Integrand size = 14, antiderivative size = 64 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{6} b c^3 \sqrt {1-\frac {c^2}{x^2}} x+\frac {1}{12} b c \sqrt {1-\frac {c^2}{x^2}} x^3+\frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \] Output:
1/6*b*c^3*(1-c^2/x^2)^(1/2)*x+1/12*b*c*(1-c^2/x^2)^(1/2)*x^3+1/4*x^4*(a+b* arcsin(c/x))
Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {a x^4}{4}+b \sqrt {\frac {-c^2+x^2}{x^2}} \left (\frac {c^3 x}{6}+\frac {c x^3}{12}\right )+\frac {1}{4} b x^4 \arcsin \left (\frac {c}{x}\right ) \] Input:
Integrate[x^3*(a + b*ArcSin[c/x]),x]
Output:
(a*x^4)/4 + b*Sqrt[(-c^2 + x^2)/x^2]*((c^3*x)/6 + (c*x^3)/12) + (b*x^4*Arc Sin[c/x])/4
Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5341, 25, 27, 803, 746}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 5341 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right )-\frac {1}{4} b \int -\frac {c x^2}{\sqrt {1-\frac {c^2}{x^2}}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} b \int \frac {c x^2}{\sqrt {1-\frac {c^2}{x^2}}}dx+\frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} b c \int \frac {x^2}{\sqrt {1-\frac {c^2}{x^2}}}dx+\frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {1}{4} b c \left (\frac {2}{3} c^2 \int \frac {1}{\sqrt {1-\frac {c^2}{x^2}}}dx+\frac {1}{3} x^3 \sqrt {1-\frac {c^2}{x^2}}\right )+\frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right )\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \arcsin \left (\frac {c}{x}\right )\right )+\frac {1}{4} b c \left (\frac {2}{3} c^2 x \sqrt {1-\frac {c^2}{x^2}}+\frac {1}{3} x^3 \sqrt {1-\frac {c^2}{x^2}}\right )\) |
Input:
Int[x^3*(a + b*ArcSin[c/x]),x]
Output:
(b*c*((2*c^2*Sqrt[1 - c^2/x^2]*x)/3 + (Sqrt[1 - c^2/x^2]*x^3)/3))/4 + (x^4 *(a + b*ArcSin[c/x]))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[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] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcSin[u])/(d*(m + 1))), x] - Simp[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]
Time = 0.06 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.05
method | result | size |
parts | \(\frac {a \,x^{4}}{4}-b \,c^{4} \left (-\frac {x^{4} \arcsin \left (\frac {c}{x}\right )}{4 c^{4}}-\frac {x^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{12 c^{3}}-\frac {x \sqrt {1-\frac {c^{2}}{x^{2}}}}{6 c}\right )\) | \(67\) |
derivativedivides | \(-c^{4} \left (-\frac {a \,x^{4}}{4 c^{4}}+b \left (-\frac {x^{4} \arcsin \left (\frac {c}{x}\right )}{4 c^{4}}-\frac {x^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{12 c^{3}}-\frac {x \sqrt {1-\frac {c^{2}}{x^{2}}}}{6 c}\right )\right )\) | \(71\) |
default | \(-c^{4} \left (-\frac {a \,x^{4}}{4 c^{4}}+b \left (-\frac {x^{4} \arcsin \left (\frac {c}{x}\right )}{4 c^{4}}-\frac {x^{3} \sqrt {1-\frac {c^{2}}{x^{2}}}}{12 c^{3}}-\frac {x \sqrt {1-\frac {c^{2}}{x^{2}}}}{6 c}\right )\right )\) | \(71\) |
orering | \(-\frac {\left (2 c^{4}-c^{2} x^{2}-2 x^{4}\right ) \left (a +b \arcsin \left (\frac {c}{x}\right )\right )}{4}+\frac {\left (2 c^{2}+x^{2}\right ) \left (-x +c \right ) \left (x +c \right ) \left (3 x^{2} \left (a +b \arcsin \left (\frac {c}{x}\right )\right )-\frac {x b c}{\sqrt {1-\frac {c^{2}}{x^{2}}}}\right )}{12 x^{2}}\) | \(88\) |
Input:
int(x^3*(a+b*arcsin(c/x)),x,method=_RETURNVERBOSE)
Output:
1/4*a*x^4-b*c^4*(-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))
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \arcsin \left (\frac {c}{x}\right ) + \frac {1}{4} \, a x^{4} + \frac {1}{12} \, {\left (2 \, b c^{3} x + b c x^{3}\right )} \sqrt {-\frac {c^{2} - x^{2}}{x^{2}}} \] Input:
integrate(x^3*(a+b*arcsin(c/x)),x, algorithm="fricas")
Output:
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)
Time = 1.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.67 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {a x^{4}}{4} + \frac {b c \left (\begin {cases} \frac {2 c^{3} \sqrt {-1 + \frac {x^{2}}{c^{2}}}}{3} + \frac {c x^{2} \sqrt {-1 + \frac {x^{2}}{c^{2}}}}{3} & \text {for}\: \left |{\frac {x^{2}}{c^{2}}}\right | > 1 \\\frac {2 i c^{3} \sqrt {1 - \frac {x^{2}}{c^{2}}}}{3} + \frac {i c x^{2} \sqrt {1 - \frac {x^{2}}{c^{2}}}}{3} & \text {otherwise} \end {cases}\right )}{4} + \frac {b x^{4} \operatorname {asin}{\left (\frac {c}{x} \right )}}{4} \] Input:
integrate(x**3*(a+b*asin(c/x)),x)
Output:
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/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
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.92 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{12} \, {\left (3 \, x^{4} \arcsin \left (\frac {c}{x}\right ) + {\left (x^{3} {\left (-\frac {c^{2}}{x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, c^{2} x \sqrt {-\frac {c^{2}}{x^{2}} + 1}\right )} c\right )} b \] Input:
integrate(x^3*(a+b*arcsin(c/x)),x, algorithm="maxima")
Output:
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
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (54) = 108\).
