Integrand size = 21, antiderivative size = 90 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {3 b d x \sqrt {1-c^2 x^2}}{32 c}+\frac {b d x \left (1-c^2 x^2\right )^{3/2}}{16 c}+\frac {3 b d \arcsin (c x)}{32 c^2}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{4 c^2} \]
1/16*b*d*x*(-c^2*x^2+1)^(3/2)/c+3/32*b*d*arcsin(c*x)/c^2-1/4*d*(-c^2*x^2+1 )^2*(a+b*arcsin(c*x))/c^2+3/32*b*d*x*(-c^2*x^2+1)^(1/2)/c
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {d \left (c x \left (8 a c x \left (-2+c^2 x^2\right )+b \sqrt {1-c^2 x^2} \left (-5+2 c^2 x^2\right )\right )+b \left (5-16 c^2 x^2+8 c^4 x^4\right ) \arcsin (c x)\right )}{32 c^2} \]
-1/32*(d*(c*x*(8*a*c*x*(-2 + c^2*x^2) + b*Sqrt[1 - c^2*x^2]*(-5 + 2*c^2*x^ 2)) + b*(5 - 16*c^2*x^2 + 8*c^4*x^4)*ArcSin[c*x]))/c^2
Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5182, 211, 211, 223}
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.
\(\displaystyle \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5182 |
\(\displaystyle \frac {b d \int \left (1-c^2 x^2\right )^{3/2}dx}{4 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{4 c^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {b d \left (\frac {3}{4} \int \sqrt {1-c^2 x^2}dx+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2}\right )}{4 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{4 c^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {b d \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-c^2 x^2}}dx+\frac {1}{2} x \sqrt {1-c^2 x^2}\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2}\right )}{4 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{4 c^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {b d \left (\frac {3}{4} \left (\frac {\arcsin (c x)}{2 c}+\frac {1}{2} x \sqrt {1-c^2 x^2}\right )+\frac {1}{4} x \left (1-c^2 x^2\right )^{3/2}\right )}{4 c}-\frac {d \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{4 c^2}\) |
-1/4*(d*(1 - c^2*x^2)^2*(a + b*ArcSin[c*x]))/c^2 + (b*d*((x*(1 - c^2*x^2)^ (3/2))/4 + (3*((x*Sqrt[1 - c^2*x^2])/2 + ArcSin[c*x]/(2*c)))/4))/(4*c)
3.1.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_ .), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] I nt[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]
Time = 0.07 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {-\frac {d a \left (c^{2} x^{2}-1\right )^{2}}{4}-d b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(92\) |
default | \(\frac {-\frac {d a \left (c^{2} x^{2}-1\right )^{2}}{4}-d b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(92\) |
parts | \(-\frac {d a \left (c^{2} x^{2}-1\right )^{2}}{4 c^{2}}-\frac {d b \left (\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}-\frac {c^{2} x^{2} \arcsin \left (c x \right )}{2}+\frac {5 \arcsin \left (c x \right )}{32}+\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{16}-\frac {5 c x \sqrt {-c^{2} x^{2}+1}}{32}\right )}{c^{2}}\) | \(94\) |
1/c^2*(-1/4*d*a*(c^2*x^2-1)^2-d*b*(1/4*c^4*x^4*arcsin(c*x)-1/2*c^2*x^2*arc sin(c*x)+5/32*arcsin(c*x)+1/16*c^3*x^3*(-c^2*x^2+1)^(1/2)-5/32*c*x*(-c^2*x ^2+1)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {8 \, a c^{4} d x^{4} - 16 \, a c^{2} d x^{2} + {\left (8 \, b c^{4} d x^{4} - 16 \, b c^{2} d x^{2} + 5 \, b d\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{3} d x^{3} - 5 \, b c d x\right )} \sqrt {-c^{2} x^{2} + 1}}{32 \, c^{2}} \]
-1/32*(8*a*c^4*d*x^4 - 16*a*c^2*d*x^2 + (8*b*c^4*d*x^4 - 16*b*c^2*d*x^2 + 5*b*d)*arcsin(c*x) + (2*b*c^3*d*x^3 - 5*b*c*d*x)*sqrt(-c^2*x^2 + 1))/c^2
Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.30 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\begin {cases} - \frac {a c^{2} d x^{4}}{4} + \frac {a d x^{2}}{2} - \frac {b c^{2} d x^{4} \operatorname {asin}{\left (c x \right )}}{4} - \frac {b c d x^{3} \sqrt {- c^{2} x^{2} + 1}}{16} + \frac {b d x^{2} \operatorname {asin}{\left (c x \right )}}{2} + \frac {5 b d x \sqrt {- c^{2} x^{2} + 1}}{32 c} - \frac {5 b d \operatorname {asin}{\left (c x \right )}}{32 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d x^{2}}{2} & \text {otherwise} \end {cases} \]
Piecewise((-a*c**2*d*x**4/4 + a*d*x**2/2 - b*c**2*d*x**4*asin(c*x)/4 - b*c *d*x**3*sqrt(-c**2*x**2 + 1)/16 + b*d*x**2*asin(c*x)/2 + 5*b*d*x*sqrt(-c** 2*x**2 + 1)/(32*c) - 5*b*d*asin(c*x)/(32*c**2), Ne(c, 0)), (a*d*x**2/2, Tr ue))
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.42 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {1}{4} \, a c^{2} d x^{4} - \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b c^{2} d + \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} b d \]
-1/4*a*c^2*d*x^4 - 1/32*(8*x^4*arcsin(c*x) + (2*sqrt(-c^2*x^2 + 1)*x^3/c^2 + 3*sqrt(-c^2*x^2 + 1)*x/c^4 - 3*arcsin(c*x)/c^5)*c)*b*c^2*d + 1/2*a*d*x^ 2 + 1/4*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3 ))*b*d
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {1}{4} \, a c^{2} d x^{4} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d x}{16 \, c} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b d \arcsin \left (c x\right )}{4 \, c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} b d x}{32 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} a d}{2 \, c^{2}} + \frac {3 \, b d \arcsin \left (c x\right )}{32 \, c^{2}} \]
-1/4*a*c^2*d*x^4 + 1/16*(-c^2*x^2 + 1)^(3/2)*b*d*x/c - 1/4*(c^2*x^2 - 1)^2 *b*d*arcsin(c*x)/c^2 + 3/32*sqrt(-c^2*x^2 + 1)*b*d*x/c + 1/2*(c^2*x^2 - 1) *a*d/c^2 + 3/32*b*d*arcsin(c*x)/c^2
Timed out. \[ \int x \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int x\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right ) \,d x \]