Integrand size = 23, antiderivative size = 123 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {b d x \sqrt {1-c^2 x^2}}{24 c^3}+\frac {b d x^3 \sqrt {1-c^2 x^2}}{36 c}-\frac {1}{36} b c d x^5 \sqrt {1-c^2 x^2}-\frac {b d \arcsin (c x)}{24 c^4}+\frac {1}{4} d x^4 (a+b \arcsin (c x))-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x)) \]
-1/24*b*d*arcsin(c*x)/c^4+1/4*d*x^4*(a+b*arcsin(c*x))-1/6*c^2*d*x^6*(a+b*a rcsin(c*x))+1/24*b*d*x*(-c^2*x^2+1)^(1/2)/c^3+1/36*b*d*x^3*(-c^2*x^2+1)^(1 /2)/c-1/36*b*c*d*x^5*(-c^2*x^2+1)^(1/2)
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.72 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\frac {d \left (-6 a c^4 x^4 \left (-3+2 c^2 x^2\right )+b c x \sqrt {1-c^2 x^2} \left (3+2 c^2 x^2-2 c^4 x^4\right )-3 b \left (1-6 c^4 x^4+4 c^6 x^6\right ) \arcsin (c x)\right )}{72 c^4} \]
(d*(-6*a*c^4*x^4*(-3 + 2*c^2*x^2) + b*c*x*Sqrt[1 - c^2*x^2]*(3 + 2*c^2*x^2 - 2*c^4*x^4) - 3*b*(1 - 6*c^4*x^4 + 4*c^6*x^6)*ArcSin[c*x]))/(72*c^4)
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5192, 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.
\(\displaystyle \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx\) |
\(\Big \downarrow \) 5192 |
\(\displaystyle -b c \int \frac {d x^4 \left (3-2 c^2 x^2\right )}{12 \sqrt {1-c^2 x^2}}dx-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{12} b c d \int \frac {x^4 \left (3-2 c^2 x^2\right )}{\sqrt {1-c^2 x^2}}dx-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \int \frac {x^4}{\sqrt {1-c^2 x^2}}dx+\frac {1}{3} x^5 \sqrt {1-c^2 x^2}\right )-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {3 \int \frac {x^2}{\sqrt {1-c^2 x^2}}dx}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {1-c^2 x^2}\right )-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 262 |
\(\displaystyle -\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^2}}dx}{2 c^2}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {1-c^2 x^2}\right )-\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))\) |
\(\Big \downarrow \) 223 |
\(\displaystyle -\frac {1}{6} c^2 d x^6 (a+b \arcsin (c x))+\frac {1}{4} d x^4 (a+b \arcsin (c x))-\frac {1}{12} b c d \left (\frac {4}{3} \left (\frac {3 \left (\frac {\arcsin (c x)}{2 c^3}-\frac {x \sqrt {1-c^2 x^2}}{2 c^2}\right )}{4 c^2}-\frac {x^3 \sqrt {1-c^2 x^2}}{4 c^2}\right )+\frac {1}{3} x^5 \sqrt {1-c^2 x^2}\right )\) |
(d*x^4*(a + b*ArcSin[c*x]))/4 - (c^2*d*x^6*(a + b*ArcSin[c*x]))/6 - (b*c*d *((x^5*Sqrt[1 - c^2*x^2])/3 + (4*(-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)))/3))/12
3.1.2.3.1 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[((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 = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.93
method | result | size |
parts | \(-d a \left (\frac {1}{6} c^{2} x^{6}-\frac {1}{4} x^{4}\right )-\frac {d b \left (\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{6}-\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{24}+\frac {\arcsin \left (c x \right )}{24}\right )}{c^{4}}\) | \(114\) |
derivativedivides | \(\frac {-d a \left (\frac {1}{6} c^{6} x^{6}-\frac {1}{4} c^{4} x^{4}\right )-d b \left (\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{6}-\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{24}+\frac {\arcsin \left (c x \right )}{24}\right )}{c^{4}}\) | \(118\) |
default | \(\frac {-d a \left (\frac {1}{6} c^{6} x^{6}-\frac {1}{4} c^{4} x^{4}\right )-d b \left (\frac {\arcsin \left (c x \right ) c^{6} x^{6}}{6}-\frac {c^{4} x^{4} \arcsin \left (c x \right )}{4}+\frac {c^{5} x^{5} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}{36}-\frac {c x \sqrt {-c^{2} x^{2}+1}}{24}+\frac {\arcsin \left (c x \right )}{24}\right )}{c^{4}}\) | \(118\) |