Time = 0.14 (sec) , antiderivative size = 340, normalized size of antiderivative = 5.31 \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\frac {3 \, b c x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4} \arcsin \left (\frac {c}{x}\right ) + 3 \, a c x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4} + 2 \, b c^{2} x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3} + 12 \, b c^{3} x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2} \arcsin \left (\frac {c}{x}\right ) + 12 \, a c^{3} x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2} + 18 \, b c^{4} x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )} + 18 \, b c^{5} \arcsin \left (\frac {c}{x}\right ) + 18 \, a c^{5} - \frac {18 \, b c^{6}}{x {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}} + \frac {12 \, b c^{7} \arcsin \left (\frac {c}{x}\right )}{x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2}} + \frac {12 \, a c^{7}}{x^{2} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{2}} - \frac {2 \, b c^{8}}{x^{3} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{3}} + \frac {3 \, b c^{9} \arcsin \left (\frac {c}{x}\right )}{x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4}} + \frac {3 \, a c^{9}}{x^{4} {\left (\sqrt {-\frac {c^{2}}{x^{2}} + 1} + 1\right )}^{4}}}{192 \, c} \] Input:
integrate(x^3*(a+b*arcsin(c/x)),x, algorithm="giac")
Output:
1/192*(3*b*c*x^4*(sqrt(-c^2/x^2 + 1) + 1)^4*arcsin(c/x) + 3*a*c*x^4*(sqrt( -c^2/x^2 + 1) + 1)^4 + 2*b*c^2*x^3*(sqrt(-c^2/x^2 + 1) + 1)^3 + 12*b*c^3*x ^2*(sqrt(-c^2/x^2 + 1) + 1)^2*arcsin(c/x) + 12*a*c^3*x^2*(sqrt(-c^2/x^2 + 1) + 1)^2 + 18*b*c^4*x*(sqrt(-c^2/x^2 + 1) + 1) + 18*b*c^5*arcsin(c/x) + 1 8*a*c^5 - 18*b*c^6/(x*(sqrt(-c^2/x^2 + 1) + 1)) + 12*b*c^7*arcsin(c/x)/(x^ 2*(sqrt(-c^2/x^2 + 1) + 1)^2) + 12*a*c^7/(x^2*(sqrt(-c^2/x^2 + 1) + 1)^2) - 2*b*c^8/(x^3*(sqrt(-c^2/x^2 + 1) + 1)^3) + 3*b*c^9*arcsin(c/x)/(x^4*(sqr t(-c^2/x^2 + 1) + 1)^4) + 3*a*c^9/(x^4*(sqrt(-c^2/x^2 + 1) + 1)^4))/c
Timed out. \[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\int x^3\,\left (a+b\,\mathrm {asin}\left (\frac {c}{x}\right )\right ) \,d x \] Input:
int(x^3*(a + b*asin(c/x)),x)
Output:
int(x^3*(a + b*asin(c/x)), x)
\[ \int x^3 \left (a+b \arcsin \left (\frac {c}{x}\right )\right ) \, dx=\left (\int \mathit {asin} \left (\frac {c}{x}\right ) x^{3}d x \right ) b +\frac {a \,x^{4}}{4} \] Input:
int(x^3*(a+b*asin(c/x)),x)
Output:
(4*int(asin(c/x)*x**3,x)*b + a*x**4)/4