3.341 \(\int x^7 (a+b \sin ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=82 \[ \frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}+\frac{3 b x^2 \sqrt{1-c^2 x^4}}{64 c^3}-\frac{3 b \sin ^{-1}\left (c x^2\right )}{64 c^4} \]

[Out]

(3*b*x^2*Sqrt[1 - c^2*x^4])/(64*c^3) + (b*x^6*Sqrt[1 - c^2*x^4])/(32*c) - (3*b*ArcSin[c*x^2])/(64*c^4) + (x^8*
(a + b*ArcSin[c*x^2]))/8

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Rubi [A]  time = 0.0609173, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4842, 12, 275, 321, 216} \[ \frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}+\frac{3 b x^2 \sqrt{1-c^2 x^4}}{64 c^3}-\frac{3 b \sin ^{-1}\left (c x^2\right )}{64 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcSin[c*x^2]),x]

[Out]

(3*b*x^2*Sqrt[1 - c^2*x^4])/(64*c^3) + (b*x^6*Sqrt[1 - c^2*x^4])/(32*c) - (3*b*ArcSin[c*x^2])/(64*c^4) + (x^8*
(a + b*ArcSin[c*x^2]))/8

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int x^7 \left (a+b \sin ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{8} b \int \frac{2 c x^9}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{4} (b c) \int \frac{x^9}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-c^2 x^2}} \, dx,x,x^2\right )\\ &=\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}+\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx,x,x^2\right )}{32 c}\\ &=\frac{3 b x^2 \sqrt{1-c^2 x^4}}{64 c^3}+\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}+\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2 x^2}} \, dx,x,x^2\right )}{64 c^3}\\ &=\frac{3 b x^2 \sqrt{1-c^2 x^4}}{64 c^3}+\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}-\frac{3 b \sin ^{-1}\left (c x^2\right )}{64 c^4}+\frac{1}{8} x^8 \left (a+b \sin ^{-1}\left (c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0323159, size = 87, normalized size = 1.06 \[ \frac{a x^8}{8}+\frac{b x^6 \sqrt{1-c^2 x^4}}{32 c}+\frac{3 b x^2 \sqrt{1-c^2 x^4}}{64 c^3}-\frac{3 b \sin ^{-1}\left (c x^2\right )}{64 c^4}+\frac{1}{8} b x^8 \sin ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*ArcSin[c*x^2]),x]

[Out]

(a*x^8)/8 + (3*b*x^2*Sqrt[1 - c^2*x^4])/(64*c^3) + (b*x^6*Sqrt[1 - c^2*x^4])/(32*c) - (3*b*ArcSin[c*x^2])/(64*
c^4) + (b*x^8*ArcSin[c*x^2])/8

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Maple [A]  time = 0.024, size = 95, normalized size = 1.2 \begin{align*}{\frac{{x}^{8}a}{8}}+{\frac{b{x}^{8}\arcsin \left ( c{x}^{2} \right ) }{8}}+{\frac{b{x}^{6}}{32\,c}\sqrt{-{c}^{2}{x}^{4}+1}}+{\frac{3\,b{x}^{2}}{64\,{c}^{3}}\sqrt{-{c}^{2}{x}^{4}+1}}-{\frac{3\,b}{64\,{c}^{3}}\arctan \left ({{x}^{2}\sqrt{{c}^{2}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ){\frac{1}{\sqrt{{c}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a+b*arcsin(c*x^2)),x)

[Out]

1/8*x^8*a+1/8*b*x^8*arcsin(c*x^2)+1/32*b*x^6*(-c^2*x^4+1)^(1/2)/c+3/64*b*x^2*(-c^2*x^4+1)^(1/2)/c^3-3/64*b/c^3
/(c^2)^(1/2)*arctan((c^2)^(1/2)*x^2/(-c^2*x^4+1)^(1/2))

