3.342 \(\int x^5 (a+b \sin ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=62 \[ \frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{b \left (1-c^2 x^4\right )^{3/2}}{18 c^3}+\frac{b \sqrt{1-c^2 x^4}}{6 c^3} \]

[Out]

(b*Sqrt[1 - c^2*x^4])/(6*c^3) - (b*(1 - c^2*x^4)^(3/2))/(18*c^3) + (x^6*(a + b*ArcSin[c*x^2]))/6

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Rubi [A]  time = 0.0479316, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4842, 12, 266, 43} \[ \frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{b \left (1-c^2 x^4\right )^{3/2}}{18 c^3}+\frac{b \sqrt{1-c^2 x^4}}{6 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcSin[c*x^2]),x]

[Out]

(b*Sqrt[1 - c^2*x^4])/(6*c^3) - (b*(1 - c^2*x^4)^(3/2))/(18*c^3) + (x^6*(a + b*ArcSin[c*x^2]))/6

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^5 \left (a+b \sin ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{6} b \int \frac{2 c x^7}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{3} (b c) \int \frac{x^7}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-c^2 x}} \, dx,x,x^4\right )\\ &=\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{12} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{c^2 \sqrt{1-c^2 x}}-\frac{\sqrt{1-c^2 x}}{c^2}\right ) \, dx,x,x^4\right )\\ &=\frac{b \sqrt{1-c^2 x^4}}{6 c^3}-\frac{b \left (1-c^2 x^4\right )^{3/2}}{18 c^3}+\frac{1}{6} x^6 \left (a+b \sin ^{-1}\left (c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.039535, size = 70, normalized size = 1.13 \[ \frac{a x^6}{6}+\frac{b x^4 \sqrt{1-c^2 x^4}}{18 c}+\frac{b \sqrt{1-c^2 x^4}}{9 c^3}+\frac{1}{6} b x^6 \sin ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcSin[c*x^2]),x]

[Out]

(a*x^6)/6 + (b*Sqrt[1 - c^2*x^4])/(9*c^3) + (b*x^4*Sqrt[1 - c^2*x^4])/(18*c) + (b*x^6*ArcSin[c*x^2])/6

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Maple [A]  time = 0.012, size = 62, normalized size = 1. \begin{align*}{\frac{{x}^{6}a}{6}}+b \left ({\frac{{x}^{6}\arcsin \left ( c{x}^{2} \right ) }{6}}-{\frac{ \left ( c{x}^{2}-1 \right ) \left ( c{x}^{2}+1 \right ) \left ({c}^{2}{x}^{4}+2 \right ) }{18\,{c}^{3}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsin(c*x^2)),x)

[Out]

1/6*x^6*a+b*(1/6*x^6*arcsin(c*x^2)-1/18/c^3*(c*x^2-1)*(c*x^2+1)*(c^2*x^4+2)/(-c^2*x^4+1)^(1/2))

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Maxima [A]  time = 1.4572, size = 80, normalized size = 1.29 \begin{align*} \frac{1}{6} \, a x^{6} + \frac{1}{18} \,{\left (3 \, x^{6} \arcsin \left (c x^{2}\right ) - c{\left (\frac{{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{c^{4}} - \frac{3 \, \sqrt{-c^{2} x^{4} + 1}}{c^{4}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x^2)),x, algorithm="maxima")

[Out]

1/6*a*x^6 + 1/18*(3*x^6*arcsin(c*x^2) - c*((-c^2*x^4 + 1)^(3/2)/c^4 - 3*sqrt(-c^2*x^4 + 1)/c^4))*b

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Fricas [A]  time = 2.39229, size = 123, normalized size = 1.98 \begin{align*} \frac{3 \, b c^{3} x^{6} \arcsin \left (c x^{2}\right ) + 3 \, a c^{3} x^{6} +{\left (b c^{2} x^{4} + 2 \, b\right )} \sqrt{-c^{2} x^{4} + 1}}{18 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x^2)),x, algorithm="fricas")

[Out]

1/18*(3*b*c^3*x^6*arcsin(c*x^2) + 3*a*c^3*x^6 + (b*c^2*x^4 + 2*b)*sqrt(-c^2*x^4 + 1))/c^3

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Sympy [A]  time = 4.02234, size = 65, normalized size = 1.05 \begin{align*} \begin{cases} \frac{a x^{6}}{6} + \frac{b x^{6} \operatorname{asin}{\left (c x^{2} \right )}}{6} + \frac{b x^{4} \sqrt{- c^{2} x^{4} + 1}}{18 c} + \frac{b \sqrt{- c^{2} x^{4} + 1}}{9 c^{3}} & \text{for}\: c \neq 0 \\\frac{a x^{6}}{6} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asin(c*x**2)),x)

[Out]

Piecewise((a*x**6/6 + b*x**6*asin(c*x**2)/6 + b*x**4*sqrt(-c**2*x**4 + 1)/(18*c) + b*sqrt(-c**2*x**4 + 1)/(9*c
**3), Ne(c, 0)), (a*x**6/6, True))

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Giac [A]  time = 1.15378, size = 117, normalized size = 1.89 \begin{align*} \frac{3 \, a c x^{6} +{\left (\frac{3 \,{\left (c^{2} x^{4} - 1\right )} x^{2} \arcsin \left (c x^{2}\right )}{c} + \frac{3 \, x^{2} \arcsin \left (c x^{2}\right )}{c} - \frac{{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{4} + 1}}{c^{2}}\right )} b}{18 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsin(c*x^2)),x, algorithm="giac")

[Out]

1/18*(3*a*c*x^6 + (3*(c^2*x^4 - 1)*x^2*arcsin(c*x^2)/c + 3*x^2*arcsin(c*x^2)/c - (-c^2*x^4 + 1)^(3/2)/c^2 + 3*
sqrt(-c^2*x^4 + 1)/c^2)*b)/c