Integrand size = 29, antiderivative size = 29 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\frac {2 b^2 c^2 d x^{3+m} \sqrt {d-c^2 d x^2}}{(4+m)^3}-\frac {6 b c d x^{2+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{(2+m)^2 (4+m) \sqrt {1-c^2 x^2}}-\frac {2 b c d x^{2+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{\left (8+6 m+m^2\right ) \sqrt {1-c^2 x^2}}+\frac {2 b c^3 d x^{4+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))}{(4+m)^2 \sqrt {1-c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{8+6 m+m^2}+\frac {x^{1+m} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{4+m}+\frac {6 b^2 c^2 d x^{3+m} \sqrt {d-c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(2+m)^2 (3+m) (4+m) \sqrt {1-c^2 x^2}}+\frac {2 b^2 c^2 d (10+3 m) x^{3+m} \sqrt {d-c^2 d x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{(2+m) (3+m) (4+m)^3 \sqrt {1-c^2 x^2}}+\frac {3 d^2 \text {Int}\left (\frac {x^m (a+b \arcsin (c x))^2}{\sqrt {d-c^2 d x^2}},x\right )}{8+6 m+m^2} \]
x^(1+m)*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^2/(4+m)+2*b^2*c^2*d*x^(3+m) *(-c^2*d*x^2+d)^(1/2)/(4+m)^3+3*d*x^(1+m)*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+ d)^(1/2)/(m^2+6*m+8)-6*b*c*d*x^(2+m)*(a+b*arcsin(c*x))*(-c^2*d*x^2+d)^(1/2 )/(2+m)^2/(4+m)/(-c^2*x^2+1)^(1/2)-2*b*c*d*x^(2+m)*(a+b*arcsin(c*x))*(-c^2 *d*x^2+d)^(1/2)/(m^2+6*m+8)/(-c^2*x^2+1)^(1/2)+2*b*c^3*d*x^(4+m)*(a+b*arcs in(c*x))*(-c^2*d*x^2+d)^(1/2)/(4+m)^2/(-c^2*x^2+1)^(1/2)+2*b^2*c^2*d*(10+3 *m)*x^(3+m)*hypergeom([1/2, 3/2+1/2*m],[5/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d) ^(1/2)/(4+m)^3/(m^2+5*m+6)/(-c^2*x^2+1)^(1/2)+6*b^2*c^2*d*x^(3+m)*hypergeo m([1/2, 3/2+1/2*m],[5/2+1/2*m],c^2*x^2)*(-c^2*d*x^2+d)^(1/2)/(2+m)^2/(m^2+ 7*m+12)/(-c^2*x^2+1)^(1/2)+3*d^2*Unintegrable(x^m*(a+b*arcsin(c*x))^2/(-c^ 2*d*x^2+d)^(1/2),x)/(m^2+6*m+8)
Not integrable
Time = 0.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx \]
Not integrable
Time = 1.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {5202, 5192, 363, 278, 5202, 5138, 278, 5234}
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.
\(\displaystyle \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx\) |
\(\Big \downarrow \) 5202 |
\(\displaystyle -\frac {2 b c d \sqrt {d-c^2 d x^2} \int x^{m+1} \left (1-c^2 x^2\right ) (a+b \arcsin (c x))dx}{(m+4) \sqrt {1-c^2 x^2}}+\frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2dx}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 5192 |
\(\displaystyle -\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-b c \int \frac {x^{m+2} \left (\frac {1}{m+2}-\frac {c^2 x^2}{m+4}\right )}{\sqrt {1-c^2 x^2}}dx-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2dx}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle -\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-b c \left (\frac {(3 m+10) \int \frac {x^{m+2}}{\sqrt {1-c^2 x^2}}dx}{(m+2) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2dx}{m+4}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {3 d \int x^m \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2dx}{m+4}-\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 5202 |
\(\displaystyle \frac {3 d \left (-\frac {2 b c \sqrt {d-c^2 d x^2} \int x^{m+1} (a+b \arcsin (c x))dx}{(m+2) \sqrt {1-c^2 x^2}}+\frac {d \int \frac {x^m (a+b \arcsin (c x))^2}{\sqrt {d-c^2 d x^2}}dx}{m+2}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 5138 |
\(\displaystyle \frac {3 d \left (-\frac {2 b c \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-\frac {b c \int \frac {x^{m+2}}{\sqrt {1-c^2 x^2}}dx}{m+2}\right )}{(m+2) \sqrt {1-c^2 x^2}}+\frac {d \int \frac {x^m (a+b \arcsin (c x))^2}{\sqrt {d-c^2 d x^2}}dx}{m+2}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle \frac {3 d \left (\frac {d \int \frac {x^m (a+b \arcsin (c x))^2}{\sqrt {d-c^2 d x^2}}dx}{m+2}-\frac {2 b c \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3)}\right )}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
\(\Big \downarrow \) 5234 |
\(\displaystyle \frac {3 d \left (\frac {d \int \frac {x^m (a+b \arcsin (c x))^2}{\sqrt {d-c^2 d x^2}}dx}{m+2}-\frac {2 b c \sqrt {d-c^2 d x^2} \left (\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3)}\right )}{(m+2) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^2}{m+2}\right )}{m+4}-\frac {2 b c d \sqrt {d-c^2 d x^2} \left (-\frac {c^2 x^{m+4} (a+b \arcsin (c x))}{m+4}+\frac {x^{m+2} (a+b \arcsin (c x))}{m+2}-b c \left (\frac {(3 m+10) x^{m+3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{(m+2) (m+3) (m+4)^2}+\frac {\sqrt {1-c^2 x^2} x^{m+3}}{(m+4)^2}\right )\right )}{(m+4) \sqrt {1-c^2 x^2}}+\frac {x^{m+1} \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2}{m+4}\) |
3.3.83.3.1 Defintions of rubi rules used
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_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
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[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*ArcS in[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1)) Int[(f*x) ^m*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Simp[b*c*(n/(f*(m + 2*p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m + 1)*(1 - c^2 *x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && !LtQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_)^2)^(p_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^2)^p*(a + b*Ar cSin[c*x])^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
Not integrable
Time = 1.99 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
\[\int x^{m} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}d x\]
Not integrable
Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.93 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]
integral(-(a^2*c^2*d*x^2 - a^2*d + (b^2*c^2*d*x^2 - b^2*d)*arcsin(c*x)^2 + 2*(a*b*c^2*d*x^2 - a*b*d)*arcsin(c*x))*sqrt(-c^2*d*x^2 + d)*x^m, x)
Timed out. \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Timed out} \]
Not integrable
Time = 0.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m} \,d x } \]
Exception generated. \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Not integrable
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int x^m \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^2 \, dx=\int x^m\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]