3.369 \(\int x^4 (a+b \sin ^{-1}(\frac{c}{x})) \, dx\)

Optimal. Leaf size=89 \[ \frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{3}{40} b c^3 x^2 \sqrt{1-\frac{c^2}{x^2}}+\frac{1}{20} b c x^4 \sqrt{1-\frac{c^2}{x^2}}+\frac{3}{40} b c^5 \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{x^2}}\right ) \]

[Out]

(3*b*c^3*Sqrt[1 - c^2/x^2]*x^2)/40 + (b*c*Sqrt[1 - c^2/x^2]*x^4)/20 + (x^5*(a + b*ArcSin[c/x]))/5 + (3*b*c^5*A
rcTanh[Sqrt[1 - c^2/x^2]])/40

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Rubi [A]  time = 0.0587458, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{3}{40} b c^3 x^2 \sqrt{1-\frac{c^2}{x^2}}+\frac{1}{20} b c x^4 \sqrt{1-\frac{c^2}{x^2}}+\frac{3}{40} b c^5 \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcSin[c/x]),x]

[Out]

(3*b*c^3*Sqrt[1 - c^2/x^2]*x^2)/40 + (b*c*Sqrt[1 - c^2/x^2]*x^4)/20 + (x^5*(a + b*ArcSin[c/x]))/5 + (3*b*c^5*A
rcTanh[Sqrt[1 - c^2/x^2]])/40

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^4 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{5} b \int \frac{c x^3}{\sqrt{1-\frac{c^2}{x^2}}} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{5} (b c) \int \frac{x^3}{\sqrt{1-\frac{c^2}{x^2}}} \, dx\\ &=\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{10} (b c) \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1-c^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{1}{20} b c \sqrt{1-\frac{c^2}{x^2}} x^4+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{40} \left (3 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-c^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{3}{40} b c^3 \sqrt{1-\frac{c^2}{x^2}} x^2+\frac{1}{20} b c \sqrt{1-\frac{c^2}{x^2}} x^4+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )-\frac{1}{80} \left (3 b c^5\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{3}{40} b c^3 \sqrt{1-\frac{c^2}{x^2}} x^2+\frac{1}{20} b c \sqrt{1-\frac{c^2}{x^2}} x^4+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{1}{40} \left (3 b c^3\right ) \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{3}{40} b c^3 \sqrt{1-\frac{c^2}{x^2}} x^2+\frac{1}{20} b c \sqrt{1-\frac{c^2}{x^2}} x^4+\frac{1}{5} x^5 \left (a+b \sin ^{-1}\left (\frac{c}{x}\right )\right )+\frac{3}{40} b c^5 \tanh ^{-1}\left (\sqrt{1-\frac{c^2}{x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0711722, size = 91, normalized size = 1.02 \[ \frac{a x^5}{5}+b \sqrt{\frac{x^2-c^2}{x^2}} \left (\frac{3 c^3 x^2}{40}+\frac{c x^4}{20}\right )+\frac{3}{40} b c^5 \log \left (x \left (\sqrt{\frac{x^2-c^2}{x^2}}+1\right )\right )+\frac{1}{5} b x^5 \sin ^{-1}\left (\frac{c}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcSin[c/x]),x]

[Out]

(a*x^5)/5 + b*Sqrt[(-c^2 + x^2)/x^2]*((3*c^3*x^2)/40 + (c*x^4)/20) + (b*x^5*ArcSin[c/x])/5 + (3*b*c^5*Log[x*(1
+ Sqrt[(-c^2 + x^2)/x^2])])/40

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Maple [A]  time = 0.018, size = 88, normalized size = 1. \begin{align*} -{c}^{5} \left ( -{\frac{a{x}^{5}}{5\,{c}^{5}}}+b \left ( -{\frac{{x}^{5}}{5\,{c}^{5}}\arcsin \left ({\frac{c}{x}} \right ) }-{\frac{{x}^{4}}{20\,{c}^{4}}\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}}-{\frac{3\,{x}^{2}}{40\,{c}^{2}}\sqrt{1-{\frac{{c}^{2}}{{x}^{2}}}}}-{\frac{3}{40}{\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^4*(a+b*arcsin(c/x)),x)

[Out]

-c^5*(-1/5*a/c^5*x^5+b*(-1/5*arcsin(c/x)/c^5*x^5-1/20/c^4*x^4*(1-c^2/x^2)^(1/2)-3/40/c^2*x^2*(1-c^2/x^2)^(1/2)
-3/40*arctanh(1/(1-c^2/x^2)^(1/2))))

