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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 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^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*}

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Mathematica [A]  time = 0.09, size = 91, normalized size = 1.02 \[ \frac {a x^5}{5}+\frac {3}{40} b c^5 \log \left (x \left (\sqrt {\frac {x^2-c^2}{x^2}}+1\right )\right )+b \sqrt {\frac {x^2-c^2}{x^2}} \left (\frac {3 c^3 x^2}{40}+\frac {c x^4}{20}\right )+\frac {1}{5} b x^5 \sin ^{-1}\left (\frac {c}{x}\right ) \]

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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fricas [A]  time = 0.73, size = 118, normalized size = 1.33 \[ -\frac {3}{40} \, b c^{5} \log \left (x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x\right ) + \frac {1}{5} \, a x^{5} + \frac {1}{5} \, {\left (b x^{5} - b\right )} \arcsin \left (\frac {c}{x}\right ) - \frac {2}{5} \, b \arctan \left (\frac {x \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} - x}{c}\right ) + \frac {1}{40} \, {\left (3 \, b c^{3} x^{2} + 2 \, b c x^{4}\right )} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} \]

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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giac [B]  time = 2.31, size = 464, normalized size = 5.21 \[ \frac {2 \, b c x^{5} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{5} \arcsin \left (\frac {c}{x}\right ) + 2 \, a c x^{5} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{5} + b c^{2} x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4} + 10 \, b c^{3} x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3} \arcsin \left (\frac {c}{x}\right ) + 10 \, a c^{3} x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3} + 8 \, b c^{4} x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2} + 20 \, b c^{5} x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )} \arcsin \left (\frac {c}{x}\right ) + 20 \, a c^{5} x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )} + 24 \, b c^{6} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - 24 \, b c^{6} \log \left (\frac {{\left | c \right |}}{{\left | x \right |}}\right ) + \frac {20 \, b c^{7} \arcsin \left (\frac {c}{x}\right )}{x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}} + \frac {20 \, a c^{7}}{x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}} - \frac {8 \, b c^{8}}{x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2}} + \frac {10 \, b c^{9} \arcsin \left (\frac {c}{x}\right )}{x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3}} + \frac {10 \, a c^{9}}{x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3}} - \frac {b c^{10}}{x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4}} + \frac {2 \, b c^{11} \arcsin \left (\frac {c}{x}\right )}{x^{5} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{5}} + \frac {2 \, a c^{11}}{x^{5} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{5}}}{320 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*(2*b*c*x^5*(sqrt(-c^2/x^2 + 1) + 1)^5*arcsin(c/x) + 2*a*c*x^5*(sqrt(-c^2/x^2 + 1) + 1)^5 + b*c^2*x^4*(sq
rt(-c^2/x^2 + 1) + 1)^4 + 10*b*c^3*x^3*(sqrt(-c^2/x^2 + 1) + 1)^3*arcsin(c/x) + 10*a*c^3*x^3*(sqrt(-c^2/x^2 +
1) + 1)^3 + 8*b*c^4*x^2*(sqrt(-c^2/x^2 + 1) + 1)^2 + 20*b*c^5*x*(sqrt(-c^2/x^2 + 1) + 1)*arcsin(c/x) + 20*a*c^
5*x*(sqrt(-c^2/x^2 + 1) + 1) + 24*b*c^6*log(sqrt(-c^2/x^2 + 1) + 1) - 24*b*c^6*log(abs(c)/abs(x)) + 20*b*c^7*a
rcsin(c/x)/(x*(sqrt(-c^2/x^2 + 1) + 1)) + 20*a*c^7/(x*(sqrt(-c^2/x^2 + 1) + 1)) - 8*b*c^8/(x^2*(sqrt(-c^2/x^2
+ 1) + 1)^2) + 10*b*c^9*arcsin(c/x)/(x^3*(sqrt(-c^2/x^2 + 1) + 1)^3) + 10*a*c^9/(x^3*(sqrt(-c^2/x^2 + 1) + 1)^
3) - b*c^10/(x^4*(sqrt(-c^2/x^2 + 1) + 1)^4) + 2*b*c^11*arcsin(c/x)/(x^5*(sqrt(-c^2/x^2 + 1) + 1)^5) + 2*a*c^1
1/(x^5*(sqrt(-c^2/x^2 + 1) + 1)^5))/c

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maple [A]  time = 0.02, size = 88, normalized size = 0.99 \[ -c^{5} \left (-\frac {a \,x^{5}}{5 c^{5}}+b \left (-\frac {\arcsin \left (\frac {c}{x}\right ) x^{5}}{5 c^{5}}-\frac {x^{4} \sqrt {1-\frac {c^{2}}{x^{2}}}}{20 c^{4}}-\frac {3 x^{2} \sqrt {1-\frac {c^{2}}{x^{2}}}}{40 c^{2}}-\frac {3 \arctanh \left (\frac {1}{\sqrt {1-\frac {c^{2}}{x^{2}}}}\right )}{40}\right )\right ) \]

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 = 0.61, size = 125, normalized size = 1.40 \[ \frac {1}{5} \, a x^{5} + \frac {1}{80} \, {\left (16 \, x^{5} \arcsin \left (\frac {c}{x}\right ) + {\left (3 \, c^{4} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right ) - 3 \, c^{4} \log \left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (3 \, c^{4} {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}} - 5 \, c^{4} \sqrt {-\frac {c^{2}}{x^{2}} + 1}\right )}}{{\left (\frac {c^{2}}{x^{2}} - 1\right )}^{2} + \frac {2 \, c^{2}}{x^{2}} - 1}\right )} c\right )} b \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\left (a+b\,\mathrm {asin}\left (\frac {c}{x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*asin(c/x)),x)

[Out]

int(x^4*(a + b*asin(c/x)), x)

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sympy [A]  time = 5.26, size = 175, normalized size = 1.97 \[ \frac {a x^{5}}{5} + \frac {b c \left (\begin {cases} \frac {3 c^{4} \operatorname {acosh}{\left (\frac {x}{c} \right )}}{8} - \frac {3 c^{3} x}{8 \sqrt {-1 + \frac {x^{2}}{c^{2}}}} + \frac {c x^{3}}{8 \sqrt {-1 + \frac {x^{2}}{c^{2}}}} + \frac {x^{5}}{4 c \sqrt {-1 + \frac {x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {x^{2}}{c^{2}}}\right | > 1 \\- \frac {3 i c^{4} \operatorname {asin}{\left (\frac {x}{c} \right )}}{8} + \frac {3 i c^{3} x}{8 \sqrt {1 - \frac {x^{2}}{c^{2}}}} - \frac {i c x^{3}}{8 \sqrt {1 - \frac {x^{2}}{c^{2}}}} - \frac {i x^{5}}{4 c \sqrt {1 - \frac {x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right )}{5} + \frac {b x^{5} \operatorname {asin}{\left (\frac {c}{x} \right )}}{5} \]

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/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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