3.340 \(\int x^9 (a+b \sin ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=84 \[ \frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b \left (1-c^2 x^4\right )^{5/2}}{50 c^5}-\frac{b \left (1-c^2 x^4\right )^{3/2}}{15 c^5}+\frac{b \sqrt{1-c^2 x^4}}{10 c^5} \]

[Out]

(b*Sqrt[1 - c^2*x^4])/(10*c^5) - (b*(1 - c^2*x^4)^(3/2))/(15*c^5) + (b*(1 - c^2*x^4)^(5/2))/(50*c^5) + (x^10*(
a + b*ArcSin[c*x^2]))/10

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Rubi [A]  time = 0.064852, antiderivative size = 84, 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} \[ \frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b \left (1-c^2 x^4\right )^{5/2}}{50 c^5}-\frac{b \left (1-c^2 x^4\right )^{3/2}}{15 c^5}+\frac{b \sqrt{1-c^2 x^4}}{10 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^9*(a + b*ArcSin[c*x^2]),x]

[Out]

(b*Sqrt[1 - c^2*x^4])/(10*c^5) - (b*(1 - c^2*x^4)^(3/2))/(15*c^5) + (b*(1 - c^2*x^4)^(5/2))/(50*c^5) + (x^10*(
a + b*ArcSin[c*x^2]))/10

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^9 \left (a+b \sin ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{10} b \int \frac{2 c x^{11}}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{5} (b c) \int \frac{x^{11}}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{20} (b c) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-c^2 x}} \, dx,x,x^4\right )\\ &=\frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{20} (b c) \operatorname{Subst}\left (\int \left (\frac{1}{c^4 \sqrt{1-c^2 x}}-\frac{2 \sqrt{1-c^2 x}}{c^4}+\frac{\left (1-c^2 x\right )^{3/2}}{c^4}\right ) \, dx,x,x^4\right )\\ &=\frac{b \sqrt{1-c^2 x^4}}{10 c^5}-\frac{b \left (1-c^2 x^4\right )^{3/2}}{15 c^5}+\frac{b \left (1-c^2 x^4\right )^{5/2}}{50 c^5}+\frac{1}{10} x^{10} \left (a+b \sin ^{-1}\left (c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0550731, size = 60, normalized size = 0.71 \[ \frac{1}{150} \left (15 a x^{10}+\frac{b \sqrt{1-c^2 x^4} \left (3 c^4 x^8+4 c^2 x^4+8\right )}{c^5}+15 b x^{10} \sin ^{-1}\left (c x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^9*(a + b*ArcSin[c*x^2]),x]

[Out]

(15*a*x^10 + (b*Sqrt[1 - c^2*x^4]*(8 + 4*c^2*x^4 + 3*c^4*x^8))/c^5 + 15*b*x^10*ArcSin[c*x^2])/150

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Maple [A]  time = 0.025, size = 71, normalized size = 0.9 \begin{align*}{\frac{{x}^{10}a}{10}}+b \left ({\frac{{x}^{10}\arcsin \left ( c{x}^{2} \right ) }{10}}-{\frac{ \left ( c{x}^{2}-1 \right ) \left ( c{x}^{2}+1 \right ) \left ( 3\,{c}^{4}{x}^{8}+4\,{c}^{2}{x}^{4}+8 \right ) }{150\,{c}^{5}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(a+b*arcsin(c*x^2)),x)

[Out]

1/10*x^10*a+b*(1/10*x^10*arcsin(c*x^2)-1/150/c^5*(c*x^2-1)*(c*x^2+1)*(3*c^4*x^8+4*c^2*x^4+8)/(-c^2*x^4+1)^(1/2
))

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Maxima [A]  time = 1.60451, size = 103, normalized size = 1.23 \begin{align*} \frac{1}{10} \, a x^{10} + \frac{1}{150} \,{\left (15 \, x^{10} \arcsin \left (c x^{2}\right ) + c{\left (\frac{3 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{5}{2}}}{c^{6}} - \frac{10 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{c^{6}} + \frac{15 \, \sqrt{-c^{2} x^{4} + 1}}{c^{6}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(a+b*arcsin(c*x^2)),x, algorithm="maxima")

