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

Optimal. Leaf size=64 \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{12} b c x^3 \sqrt{1-\frac{c^2}{x^2}}+\frac{1}{6} b c^3 x \sqrt{1-\frac{c^2}{x^2}} \]

[Out]

(b*c^3*Sqrt[1 - c^2/x^2]*x)/6 + (b*c*Sqrt[1 - c^2/x^2]*x^3)/12 + (x^4*(a + b*ArcSin[c/x]))/4

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Rubi [A]  time = 0.0380079, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4842, 12, 271, 191} \[ \frac{1}{4} x^4 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{12} b c x^3 \sqrt{1-\frac{c^2}{x^2}}+\frac{1}{6} b c^3 x \sqrt{1-\frac{c^2}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcSin[c/x]),x]

[Out]

(b*c^3*Sqrt[1 - c^2/x^2]*x)/6 + (b*c*Sqrt[1 - c^2/x^2]*x^3)/12 + (x^4*(a + b*ArcSin[c/x]))/4

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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.0411473, size = 59, normalized size = 0.92 \[ \frac{a x^4}{4}+b \sqrt{\frac{x^2-c^2}{x^2}} \left (\frac{c^3 x}{6}+\frac{c x^3}{12}\right )+\frac{1}{4} b x^4 \sin ^{-1}\left (\frac{c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcSin[c/x]),x]

[Out]

(a*x^4)/4 + b*Sqrt[(-c^2 + x^2)/x^2]*((c^3*x)/6 + (c*x^3)/12) + (b*x^4*ArcSin[c/x])/4

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Maple [A]  time = 0.006, size = 71, normalized size = 1.1 \begin{align*} -{c}^{4} \left ( -{\frac{{x}^{4}a}{4\,{c}^{4}}}+b \left ( -{\frac{{x}^{4}}{4\,{c}^{4}}\arcsin \left ({\frac{c}{x}} \right ) }-{\frac{{x}^{3}}{12\,{c}^{3}}\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}}-{\frac{x}{6\,c}\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}} \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]

-c^4*(-1/4*a/c^4*x^4+b*(-1/4/c^4*x^4*arcsin(c/x)-1/12/c^3*x^3*(1-c^2/x^2)^(1/2)-1/6/c*x*(1-c^2/x^2)^(1/2)))

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Maxima [A]  time = 1.41902, size = 80, normalized size = 1.25 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{12} \,{\left (3 \, x^{4} \arcsin \left (\frac{c}{x}\right ) +{\left (x^{3}{\left (-\frac{c^{2}}{x^{2}} + 1\right )}^{\frac{3}{2}} + 3 \, c^{2} x \sqrt{-\frac{c^{2}}{x^{2}} + 1}\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 + 1/12*(3*x^4*arcsin(c/x) + (x^3*(-c^2/x^2 + 1)^(3/2) + 3*c^2*x*sqrt(-c^2/x^2 + 1))*c)*b

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Fricas [A]  time = 2.44223, size = 117, normalized size = 1.83 \begin{align*} \frac{1}{4} \, b x^{4} \arcsin \left (\frac{c}{x}\right ) + \frac{1}{4} \, a x^{4} + \frac{1}{12} \,{\left (2 \, b c^{3} x + b c x^{3}\right )} \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} \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]

1/4*b*x^4*arcsin(c/x) + 1/4*a*x^4 + 1/12*(2*b*c^3*x + b*c*x^3)*sqrt(-(c^2 - x^2)/x^2)

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Sympy [A]  time = 5.07008, size = 109, normalized size = 1.7 \begin{align*} \frac{a x^{4}}{4} + \frac{b c \left (\begin{cases} \frac{2 c^{3} \sqrt{-1 + \frac{x^{2}}{c^{2}}}}{3} + \frac{c x^{2} \sqrt{-1 + \frac{x^{2}}{c^{2}}}}{3} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{c^{2}}\right |} > 1 \\\frac{2 i c^{3} \sqrt{1 - \frac{x^{2}}{c^{2}}}}{3} + \frac{i c x^{2} \sqrt{1 - \frac{x^{2}}{c^{2}}}}{3} & \text{otherwise} \end{cases}\right )}{4} + \frac{b x^{4} \operatorname{asin}{\left (\frac{c}{x} \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**4/4 + b*c*Piecewise((2*c**3*sqrt(-1 + x**2/c**2)/3 + c*x**2*sqrt(-1 + x**2/c**2)/3, Abs(x**2)/Abs(c**2) >
1), (2*I*c**3*sqrt(1 - x**2/c**2)/3 + I*c*x**2*sqrt(1 - x**2/c**2)/3, True))/4 + b*x**4*asin(c/x)/4

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (\frac{c}{x}\right ) + a\right )} x^{3}\,{d x} \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]

integrate((b*arcsin(c/x) + a)*x^3, x)