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]

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

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Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.51, size = 106, normalized size = 1.66 \[ -\frac {1}{6} \, b c^{3} \log \left (x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x\right ) + \frac {1}{6} \, b c x^{2} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} + \frac {1}{3} \, a x^{3} + \frac {1}{3} \, {\left (b x^{3} - b\right )} \arcsin \left (\frac {c}{x}\right ) - \frac {2}{3} \, b \arctan \left (\frac {x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x}{c}\right ) \]

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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giac [B]  time = 0.69, size = 298, normalized size = 4.66 \[ \frac {b c x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3} \arcsin \left (\frac {c}{x}\right ) + a c x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3} + b c^{2} x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2} + 3 \, b c^{3} x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )} \arcsin \left (\frac {c}{x}\right ) + 3 \, a c^{3} x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )} + 4 \, b c^{4} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - 4 \, b c^{4} \log \left (\frac {{\left | c \right |}}{{\left | x \right |}}\right ) + \frac {3 \, b c^{5} \arcsin \left (\frac {c}{x}\right )}{x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}} + \frac {3 \, a c^{5}}{x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}} - \frac {b c^{6}}{x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2}} + \frac {b c^{7} \arcsin \left (\frac {c}{x}\right )}{x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3}} + \frac {a c^{7}}{x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3}}}{24 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(b*c*x^3*(sqrt(-c^2/x^2 + 1) + 1)^3*arcsin(c/x) + a*c*x^3*(sqrt(-c^2/x^2 + 1) + 1)^3 + b*c^2*x^2*(sqrt(-c
^2/x^2 + 1) + 1)^2 + 3*b*c^3*x*(sqrt(-c^2/x^2 + 1) + 1)*arcsin(c/x) + 3*a*c^3*x*(sqrt(-c^2/x^2 + 1) + 1) + 4*b
*c^4*log(sqrt(-c^2/x^2 + 1) + 1) - 4*b*c^4*log(abs(c)/abs(x)) + 3*b*c^5*arcsin(c/x)/(x*(sqrt(-c^2/x^2 + 1) + 1
)) + 3*a*c^5/(x*(sqrt(-c^2/x^2 + 1) + 1)) - b*c^6/(x^2*(sqrt(-c^2/x^2 + 1) + 1)^2) + b*c^7*arcsin(c/x)/(x^3*(s
qrt(-c^2/x^2 + 1) + 1)^3) + a*c^7/(x^3*(sqrt(-c^2/x^2 + 1) + 1)^3))/c

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maple [A]  time = 0.01, size = 68, normalized size = 1.06 \[ -c^{3} \left (-\frac {a \,x^{3}}{3 c^{3}}+b \left (-\frac {x^{3} \arcsin \left (\frac {c}{x}\right )}{3 c^{3}}-\frac {x^{2} \sqrt {1-\frac {c^{2}}{x^{2}}}}{6 c^{2}}-\frac {\arctanh \left (\frac {1}{\sqrt {1-\frac {c^{2}}{x^{2}}}}\right )}{6}\right )\right ) \]

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 = 0.64, size = 81, normalized size = 1.27 \[ \frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \arcsin \left (\frac {c}{x}\right ) + {\left (c^{2} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - c^{2} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} - 1\right ) + 2 \, x^{2} \sqrt {-\frac {c^{2}}{x^{2}} + 1}\right )} c\right )} b \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^2\,\left (a+b\,\mathrm {asin}\left (\frac {c}{x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asin(c/x)),x)

[Out]

int(x^2*(a + b*asin(c/x)), x)

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sympy [A]  time = 3.18, size = 107, normalized size = 1.67 \[ \frac {a x^{3}}{3} + \frac {b c \left (\begin {cases} \frac {c^{2} \operatorname {acosh}{\left (\frac {x}{c} \right )}}{2} + \frac {c x \sqrt {-1 + \frac {x^{2}}{c^{2}}}}{2} & \text {for}\: \left |{\frac {x^{2}}{c^{2}}}\right | > 1 \\- \frac {i c^{2} \operatorname {asin}{\left (\frac {x}{c} \right )}}{2} + \frac {i c x}{2 \sqrt {1 - \frac {x^{2}}{c^{2}}}} - \frac {i x^{3}}{2 c \sqrt {1 - \frac {x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right )}{3} + \frac {b x^{3} \operatorname {asin}{\left (\frac {c}{x} \right )}}{3} \]

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/c**2) > 1), (-I*c**2*asin(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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