Integrand size = 25, antiderivative size = 217 \[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=-\frac {b c d^2 \left (38+13 m+m^2\right ) x^{2+m} \sqrt {1-c^2 x^2}}{(3+m)^2 (5+m)^2}+\frac {b c^3 d^2 x^{4+m} \sqrt {1-c^2 x^2}}{(5+m)^2}+\frac {d^2 x^{1+m} (a+b \arcsin (c x))}{1+m}-\frac {2 c^2 d^2 x^{3+m} (a+b \arcsin (c x))}{3+m}+\frac {c^4 d^2 x^{5+m} (a+b \arcsin (c x))}{5+m}-\frac {b c d^2 \left (149+100 m+15 m^2\right ) x^{2+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{(1+m) (2+m) (3+m)^2 (5+m)^2} \]
d^2*x^(1+m)*(a+b*arcsin(c*x))/(1+m)-2*c^2*d^2*x^(3+m)*(a+b*arcsin(c*x))/(3 +m)+c^4*d^2*x^(5+m)*(a+b*arcsin(c*x))/(5+m)-b*c*d^2*(15*m^2+100*m+149)*x^( 2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/(m^2+3*m+2)/(m^2+8*m+15)^ 2-b*c*d^2*(m^2+13*m+38)*x^(2+m)*(-c^2*x^2+1)^(1/2)/(3+m)^2/(5+m)^2+b*c^3*d ^2*x^(4+m)*(-c^2*x^2+1)^(1/2)/(5+m)^2
Time = 0.01 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.86 \[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\frac {x^{1+m} \left (\left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x))-\frac {b c d^2 x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )}{2+m}-\frac {4 d^2 \left ((2+m) \left (-3+c^2 x^2+m \left (-1+c^2 x^2\right )\right ) (a+b \arcsin (c x))+b c (1+m) x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )+2 b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )\right )}{(1+m) (2+m) (3+m)}\right )}{5+m} \]
(x^(1 + m)*((d - c^2*d*x^2)^2*(a + b*ArcSin[c*x]) - (b*c*d^2*x*Hypergeomet ric2F1[-3/2, 1 + m/2, 2 + m/2, c^2*x^2])/(2 + m) - (4*d^2*((2 + m)*(-3 + c ^2*x^2 + m*(-1 + c^2*x^2))*(a + b*ArcSin[c*x]) + b*c*(1 + m)*x*Hypergeomet ric2F1[-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))))/(5 + m)
Time = 0.50 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {5192, 27, 1590, 25, 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.
| \(\displaystyle \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx\) |
| \(\Big \downarrow \) 5192 |
| \(\displaystyle -b c \int \frac {d^2 x^{m+1} \left (\frac {c^4 x^4}{m+5}-\frac {2 c^2 x^2}{m+3}+\frac {1}{m+1}\right )}{\sqrt {1-c^2 x^2}}dx+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -b c d^2 \int \frac {x^{m+1} \left (\frac {c^4 x^4}{m+5}-\frac {2 c^2 x^2}{m+3}+\frac {1}{m+1}\right )}{\sqrt {1-c^2 x^2}}dx+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 1590 |
| \(\displaystyle -b c d^2 \left (-\frac {\int -\frac {c^2 x^{m+1} \left (\frac {m+5}{m+1}-\frac {c^2 \left (m^2+13 m+38\right ) x^2}{(m+3) (m+5)}\right )}{\sqrt {1-c^2 x^2}}dx}{c^2 (m+5)}-\frac {c^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2}\right )+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -b c d^2 \left (\frac {\int \frac {c^2 x^{m+1} \left (\frac {m+5}{m+1}-\frac {c^2 \left (m^2+13 m+38\right ) x^2}{(m+3) (m+5)}\right )}{\sqrt {1-c^2 x^2}}dx}{c^2 (m+5)}-\frac {c^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2}\right )+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -b c d^2 \left (\frac {\int \frac {x^{m+1} \left (\frac {m+5}{m+1}-\frac {c^2 \left (m^2+13 m+38\right ) x^2}{(m+3) (m+5)}\right )}{\sqrt {1-c^2 x^2}}dx}{m+5}-\frac {c^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2}\right )+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle -b c d^2 \left (\frac {\frac {\left (15 m^2+100 m+149\right ) \int \frac {x^{m+1}}{\sqrt {1-c^2 x^2}}dx}{(m+1) (m+3)^2 (m+5)}+\frac {\left (m^2+13 m+38\right ) \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)}}{m+5}-\frac {c^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2}\right )+\frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle \frac {c^4 d^2 x^{m+5} (a+b \arcsin (c x))}{m+5}-\frac {2 c^2 d^2 x^{m+3} (a+b \arcsin (c x))}{m+3}+\frac {d^2 x^{m+1} (a+b \arcsin (c x))}{m+1}-b c d^2 \left (\frac {\frac {\left (15 m^2+100 m+149\right ) x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{(m+1) (m+2) (m+3)^2 (m+5)}+\frac {\left (m^2+13 m+38\right ) \sqrt {1-c^2 x^2} x^{m+2}}{(m+3)^2 (m+5)}}{m+5}-\frac {c^2 \sqrt {1-c^2 x^2} x^{m+4}}{(m+5)^2}\right )\) |
(d^2*x^(1 + m)*(a + b*ArcSin[c*x]))/(1 + m) - (2*c^2*d^2*x^(3 + m)*(a + b* ArcSin[c*x]))/(3 + m) + (c^4*d^2*x^(5 + m)*(a + b*ArcSin[c*x]))/(5 + m) - b*c*d^2*(-((c^2*x^(4 + m)*Sqrt[1 - c^2*x^2])/(5 + m)^2) + (((38 + 13*m + m ^2)*x^(2 + m)*Sqrt[1 - c^2*x^2])/((3 + m)^2*(5 + m)) + ((149 + 100*m + 15* m^2)*x^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, c^2*x^2])/((1 + m)*(2 + m)*(3 + m)^2*(5 + m)))/(5 + m))
3.2.44.3.1 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[((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]
\[\int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{2} \left (a +b \arcsin \left (c x \right )\right )d x\]
\[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int { {\left (c^{2} d x^{2} - d\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
integral((a*c^4*d^2*x^4 - 2*a*c^2*d^2*x^2 + a*d^2 + (b*c^4*d^2*x^4 - 2*b*c ^2*d^2*x^2 + b*d^2)*arcsin(c*x))*x^m, x)
\[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=d^{2} \left (\int a x^{m}\, dx + \int b x^{m} \operatorname {asin}{\left (c x \right )}\, dx + \int \left (- 2 a c^{2} x^{2} x^{m}\right )\, dx + \int a c^{4} x^{4} x^{m}\, dx + \int \left (- 2 b c^{2} x^{2} x^{m} \operatorname {asin}{\left (c x \right )}\right )\, dx + \int b c^{4} x^{4} x^{m} \operatorname {asin}{\left (c x \right )}\, dx\right ) \]
d**2*(Integral(a*x**m, x) + Integral(b*x**m*asin(c*x), x) + Integral(-2*a* c**2*x**2*x**m, x) + Integral(a*c**4*x**4*x**m, x) + Integral(-2*b*c**2*x* *2*x**m*asin(c*x), x) + Integral(b*c**4*x**4*x**m*asin(c*x), x))
\[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int { {\left (c^{2} d x^{2} - d\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
a*c^4*d^2*x^(m + 5)/(m + 5) - 2*a*c^2*d^2*x^(m + 3)/(m + 3) + a*d^2*x^(m + 1)/(m + 1) + (((b*c^4*d^2*m^2 + 4*b*c^4*d^2*m + 3*b*c^4*d^2)*x^5 - 2*(b*c ^2*d^2*m^2 + 6*b*c^2*d^2*m + 5*b*c^2*d^2)*x^3 + (b*d^2*m^2 + 8*b*d^2*m + 1 5*b*d^2)*x)*x^m*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (m^3 + 9*m^2 + 23*m + 15)*integrate(-((b*c^5*d^2*m^2 + 4*b*c^5*d^2*m + 3*b*c^5*d^2)*x^5 - 2*(b*c^3*d^2*m^2 + 6*b*c^3*d^2*m + 5*b*c^3*d^2)*x^3 + (b*c*d^2*m^2 + 8* b*c*d^2*m + 15*b*c*d^2)*x)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^m/(m^3 - (c^2*m^ 3 + 9*c^2*m^2 + 23*c^2*m + 15*c^2)*x^2 + 9*m^2 + 23*m + 15), x))/(m^3 + 9* m^2 + 23*m + 15)
\[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int { {\left (c^{2} d x^{2} - d\right )}^{2} {\left (b \arcsin \left (c x\right ) + a\right )} x^{m} \,d x } \]
Timed out. \[ \int x^m \left (d-c^2 d x^2\right )^2 (a+b \arcsin (c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^2 \,d x \]