Integrand size = 23, antiderivative size = 129 \[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {b c d x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2}+\frac {d x^{1+m} (a+b \arcsin (c x))}{1+m}-\frac {c^2 d x^{3+m} (a+b \arcsin (c x))}{3+m}-\frac {b c d (7+3 m) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{(1+m) (2+m) (3+m)^2} \]
d*x^(1+m)*(a+b*arcsin(c*x))/(1+m)-c^2*d*x^(3+m)*(a+b*arcsin(c*x))/(3+m)-b* c*d*(7+3*m)*x^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/(3+m)^2/(m ^2+3*m+2)-b*c*d*x^(2+m)*(-c^2*x^2+1)^(1/2)/(3+m)^2
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91 \[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=-\frac {d x^{1+m} \left ((2+m) \left (-3+c^2 x^2+m \left (-1+c^2 x^2\right )\right ) (a+b \arcsin (c x))+b c (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )+2 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )\right )}{(1+m) (2+m) (3+m)} \]
-((d*x^(1 + m)*((2 + m)*(-3 + c^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcSin[c* x]) + b*c*(1 + m)*x*Hypergeometric2F1[-1/2, 1 + m/2, 2 + m/2, c^2*x^2] + 2 *b*c*x*Hypergeometric2F1[1/2, 1 + m/2, 2 + m/2, c^2*x^2]))/((1 + m)*(2 + m )*(3 + m)))
Time = 0.32 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {5192, 27, 363, 278}
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.
| \(\displaystyle \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5192 |
| \(\displaystyle -b c \int \frac {d x^{m+1} \left (\frac {1}{m+1}-\frac {c^2 x^2}{m+3}\right )}{\sqrt {1-c^2 x^2}}dx-\frac {c^2 d x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -b c d \int \frac {x^{m+1} \left (\frac {1}{m+1}-\frac {c^2 x^2}{m+3}\right )}{\sqrt {1-c^2 x^2}}dx-\frac {c^2 d x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle -b c d \left (\frac {(3 m+7) \int \frac {x^{m+1}}{\sqrt {1-c^2 x^2}}dx}{(m+1) (m+3)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2}\right )-\frac {c^2 d x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle -\frac {c^2 d x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d x^{m+1} (a+b \arcsin (c x))}{m+1}-b c d \left (\frac {(3 m+7) x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{(m+1) (m+2) (m+3)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2}\right )\) |
(d*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (c^2*d*x^(3 + m)*(a + b*ArcSin [c*x]))/(3 + m) - b*c*d*((x^(2 + m)*Sqrt[1 - c^2*x^2])/(3 + m)^2 + ((7 + 3 *m)*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + m)*(2 + m)*(3 + m)^2))
3.2.45.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
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]
\[\int x^{m} \left (-c^{2} d \,x^{2}+d \right ) \left (a +b \arcsin \left (c x \right )\right )d x\]
\[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int { -{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
\[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=- d \left (\int \left (- a x^{m}\right )\, dx + \int \left (- b x^{m} \operatorname {asin}{\left (c x \right )}\right )\, dx + \int a c^{2} x^{2} x^{m}\, dx + \int b c^{2} x^{2} x^{m} \operatorname {asin}{\left (c x \right )}\, dx\right ) \]
-d*(Integral(-a*x**m, x) + Integral(-b*x**m*asin(c*x), x) + Integral(a*c** 2*x**2*x**m, x) + Integral(b*c**2*x**2*x**m*asin(c*x), x))
\[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int { -{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
-a*c^2*d*x^(m + 3)/(m + 3) + a*d*x^(m + 1)/(m + 1) - (((b*c^2*d*m + b*c^2* d)*x^3 - (b*d*m + 3*b*d)*x)*x^m*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (m^2 + 4*m + 3)*integrate(((b*c^3*d*m + b*c^3*d)*x^3 - (b*c*d*m + 3*b*c *d)*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^m/((c^2*m^2 + 4*c^2*m + 3*c^2)*x^2 - m^2 - 4*m - 3), x))/(m^2 + 4*m + 3)
\[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int { -{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
Timed out. \[ \int x^m \left (d-c^2 d x^2\right ) (a+b \arcsin (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right ) \,d x \]