Integrand size = 25, antiderivative size = 184 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {73 b d^2 x \sqrt {1-c^2 x^2}}{3072 c^3}+\frac {73 b d^2 x^3 \sqrt {1-c^2 x^2}}{4608 c}-\frac {43 b c d^2 x^5 \sqrt {1-c^2 x^2}}{1152}+\frac {1}{64} b c^3 d^2 x^7 \sqrt {1-c^2 x^2}-\frac {73 b d^2 \arcsin (c x)}{3072 c^4}+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x)) \]
-73/3072*b*d^2*arcsin(c*x)/c^4+1/4*d^2*x^4*(a+b*arcsin(c*x))-1/3*c^2*d^2*x ^6*(a+b*arcsin(c*x))+1/8*c^4*d^2*x^8*(a+b*arcsin(c*x))+73/3072*b*d^2*x*(-c ^2*x^2+1)^(1/2)/c^3+73/4608*b*d^2*x^3*(-c^2*x^2+1)^(1/2)/c-43/1152*b*c*d^2 *x^5*(-c^2*x^2+1)^(1/2)+1/64*b*c^3*d^2*x^7*(-c^2*x^2+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.62 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {d^2 \left (384 a c^4 x^4 \left (6-8 c^2 x^2+3 c^4 x^4\right )+b c x \sqrt {1-c^2 x^2} \left (219+146 c^2 x^2-344 c^4 x^4+144 c^6 x^6\right )+3 b \left (-73+768 c^4 x^4-1024 c^6 x^6+384 c^8 x^8\right ) \arcsin (c x)\right )}{9216 c^4} \]
(d^2*(384*a*c^4*x^4*(6 - 8*c^2*x^2 + 3*c^4*x^4) + b*c*x*Sqrt[1 - c^2*x^2]* (219 + 146*c^2*x^2 - 344*c^4*x^4 + 144*c^6*x^6) + 3*b*(-73 + 768*c^4*x^4 - 1024*c^6*x^6 + 384*c^8*x^8)*ArcSin[c*x]))/(9216*c^4)
Time = 0.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5192, 27, 1590, 25, 27, 363, 262, 262, 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.
\(\displaystyle \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5192 |
\(\displaystyle -b c \int \frac {d^2 x^4 \left (3 c^4 x^4-8 c^2 x^2+6\right )}{24 \sqrt {1-c^2 x^2}}dx+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{24} b c d^2 \int \frac {x^4 \left (3 c^4 x^4-8 c^2 x^2+6\right )}{\sqrt {1-c^2 x^2}}dx+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (-\frac {\int -\frac {c^2 x^4 \left (48-43 c^2 x^2\right )}{\sqrt {1-c^2 x^2}}dx}{8 c^2}-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (\frac {\int \frac {c^2 x^4 \left (48-43 c^2 x^2\right )}{\sqrt {1-c^2 x^2}}dx}{8 c^2}-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (\frac {1}{8} \int \frac {x^4 \left (48-43 c^2 x^2\right )}{\sqrt {1-c^2 x^2}}dx-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (\frac {1}{8} \left (\frac {73}{6} \int \frac {x^4}{\sqrt {1-c^2 x^2}}dx+\frac {43}{6} x^5 \sqrt {1-c^2 x^2}\right )-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (\frac {1}{8} \left (\frac {73}{6} \left (\frac {3 \int \frac {x^2}{\sqrt {1-c^2 x^2}}dx}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {43}{6} x^5 \sqrt {1-c^2 x^2}\right )-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{24} b c d^2 \left (\frac {1}{8} \left (\frac {73}{6} \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {43}{6} x^5 \sqrt {1-c^2 x^2}\right )-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )+\frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{8} c^4 d^2 x^8 (a+b \arcsin (c x))-\frac {1}{3} c^2 d^2 x^6 (a+b \arcsin (c x))+\frac {1}{4} d^2 x^4 (a+b \arcsin (c x))-\frac {1}{24} b c d^2 \left (\frac {1}{8} \left (\frac {73}{6} \left (\frac {3 \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {43}{6} x^5 \sqrt {1-c^2 x^2}\right )-\frac {3}{8} c^2 x^7 \sqrt {1-c^2 x^2}\right )\) |
(d^2*x^4*(a + b*ArcSin[c*x]))/4 - (c^2*d^2*x^6*(a + b*ArcSin[c*x]))/3 + (c ^4*d^2*x^8*(a + b*ArcSin[c*x]))/8 - (b*c*d^2*((-3*c^2*x^7*Sqrt[1 - c^2*x^2 ])/8 + ((43*x^5*Sqrt[1 - c^2*x^2])/6 + (73*(-1/4*(x^3*Sqrt[1 - c^2*x^2])/c ^2 + (3*(-1/2*(x*Sqrt[1 - c^2*x^2])/c^2 + ArcSin[c*x]/(2*c^3)))/(4*c^2)))/ 6)/8))/24