-d*a*(1/6*c^2*x^6-1/4*x^4)-d*b/c^4*(1/6*arcsin(c*x)*c^6*x^6-1/4*c^4*x^4*ar csin(c*x)+1/36*c^5*x^5*(-c^2*x^2+1)^(1/2)-1/36*c^3*x^3*(-c^2*x^2+1)^(1/2)- 1/24*c*x*(-c^2*x^2+1)^(1/2)+1/24*arcsin(c*x))
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.78 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {12 \, a c^{6} d x^{6} - 18 \, a c^{4} d x^{4} + 3 \, {\left (4 \, b c^{6} d x^{6} - 6 \, b c^{4} d x^{4} + b d\right )} \arcsin \left (c x\right ) + {\left (2 \, b c^{5} d x^{5} - 2 \, b c^{3} d x^{3} - 3 \, b c d x\right )} \sqrt {-c^{2} x^{2} + 1}}{72 \, c^{4}} \]
-1/72*(12*a*c^6*d*x^6 - 18*a*c^4*d*x^4 + 3*(4*b*c^6*d*x^6 - 6*b*c^4*d*x^4 + b*d)*arcsin(c*x) + (2*b*c^5*d*x^5 - 2*b*c^3*d*x^3 - 3*b*c*d*x)*sqrt(-c^2 *x^2 + 1))/c^4
Time = 0.49 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\begin {cases} - \frac {a c^{2} d x^{6}}{6} + \frac {a d x^{4}}{4} - \frac {b c^{2} d x^{6} \operatorname {asin}{\left (c x \right )}}{6} - \frac {b c d x^{5} \sqrt {- c^{2} x^{2} + 1}}{36} + \frac {b d x^{4} \operatorname {asin}{\left (c x \right )}}{4} + \frac {b d x^{3} \sqrt {- c^{2} x^{2} + 1}}{36 c} + \frac {b d x \sqrt {- c^{2} x^{2} + 1}}{24 c^{3}} - \frac {b d \operatorname {asin}{\left (c x \right )}}{24 c^{4}} & \text {for}\: c \neq 0 \\\frac {a d x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((-a*c**2*d*x**6/6 + a*d*x**4/4 - b*c**2*d*x**6*asin(c*x)/6 - b*c *d*x**5*sqrt(-c**2*x**2 + 1)/36 + b*d*x**4*asin(c*x)/4 + b*d*x**3*sqrt(-c* *2*x**2 + 1)/(36*c) + b*d*x*sqrt(-c**2*x**2 + 1)/(24*c**3) - b*d*asin(c*x) /(24*c**4), Ne(c, 0)), (a*d*x**4/4, True))
Time = 0.32 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.37 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} - \frac {1}{288} \, {\left (48 \, x^{6} \arcsin \left (c x\right ) + {\left (\frac {8 \, \sqrt {-c^{2} x^{2} + 1} x^{5}}{c^{2}} + \frac {10 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{6}} - \frac {15 \, \arcsin \left (c x\right )}{c^{7}}\right )} c\right )} b c^{2} d + \frac {1}{32} \, {\left (8 \, x^{4} \arcsin \left (c x\right ) + {\left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {-c^{2} x^{2} + 1} x}{c^{4}} - \frac {3 \, \arcsin \left (c x\right )}{c^{5}}\right )} c\right )} b d \]
-1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 - 1/288*(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 + 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
Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.17 \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {1}{6} \, a c^{2} d x^{6} + \frac {1}{4} \, a d x^{4} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} \sqrt {-c^{2} x^{2} + 1} b d x}{36 \, c^{3}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{3} b d \arcsin \left (c x\right )}{6 \, c^{4}} + \frac {{\left (-c^{2} x^{2} + 1\right )}^{\frac {3}{2}} b d x}{36 \, c^{3}} - \frac {{\left (c^{2} x^{2} - 1\right )}^{2} b d \arcsin \left (c x\right )}{4 \, c^{4}} + \frac {\sqrt {-c^{2} x^{2} + 1} b d x}{24 \, c^{3}} + \frac {b d \arcsin \left (c x\right )}{24 \, c^{4}} \]
-1/6*a*c^2*d*x^6 + 1/4*a*d*x^4 - 1/36*(c^2*x^2 - 1)^2*sqrt(-c^2*x^2 + 1)*b *d*x/c^3 - 1/6*(c^2*x^2 - 1)^3*b*d*arcsin(c*x)/c^4 + 1/36*(-c^2*x^2 + 1)^( 3/2)*b*d*x/c^3 - 1/4*(c^2*x^2 - 1)^2*b*d*arcsin(c*x)/c^4 + 1/24*sqrt(-c^2* x^2 + 1)*b*d*x/c^3 + 1/24*b*d*arcsin(c*x)/c^4
Timed out. \[ \int x^3 \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right ) \,d x \]