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

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

[Out]

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

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Rubi [A]  time = 0.0426628, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4842, 12, 266, 51, 63, 208} \[ \frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{6} b c x^2 \sqrt{1-\frac{c^2}{x^2}}+\frac{1}{6} b c^3 \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c/x]),x]

[Out]

(b*c*Sqrt[1 - c^2/x^2]*x^2)/6 + (x^3*(a + b*ArcSin[c/x]))/3 + (b*c^3*ArcTanh[Sqrt[1 - c^2/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 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.0335098, size = 79, normalized size = 1.23 \[ \frac{a x^3}{3}+\frac{1}{6} b c x^2 \sqrt{\frac{x^2-c^2}{x^2}}+\frac{1}{6} b c^3 \log \left (x \left (\sqrt{\frac{x^2-c^2}{x^2}}+1\right )\right )+\frac{1}{3} b x^3 \sin ^{-1}\left (\frac{c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSin[c/x]),x]

[Out]

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

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Maple [A]  time = 0.005, size = 68, normalized size = 1.1 \begin{align*} -{c}^{3} \left ( -{\frac{{x}^{3}a}{3\,{c}^{3}}}+b \left ( -{\frac{{x}^{3}}{3\,{c}^{3}}\arcsin \left ({\frac{c}{x}} \right ) }-{\frac{{x}^{2}}{6\,{c}^{2}}\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}}-{\frac{1}{6}{\it Artanh} \left ({\frac{1}{\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}}} \right ) } \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43071, size = 109, normalized size = 1.7 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{12} \,{\left (4 \, x^{3} \arcsin \left (\frac{c}{x}\right ) +{\left (c^{2} \log \left (\sqrt{-\frac{c^{2}}{x^{2}} + 1} + 1\right ) - c^{2} \log \left (\sqrt{-\frac{c^{2}}{x^{2}} + 1} - 1\right ) + 2 \, x^{2} \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^2*(a+b*arcsin(c/x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.50002, size = 235, normalized size = 3.67 \begin{align*} -\frac{1}{6} \, b c^{3} \log \left (x \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} - x\right ) + \frac{1}{6} \, b c x^{2} \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} + \frac{1}{3} \, a x^{3} + \frac{1}{3} \,{\left (b x^{3} - b\right )} \arcsin \left (\frac{c}{x}\right ) - \frac{2}{3} \, b \arctan \left (\frac{x \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} - x}{c}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.79475, size = 109, normalized size = 1.7 \begin{align*} \frac{a x^{3}}{3} + \frac{b c \left (\begin{cases} \frac{c^{2} \operatorname{acosh}{\left (\frac{x}{c} \right )}}{2} + \frac{c x \sqrt{-1 + \frac{x^{2}}{c^{2}}}}{2} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{c^{2}}\right |} > 1 \\- \frac{i c^{2} \operatorname{asin}{\left (\frac{x}{c} \right )}}{2} + \frac{i c x}{2 \sqrt{1 - \frac{x^{2}}{c^{2}}}} - \frac{i x^{3}}{2 c \sqrt{1 - \frac{x^{2}}{c^{2}}}} & \text{otherwise} \end{cases}\right )}{3} + \frac{b x^{3} \operatorname{asin}{\left (\frac{c}{x} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c/x) + a)*x^2, x)