n)*hypergeom([1/2, 1/2*(3+n)/n],[3/2*(1+n)/n],c^2*x^(2*n))/(3+n)

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4842, 12, 364} \[ \frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^n\right )\right )-\frac {b c n x^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2 n};\frac {3 (n+1)}{2 n};c^2 x^{2 n}\right )}{3 (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c*x^n]),x]

[Out]

(x^3*(a + b*ArcSin[c*x^n]))/3 - (b*c*n*x^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/(2*n), (3*(1 + n))/(2*n), c^2*
x^(2*n)])/(3*(3 + n))

Rule 12

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 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^2 \left (a+b \sin ^{-1}\left (c x^n\right )\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^n\right )\right )-\frac {1}{3} b \int \frac {c n x^{2+n}}{\sqrt {1-c^2 x^{2 n}}} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^n\right )\right )-\frac {1}{3} (b c n) \int \frac {x^{2+n}}{\sqrt {1-c^2 x^{2 n}}} \, dx\\ &=\frac {1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^n\right )\right )-\frac {b c n x^{3+n} \, _2F_1\left (\frac {1}{2},\frac {3+n}{2 n};\frac {3 (1+n)}{2 n};c^2 x^{2 n}\right )}{3 (3+n)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 75, normalized size = 1.10 \[ \frac {a x^3}{3}-\frac {b c n x^{n+3} \, _2F_1\left (\frac {1}{2},\frac {n+3}{2 n};\frac {n+3}{2 n}+1;c^2 x^{2 n}\right )}{3 (n+3)}+\frac {1}{3} b x^3 \sin ^{-1}\left (c x^n\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3)/3 + (b*x^3*ArcSin[c*x^n])/3 - (b*c*n*x^(3 + n)*Hypergeometric2F1[1/2, (3 + n)/(2*n), 1 + (3 + n)/(2*n)
, c^2*x^(2*n)])/(3*(3 + n))

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +b \arcsin \left (c \,x^{n}\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a x^{3} + \frac {1}{3} \, {\left (x^{3} \arctan \left (c x^{n}, \sqrt {c x^{n} + 1} \sqrt {-c x^{n} + 1}\right ) + 3 \, c n \int \frac {\sqrt {c x^{n} + 1} \sqrt {-c x^{n} + 1} x^{2} x^{n}}{3 \, {\left (c^{2} x^{2 \, n} - 1\right )}}\,{d x}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/3*(x^3*arctan2(c*x^n, sqrt(c*x^n + 1)*sqrt(-c*x^n + 1)) + 3*c*n*integrate(1/3*sqrt(c*x^n + 1)*sq
rt(-c*x^n + 1)*x^2*x^n/(c^2*x^(2*n) - 1), x))*b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asin(c*x^n)),x)

[Out]

int(x^2*(a + b*asin(c*x^n)), x)

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sympy [C]  time = 6.83, size = 66, normalized size = 0.97 \[ \frac {a x^{3}}{3} + \frac {b x^{3} \operatorname {asin}{\left (c x^{n} \right )}}{3} + \frac {i b x^{3} \Gamma \left (\frac {3}{2 n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - \frac {3}{2 n} \\ 1 - \frac {3}{2 n} \end {matrix}\middle | {\frac {x^{- 2 n}}{c^{2}}} \right )}}{6 \Gamma \left (1 + \frac {3}{2 n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 + b*x**3*asin(c*x**n)/3 + I*b*x**3*gamma(3/(2*n))*hyper((1/2, -3/(2*n)), (1 - 3/(2*n),), x**(-2*n)/c*
*2)/(6*gamma(1 + 3/(2*n)))

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