3.140 \(\int x^3 (a+b \sin ^{-1}(c x)) \, dx\)

Optimal. Leaf size=76 \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}-\frac{3 b \sin ^{-1}(c x)}{32 c^4} \]

[Out]

(3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) - (3*b*ArcSin[c*x])/(32*c^4) + (x^4*(a +
b*ArcSin[c*x]))/4

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Rubi [A]  time = 0.0345858, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4627, 321, 216} \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}-\frac{3 b \sin ^{-1}(c x)}{32 c^4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSin[c*x]),x]

[Out]

(3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) - (3*b*ArcSin[c*x])/(32*c^4) + (x^4*(a +
b*ArcSin[c*x]))/4

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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^3 \left (a+b \sin ^{-1}(c x)\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{4} (b c) \int \frac{x^4}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac{(3 b) \int \frac{x^2}{\sqrt{1-c^2 x^2}} \, dx}{16 c}\\ &=\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}+\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac{(3 b) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{32 c^3}\\ &=\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}+\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}-\frac{3 b \sin ^{-1}(c x)}{32 c^4}+\frac{1}{4} x^4 \left (a+b \sin ^{-1}(c x)\right )\\ \end{align*}

Mathematica [A]  time = 0.029848, size = 81, normalized size = 1.07 \[ \frac{a x^4}{4}+\frac{b x^3 \sqrt{1-c^2 x^2}}{16 c}+\frac{3 b x \sqrt{1-c^2 x^2}}{32 c^3}-\frac{3 b \sin ^{-1}(c x)}{32 c^4}+\frac{1}{4} b x^4 \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSin[c*x]),x]

[Out]

(a*x^4)/4 + (3*b*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*x^3*Sqrt[1 - c^2*x^2])/(16*c) - (3*b*ArcSin[c*x])/(32*c^4)
+ (b*x^4*ArcSin[c*x])/4

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Maple [A]  time = 0.005, size = 72, normalized size = 1. \begin{align*}{\frac{1}{{c}^{4}} \left ({\frac{{c}^{4}{x}^{4}a}{4}}+b \left ({\frac{{c}^{4}{x}^{4}\arcsin \left ( cx \right ) }{4}}+{\frac{{c}^{3}{x}^{3}}{16}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{3\,cx}{32}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{3\,\arcsin \left ( cx \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsin(c*x)),x)

[Out]

1/c^4*(1/4*c^4*x^4*a+b*(1/4*c^4*x^4*arcsin(c*x)+1/16*c^3*x^3*(-c^2*x^2+1)^(1/2)+3/32*c*x*(-c^2*x^2+1)^(1/2)-3/
32*arcsin(c*x)))

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Maxima [A]  time = 1.59075, size = 111, normalized size = 1.46 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{32} \,{\left (8 \, x^{4} \arcsin \left (c x\right ) +{\left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac{3 \, \sqrt{-c^{2} x^{2} + 1} x}{c^{4}} - \frac{3 \, \arcsin \left (\frac{c^{2} x}{\sqrt{c^{2}}}\right )}{\sqrt{c^{2}} c^{4}}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.43564, size = 139, normalized size = 1.83 \begin{align*} \frac{8 \, a c^{4} x^{4} +{\left (8 \, b c^{4} x^{4} - 3 \, b\right )} \arcsin \left (c x\right ) +{\left (2 \, b c^{3} x^{3} + 3 \, b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{32 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.3713, size = 80, normalized size = 1.05 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{asin}{\left (c x \right )}}{4} + \frac{b x^{3} \sqrt{- c^{2} x^{2} + 1}}{16 c} + \frac{3 b x \sqrt{- c^{2} x^{2} + 1}}{32 c^{3}} - \frac{3 b \operatorname{asin}{\left (c x \right )}}{32 c^{4}} & \text{for}\: c \neq 0 \\\frac{a x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asin(c*x)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*asin(c*x)/4 + b*x**3*sqrt(-c**2*x**2 + 1)/(16*c) + 3*b*x*sqrt(-c**2*x**2 + 1)/(32
*c**3) - 3*b*asin(c*x)/(32*c**4), Ne(c, 0)), (a*x**4/4, True))

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Giac [A]  time = 1.37045, size = 163, normalized size = 2.14 \begin{align*} -\frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}} b x}{16 \, c^{3}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} b \arcsin \left (c x\right )}{4 \, c^{4}} + \frac{5 \, \sqrt{-c^{2} x^{2} + 1} b x}{32 \, c^{3}} + \frac{{\left (c^{2} x^{2} - 1\right )}^{2} a}{4 \, c^{4}} + \frac{{\left (c^{2} x^{2} - 1\right )} b \arcsin \left (c x\right )}{2 \, c^{4}} + \frac{{\left (c^{2} x^{2} - 1\right )} a}{2 \, c^{4}} + \frac{5 \, b \arcsin \left (c x\right )}{32 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

-1/16*(-c^2*x^2 + 1)^(3/2)*b*x/c^3 + 1/4*(c^2*x^2 - 1)^2*b*arcsin(c*x)/c^4 + 5/32*sqrt(-c^2*x^2 + 1)*b*x/c^3 +
1/4*(c^2*x^2 - 1)^2*a/c^4 + 1/2*(c^2*x^2 - 1)*b*arcsin(c*x)/c^4 + 1/2*(c^2*x^2 - 1)*a/c^4 + 5/32*b*arcsin(c*x
)/c^4