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Maxima [A]  time = 1.43042, size = 169, normalized size = 1.9 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{80} \,{\left (16 \, x^{5} \arcsin \left (\frac{c}{x}\right ) +{\left (3 \, c^{4} \log \left (\sqrt{-\frac{c^{2}}{x^{2}} + 1} + 1\right ) - 3 \, c^{4} \log \left (\sqrt{-\frac{c^{2}}{x^{2}} + 1} - 1\right ) - \frac{2 \,{\left (3 \, c^{4}{\left (-\frac{c^{2}}{x^{2}} + 1\right )}^{\frac{3}{2}} - 5 \, c^{4} \sqrt{-\frac{c^{2}}{x^{2}} + 1}\right )}}{{\left (\frac{c^{2}}{x^{2}} - 1\right )}^{2} + \frac{2 \, c^{2}}{x^{2}} - 1}\right )} c\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c/x)),x, algorithm="maxima")

[Out]

1/5*a*x^5 + 1/80*(16*x^5*arcsin(c/x) + (3*c^4*log(sqrt(-c^2/x^2 + 1) + 1) - 3*c^4*log(sqrt(-c^2/x^2 + 1) - 1)
- 2*(3*c^4*(-c^2/x^2 + 1)^(3/2) - 5*c^4*sqrt(-c^2/x^2 + 1))/((c^2/x^2 - 1)^2 + 2*c^2/x^2 - 1))*c)*b

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Fricas [A]  time = 2.59502, size = 262, normalized size = 2.94 \begin{align*} -\frac{3}{40} \, b c^{5} \log \left (x \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} - x\right ) + \frac{1}{5} \, a x^{5} + \frac{1}{5} \,{\left (b x^{5} - b\right )} \arcsin \left (\frac{c}{x}\right ) - \frac{2}{5} \, b \arctan \left (\frac{x \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} - x}{c}\right ) + \frac{1}{40} \,{\left (3 \, b c^{3} x^{2} + 2 \, b c x^{4}\right )} \sqrt{-\frac{c^{2} - x^{2}}{x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c/x)),x, algorithm="fricas")

[Out]

-3/40*b*c^5*log(x*sqrt(-(c^2 - x^2)/x^2) - x) + 1/5*a*x^5 + 1/5*(b*x^5 - b)*arcsin(c/x) - 2/5*b*arctan((x*sqrt
(-(c^2 - x^2)/x^2) - x)/c) + 1/40*(3*b*c^3*x^2 + 2*b*c*x^4)*sqrt(-(c^2 - x^2)/x^2)

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Sympy [A]  time = 10.2723, size = 177, normalized size = 1.99 \begin{align*} \frac{a x^{5}}{5} + \frac{b c \left (\begin{cases} \frac{3 c^{4} \operatorname{acosh}{\left (\frac{x}{c} \right )}}{8} - \frac{3 c^{3} x}{8 \sqrt{-1 + \frac{x^{2}}{c^{2}}}} + \frac{c x^{3}}{8 \sqrt{-1 + \frac{x^{2}}{c^{2}}}} + \frac{x^{5}}{4 c \sqrt{-1 + \frac{x^{2}}{c^{2}}}} & \text{for}\: \frac{\left |{x^{2}}\right |}{\left |{c^{2}}\right |} > 1 \\- \frac{3 i c^{4} \operatorname{asin}{\left (\frac{x}{c} \right )}}{8} + \frac{3 i c^{3} x}{8 \sqrt{1 - \frac{x^{2}}{c^{2}}}} - \frac{i c x^{3}}{8 \sqrt{1 - \frac{x^{2}}{c^{2}}}} - \frac{i x^{5}}{4 c \sqrt{1 - \frac{x^{2}}{c^{2}}}} & \text{otherwise} \end{cases}\right )}{5} + \frac{b x^{5} \operatorname{asin}{\left (\frac{c}{x} \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asin(c/x)),x)

[Out]

a*x**5/5 + b*c*Piecewise((3*c**4*acosh(x/c)/8 - 3*c**3*x/(8*sqrt(-1 + x**2/c**2)) + c*x**3/(8*sqrt(-1 + x**2/c
**2)) + x**5/(4*c*sqrt(-1 + x**2/c**2)), Abs(x**2)/Abs(c**2) > 1), (-3*I*c**4*asin(x/c)/8 + 3*I*c**3*x/(8*sqrt
(1 - x**2/c**2)) - I*c*x**3/(8*sqrt(1 - x**2/c**2)) - I*x**5/(4*c*sqrt(1 - x**2/c**2)), True))/5 + b*x**5*asin
(c/x)/5

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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^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsin(c/x)),x, algorithm="giac")

[Out]

integrate((b*arcsin(c/x) + a)*x^4, x)