3.1.11.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^ (q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Simp[1/(e*(m + 4*p + 2*q + 1)) Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p)) - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp[ (a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c ^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0 ] && IGtQ[p, 0]
Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85
method | result | size |
parts | \(d^{2} a \left (\frac {1}{8} c^{4} x^{8}-\frac {1}{3} c^{2} x^{6}+\frac {1}{4} x^{4}\right )+\frac {d^{2} b \left (\frac {\arcsin \left (c x \right ) c^{8} x^{8}}{8}-\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{3}+\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{7} x^{7} \sqrt {-c^{2} x^{2}+1}}{64}-\frac {43 c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{1152}+\frac {73 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4608}+\frac {73 c x \sqrt {-c^{2} x^{2}+1}}{3072}-\frac {73 \arcsin \left (c x \right )}{3072}\right )}{c^{4}}\) | \(156\) |
derivativedivides | \(\frac {d^{2} a \left (\frac {1}{8} c^{8} x^{8}-\frac {1}{3} c^{6} x^{6}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (\frac {\arcsin \left (c x \right ) c^{8} x^{8}}{8}-\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{3}+\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{7} x^{7} \sqrt {-c^{2} x^{2}+1}}{64}-\frac {43 c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{1152}+\frac {73 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4608}+\frac {73 c x \sqrt {-c^{2} x^{2}+1}}{3072}-\frac {73 \arcsin \left (c x \right )}{3072}\right )}{c^{4}}\) | \(160\) |
default | \(\frac {d^{2} a \left (\frac {1}{8} c^{8} x^{8}-\frac {1}{3} c^{6} x^{6}+\frac {1}{4} c^{4} x^{4}\right )+d^{2} b \left (\frac {\arcsin \left (c x \right ) c^{8} x^{8}}{8}-\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{3}+\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{7} x^{7} \sqrt {-c^{2} x^{2}+1}}{64}-\frac {43 c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{1152}+\frac {73 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{4608}+\frac {73 c x \sqrt {-c^{2} x^{2}+1}}{3072}-\frac {73 \arcsin \left (c x \right )}{3072}\right )}{c^{4}}\) | \(160\) |
d^2*a*(1/8*c^4*x^8-1/3*c^2*x^6+1/4*x^4)+d^2*b/c^4*(1/8*arcsin(c*x)*c^8*x^8 -1/3*arcsin(c*x)*c^6*x^6+1/4*c^4*x^4*arcsin(c*x)+1/64*c^7*x^7*(-c^2*x^2+1) ^(1/2)-43/1152*c^5*x^5*(-c^2*x^2+1)^(1/2)+73/4608*c^3*x^3*(-c^2*x^2+1)^(1/ 2)+73/3072*c*x*(-c^2*x^2+1)^(1/2)-73/3072*arcsin(c*x))
Time = 0.26 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.81 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1152 \, a c^{8} d^{2} x^{8} - 3072 \, a c^{6} d^{2} x^{6} + 2304 \, a c^{4} d^{2} x^{4} + 3 \, {\left (384 \, b c^{8} d^{2} x^{8} - 1024 \, b c^{6} d^{2} x^{6} + 768 \, b c^{4} d^{2} x^{4} - 73 \, b d^{2}\right )} \arcsin \left (c x\right ) + {\left (144 \, b c^{7} d^{2} x^{7} - 344 \, b c^{5} d^{2} x^{5} + 146 \, b c^{3} d^{2} x^{3} + 219 \, b c d^{2} x\right )} \sqrt {-c^{2} x^{2} + 1}}{9216 \, c^{4}} \]
1/9216*(1152*a*c^8*d^2*x^8 - 3072*a*c^6*d^2*x^6 + 2304*a*c^4*d^2*x^4 + 3*( 384*b*c^8*d^2*x^8 - 1024*b*c^6*d^2*x^6 + 768*b*c^4*d^2*x^4 - 73*b*d^2)*arc sin(c*x) + (144*b*c^7*d^2*x^7 - 344*b*c^5*d^2*x^5 + 146*b*c^3*d^2*x^3 + 21 9*b*c*d^2*x)*sqrt(-c^2*x^2 + 1))/c^4