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Maxima [A]  time = 1.43572, size = 176, normalized size = 2.15 \begin{align*} \frac{1}{8} \, a x^{8} + \frac{1}{64} \,{\left (8 \, x^{8} \arcsin \left (c x^{2}\right ) + c{\left (\frac{\frac{5 \, \sqrt{-c^{2} x^{4} + 1} c^{2}}{x^{2}} + \frac{3 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}}}{x^{6}}}{c^{8} - \frac{2 \,{\left (c^{2} x^{4} - 1\right )} c^{6}}{x^{4}} + \frac{{\left (c^{2} x^{4} - 1\right )}^{2} c^{4}}{x^{8}}} + \frac{3 \, \arctan \left (\frac{\sqrt{-c^{2} x^{4} + 1}}{c x^{2}}\right )}{c^{5}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arcsin(c*x^2)),x, algorithm="maxima")

[Out]

1/8*a*x^8 + 1/64*(8*x^8*arcsin(c*x^2) + c*((5*sqrt(-c^2*x^4 + 1)*c^2/x^2 + 3*(-c^2*x^4 + 1)^(3/2)/x^6)/(c^8 -
2*(c^2*x^4 - 1)*c^6/x^4 + (c^2*x^4 - 1)^2*c^4/x^8) + 3*arctan(sqrt(-c^2*x^4 + 1)/(c*x^2))/c^5))*b

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Fricas [A]  time = 2.28233, size = 144, normalized size = 1.76 \begin{align*} \frac{8 \, a c^{4} x^{8} +{\left (8 \, b c^{4} x^{8} - 3 \, b\right )} \arcsin \left (c x^{2}\right ) +{\left (2 \, b c^{3} x^{6} + 3 \, b c x^{2}\right )} \sqrt{-c^{2} x^{4} + 1}}{64 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arcsin(c*x^2)),x, algorithm="fricas")

[Out]

1/64*(8*a*c^4*x^8 + (8*b*c^4*x^8 - 3*b)*arcsin(c*x^2) + (2*b*c^3*x^6 + 3*b*c*x^2)*sqrt(-c^2*x^4 + 1))/c^4

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Sympy [A]  time = 12.8467, size = 85, normalized size = 1.04 \begin{align*} \begin{cases} \frac{a x^{8}}{8} + \frac{b x^{8} \operatorname{asin}{\left (c x^{2} \right )}}{8} + \frac{b x^{6} \sqrt{- c^{2} x^{4} + 1}}{32 c} + \frac{3 b x^{2} \sqrt{- c^{2} x^{4} + 1}}{64 c^{3}} - \frac{3 b \operatorname{asin}{\left (c x^{2} \right )}}{64 c^{4}} & \text{for}\: c \neq 0 \\\frac{a x^{8}}{8} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(a+b*asin(c*x**2)),x)

[Out]

Piecewise((a*x**8/8 + b*x**8*asin(c*x**2)/8 + b*x**6*sqrt(-c**2*x**4 + 1)/(32*c) + 3*b*x**2*sqrt(-c**2*x**4 +
1)/(64*c**3) - 3*b*asin(c*x**2)/(64*c**4), Ne(c, 0)), (a*x**8/8, True))

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Giac [A]  time = 1.13194, size = 149, normalized size = 1.82 \begin{align*} \frac{8 \, a c x^{8} -{\left (\frac{2 \,{\left (-c^{2} x^{4} + 1\right )}^{\frac{3}{2}} x^{2}}{c^{2}} - \frac{5 \, \sqrt{-c^{2} x^{4} + 1} x^{2}}{c^{2}} - \frac{8 \,{\left (c^{2} x^{4} - 1\right )}^{2} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac{16 \,{\left (c^{2} x^{4} - 1\right )} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac{5 \, \arcsin \left (c x^{2}\right )}{c^{3}}\right )} b}{64 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arcsin(c*x^2)),x, algorithm="giac")

[Out]

1/64*(8*a*c*x^8 - (2*(-c^2*x^4 + 1)^(3/2)*x^2/c^2 - 5*sqrt(-c^2*x^4 + 1)*x^2/c^2 - 8*(c^2*x^4 - 1)^2*arcsin(c*
x^2)/c^3 - 16*(c^2*x^4 - 1)*arcsin(c*x^2)/c^3 - 5*arcsin(c*x^2)/c^3)*b)/c