Integrand size = 12, antiderivative size = 82 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {b \sqrt {1-c^2 x^2}}{5 c^5}-\frac {2 b \left (1-c^2 x^2\right )^{3/2}}{15 c^5}+\frac {b \left (1-c^2 x^2\right )^{5/2}}{25 c^5}+\frac {1}{5} x^5 (a+b \arcsin (c x)) \] Output:
1/5*b*(-c^2*x^2+1)^(1/2)/c^5-2/15*b*(-c^2*x^2+1)^(3/2)/c^5+1/25*b*(-c^2*x^ 2+1)^(5/2)/c^5+1/5*x^5*(a+b*arcsin(c*x))
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.71 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {1}{75} \left (15 a x^5+\frac {b \sqrt {1-c^2 x^2} \left (8+4 c^2 x^2+3 c^4 x^4\right )}{c^5}+15 b x^5 \arcsin (c x)\right ) \] Input:
Integrate[x^4*(a + b*ArcSin[c*x]),x]
Output:
(15*a*x^5 + (b*Sqrt[1 - c^2*x^2]*(8 + 4*c^2*x^2 + 3*c^4*x^4))/c^5 + 15*b*x ^5*ArcSin[c*x])/75
Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5138, 243, 53, 2009}
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.
\(\displaystyle \int x^4 (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {1}{5} x^5 (a+b \arcsin (c x))-\frac {1}{5} b c \int \frac {x^5}{\sqrt {1-c^2 x^2}}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{5} x^5 (a+b \arcsin (c x))-\frac {1}{10} b c \int \frac {x^4}{\sqrt {1-c^2 x^2}}dx^2\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{5} x^5 (a+b \arcsin (c x))-\frac {1}{10} b c \int \left (\frac {\left (1-c^2 x^2\right )^{3/2}}{c^4}-\frac {2 \sqrt {1-c^2 x^2}}{c^4}+\frac {1}{c^4 \sqrt {1-c^2 x^2}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 (a+b \arcsin (c x))-\frac {1}{10} b c \left (-\frac {2 \left (1-c^2 x^2\right )^{5/2}}{5 c^6}+\frac {4 \left (1-c^2 x^2\right )^{3/2}}{3 c^6}-\frac {2 \sqrt {1-c^2 x^2}}{c^6}\right )\) |
Input:
Int[x^4*(a + b*ArcSin[c*x]),x]
Output:
-1/10*(b*c*((-2*Sqrt[1 - c^2*x^2])/c^6 + (4*(1 - c^2*x^2)^(3/2))/(3*c^6) - (2*(1 - c^2*x^2)^(5/2))/(5*c^6))) + (x^5*(a + b*ArcSin[c*x]))/5
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98
method | result | size |
parts | \(\frac {a \,x^{5}}{5}+\frac {b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{75}+\frac {8 \sqrt {-c^{2} x^{2}+1}}{75}\right )}{c^{5}}\) | \(80\) |
derivativedivides | \(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{75}+\frac {8 \sqrt {-c^{2} x^{2}+1}}{75}\right )}{c^{5}}\) | \(84\) |
default | \(\frac {\frac {a \,c^{5} x^{5}}{5}+b \left (\frac {c^{5} x^{5} \arcsin \left (c x \right )}{5}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {4 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{75}+\frac {8 \sqrt {-c^{2} x^{2}+1}}{75}\right )}{c^{5}}\) | \(84\) |
orering | \(\frac {\left (27 c^{6} x^{6}+4 c^{4} x^{4}+16 c^{2} x^{2}-32\right ) \left (a +b \arcsin \left (c x \right )\right )}{75 x \,c^{6}}-\frac {\left (3 c^{4} x^{4}+4 c^{2} x^{2}+8\right ) \left (c x -1\right ) \left (c x +1\right ) \left (4 x^{3} \left (a +b \arcsin \left (c x \right )\right )+\frac {x^{4} b c}{\sqrt {-c^{2} x^{2}+1}}\right )}{75 c^{6} x^{4}}\) | \(112\) |
Input:
int(x^4*(a+b*arcsin(c*x)),x,method=_RETURNVERBOSE)
Output:
1/5*a*x^5+b/c^5*(1/5*c^5*x^5*arcsin(c*x)+1/25*c^4*x^4*(-c^2*x^2+1)^(1/2)+4 /75*c^2*x^2*(-c^2*x^2+1)^(1/2)+8/75*(-c^2*x^2+1)^(1/2))
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {15 \, b c^{5} x^{5} \arcsin \left (c x\right ) + 15 \, a c^{5} x^{5} + {\left (3 \, b c^{4} x^{4} + 4 \, b c^{2} x^{2} + 8 \, b\right )} \sqrt {-c^{2} x^{2} + 1}}{75 \, c^{5}} \] Input:
integrate(x^4*(a+b*arcsin(c*x)),x, algorithm="fricas")
Output:
1/75*(15*b*c^5*x^5*arcsin(c*x) + 15*a*c^5*x^5 + (3*b*c^4*x^4 + 4*b*c^2*x^2 + 8*b)*sqrt(-c^2*x^2 + 1))/c^5
Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a x^{5}}{5} + \frac {b x^{5} \operatorname {asin}{\left (c x \right )}}{5} + \frac {b x^{4} \sqrt {- c^{2} x^{2} + 1}}{25 c} + \frac {4 b x^{2} \sqrt {- c^{2} x^{2} + 1}}{75 c^{3}} + \frac {8 b \sqrt {- c^{2} x^{2} + 1}}{75 c^{5}} & \text {for}\: c \neq 0 \\\frac {a x^{5}}{5} & \text {otherwise} \end {cases} \] Input:
integrate(x**4*(a+b*asin(c*x)),x)
Output:
Piecewise((a*x**5/5 + b*x**5*asin(c*x)/5 + b*x**4*sqrt(-c**2*x**2 + 1)/(25 *c) + 4*b*x**2*sqrt(-c**2*x**2 + 1)/(75*c**3) + 8*b*sqrt(-c**2*x**2 + 1)/( 75*c**5), Ne(c, 0)), (a*x**5/5, True))
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b \] Input:
integrate(x^4*(a+b*arcsin(c*x)),x, algorithm="maxima")
Output:
1/5*a*x^5 + 1/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/c^2 + 4*s qrt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(-c^2*x^2 + 1)/c^6)*c)*b
Time = 0.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.52 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {1}{5} \, a x^{5} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {2 \, {\left (c^{2} x^{2} - 1\right )} b x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {b x \arcsin \left (c x\right )}{5 \, c^{4}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b}{25 \, c^{5}} - \frac {2 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b}{15 \, c^{5}} + \frac {\sqrt {-c^{2} x^{2} + 1} b}{5 \, c^{5}} \] Input:
integrate(x^4*(a+b*arcsin(c*x)),x, algorithm="giac")
Output:
1/5*a*x^5 + 1/5*(c^2*x^2 - 1)^2*b*x*arcsin(c*x)/c^4 + 2/5*(c^2*x^2 - 1)*b* x*arcsin(c*x)/c^4 + 1/5*b*x*arcsin(c*x)/c^4 + 1/25*(c^2*x^2 - 1)^2*sqrt(-c ^2*x^2 + 1)*b/c^5 - 2/15*(-c^2*x^2 + 1)^(3/2)*b/c^5 + 1/5*sqrt(-c^2*x^2 + 1)*b/c^5
Timed out. \[ \int x^4 (a+b \arcsin (c x)) \, dx=\int x^4\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right ) \,d x \] Input:
int(x^4*(a + b*asin(c*x)),x)
Output:
int(x^4*(a + b*asin(c*x)), x)
Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int x^4 (a+b \arcsin (c x)) \, dx=\frac {15 \mathit {asin} \left (c x \right ) b \,c^{5} x^{5}+3 \sqrt {-c^{2} x^{2}+1}\, b \,c^{4} x^{4}+4 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} x^{2}+8 \sqrt {-c^{2} x^{2}+1}\, b +15 a \,c^{5} x^{5}}{75 c^{5}} \] Input:
int(x^4*(a+b*asin(c*x)),x)
Output:
(15*asin(c*x)*b*c**5*x**5 + 3*sqrt( - c**2*x**2 + 1)*b*c**4*x**4 + 4*sqrt( - c**2*x**2 + 1)*b*c**2*x**2 + 8*sqrt( - c**2*x**2 + 1)*b + 15*a*c**5*x** 5)/(75*c**5)