Time = 0.85 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.18 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\begin {cases} \frac {a c^{4} d^{2} x^{8}}{8} - \frac {a c^{2} d^{2} x^{6}}{3} + \frac {a d^{2} x^{4}}{4} + \frac {b c^{4} d^{2} x^{8} \operatorname {asin}{\left (c x \right )}}{8} + \frac {b c^{3} d^{2} x^{7} \sqrt {- c^{2} x^{2} + 1}}{64} - \frac {b c^{2} d^{2} x^{6} \operatorname {asin}{\left (c x \right )}}{3} - \frac {43 b c d^{2} x^{5} \sqrt {- c^{2} x^{2} + 1}}{1152} + \frac {b d^{2} x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {73 b d^{2} x^{3} \sqrt {- c^{2} x^{2} + 1}}{4608 c} + \frac {73 b d^{2} x \sqrt {- c^{2} x^{2} + 1}}{3072 c^{3}} - \frac {73 b d^{2} \operatorname {asin}{\left (c x \right )}}{3072 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a*c**4*d**2*x**8/8 - a*c**2*d**2*x**6/3 + a*d**2*x**4/4 + b*c** 4*d**2*x**8*asin(c*x)/8 + b*c**3*d**2*x**7*sqrt(-c**2*x**2 + 1)/64 - b*c** 2*d**2*x**6*asin(c*x)/3 - 43*b*c*d**2*x**5*sqrt(-c**2*x**2 + 1)/1152 + b*d **2*x**4*asin(c*x)/4 + 73*b*d**2*x**3*sqrt(-c**2*x**2 + 1)/(4608*c) + 73*b *d**2*x*sqrt(-c**2*x**2 + 1)/(3072*c**3) - 73*b*d**2*asin(c*x)/(3072*c**4) , Ne(c, 0)), (a*d**2*x**4/4, True))
Time = 0.28 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.62 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{8} \, a c^{4} d^{2} x^{8} - \frac {1}{3} \, a c^{2} d^{2} x^{6} + \frac {1}{3072} \, {\left (384 \, x^{8} \arcsin \left (c x\right ) + {\left (\frac {48 \, \sqrt {-c^{2} x^{2} + 1} x^{7}}{c^{2}} + \frac {56 \, \sqrt {-c^{2} x^{2} + 1} x^{5}}{c^{4}} + \frac {70 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{6}} + \frac {105 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{8}} - \frac {105 \, \arcsin \left (c x\right )}{c^{9}}\right )} c\right )} b c^{4} d^{2} + \frac {1}{4} \, a d^{2} x^{4} - \frac {1}{144} \, {\left (48 \, x^{6} \arcsin \left (c x\right ) + {\left (\frac {8 \, \sqrt {-c^{2} x^{2} + 1} x^{5}}{c^{2}} + \frac {10 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \arcsin \left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d^{2} + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b d^{2} \]
1/8*a*c^4*d^2*x^8 - 1/3*a*c^2*d^2*x^6 + 1/3072*(384*x^8*arcsin(c*x) + (48* sqrt(-c^2*x^2 + 1)*x^7/c^2 + 56*sqrt(-c^2*x^2 + 1)*x^5/c^4 + 70*sqrt(-c^2* x^2 + 1)*x^3/c^6 + 105*sqrt(-c^2*x^2 + 1)*x/c^8 - 105*arcsin(c*x)/c^9)*c)* b*c^4*d^2 + 1/4*a*d^2*x^4 - 1/144*(48*x^6*arcsin(c*x) + (8*sqrt(-c^2*x^2 + 1)*x^5/c^2 + 10*sqrt(-c^2*x^2 + 1)*x^3/c^4 + 15*sqrt(-c^2*x^2 + 1)*x/c^6 - 15*arcsin(c*x)/c^7)*c)*b*c^2*d^2 + 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* d^2
Time = 0.29 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.11 \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {1}{8} \, a c^{4} d^{2} x^{8} - \frac {1}{3} \, a c^{2} d^{2} x^{6} + \frac {1}{4} \, a d^{2} x^{4} + \frac {{\left (c^{2} x^{2} - 1\right )}^{3} \sqrt {-c^{2} x^{2} + 1} b d^{2} x}{64 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{4} b d^{2} \arcsin \left (c x\right )}{8 \, c^{4}} + \frac {11 \, {\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d^{2} x}{1152 \, c^{3}} + \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d^{2} \arcsin \left (c x\right )}{6 \, c^{4}} + \frac {55 \, {\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d^{2} x}{4608 \, c^{3}} + \frac {55 \, \sqrt {-c^{2} x^{2} + 1} b d^{2} x}{3072 \, c^{3}} + \frac {55 \, b d^{2} \arcsin \left (c x\right )}{3072 \, c^{4}} \]
1/8*a*c^4*d^2*x^8 - 1/3*a*c^2*d^2*x^6 + 1/4*a*d^2*x^4 + 1/64*(c^2*x^2 - 1) ^3*sqrt(-c^2*x^2 + 1)*b*d^2*x/c^3 + 1/8*(c^2*x^2 - 1)^4*b*d^2*arcsin(c*x)/ c^4 + 11/1152*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b*d^2*x/c^3 + 1/6*(c^2*x^ 2 - 1)^3*b*d^2*arcsin(c*x)/c^4 + 55/4608*(-c^2*x^2 + 1)^(3/2)*b*d^2*x/c^3 + 55/3072*sqrt(-c^2*x^2 + 1)*b*d^2*x/c^3 + 55/3072*b*d^2*arcsin(c*x)/c^4
Timed out. \[ \int x^3 \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2 \,d x \]