[Out]

1/10*a*x^10 + 1/150*(15*x^10*arcsin(c*x^2) + c*(3*(-c^2*x^4 + 1)^(5/2)/c^6 - 10*(-c^2*x^4 + 1)^(3/2)/c^6 + 15*
sqrt(-c^2*x^4 + 1)/c^6))*b

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Fricas [A]  time = 2.33201, size = 151, normalized size = 1.8 \begin{align*} \frac{15 \, b c^{5} x^{10} \arcsin \left (c x^{2}\right ) + 15 \, a c^{5} x^{10} +{\left (3 \, b c^{4} x^{8} + 4 \, b c^{2} x^{4} + 8 \, b\right )} \sqrt{-c^{2} x^{4} + 1}}{150 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(a+b*arcsin(c*x^2)),x, algorithm="fricas")

[Out]

1/150*(15*b*c^5*x^10*arcsin(c*x^2) + 15*a*c^5*x^10 + (3*b*c^4*x^8 + 4*b*c^2*x^4 + 8*b)*sqrt(-c^2*x^4 + 1))/c^5

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Sympy [A]  time = 31.1882, size = 90, normalized size = 1.07 \begin{align*} \begin{cases} \frac{a x^{10}}{10} + \frac{b x^{10} \operatorname{asin}{\left (c x^{2} \right )}}{10} + \frac{b x^{8} \sqrt{- c^{2} x^{4} + 1}}{50 c} + \frac{2 b x^{4} \sqrt{- c^{2} x^{4} + 1}}{75 c^{3}} + \frac{4 b \sqrt{- c^{2} x^{4} + 1}}{75 c^{5}} & \text{for}\: c \neq 0 \\\frac{a x^{10}}{10} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(a+b*asin(c*x**2)),x)

[Out]

Piecewise((a*x**10/10 + b*x**10*asin(c*x**2)/10 + b*x**8*sqrt(-c**2*x**4 + 1)/(50*c) + 2*b*x**4*sqrt(-c**2*x**
4 + 1)/(75*c**3) + 4*b*sqrt(-c**2*x**4 + 1)/(75*c**5), Ne(c, 0)), (a*x**10/10, True))

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Giac [A]  time = 1.22419, size = 189, normalized size = 2.25 \begin{align*} \frac{15 \, a c x^{10} +{\left (\frac{15 \,{\left (c^{2} x^{4} - 1\right )}^{2} x^{2} \arcsin \left (c x^{2}\right )}{c^{3}} + \frac{30 \,{\left (c^{2} x^{4} - 1\right )} x^{2} \arcsin \left (c x^{2}\right )}{c^{3}} + \frac{15 \, x^{2} \arcsin \left (c x^{2}\right )}{c^{3}} + \frac{3 \,{\left (c^{2} x^{4} - 1\right )}^{2} \sqrt{-c^{2} x^{4} + 1}}{c^{4}} - \frac{10 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{c^{4}} + \frac{15 \, \sqrt{-c^{2} x^{4} + 1}}{c^{4}}\right )} b}{150 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(a+b*arcsin(c*x^2)),x, algorithm="giac")

[Out]

1/150*(15*a*c*x^10 + (15*(c^2*x^4 - 1)^2*x^2*arcsin(c*x^2)/c^3 + 30*(c^2*x^4 - 1)*x^2*arcsin(c*x^2)/c^3 + 15*x
^2*arcsin(c*x^2)/c^3 + 3*(c^2*x^4 - 1)^2*sqrt(-c^2*x^4 + 1)/c^4 - 10*(-c^2*x^4 + 1)^(3/2)/c^4 + 15*sqrt(-c^2*x
^4 + 1)/c^4)*b)/c