Integrand size = 27, antiderivative size = 27 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=-\frac {b c x^{2+m} (a+b \arcsin (c x))}{6 d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c (1-m) x^{2+m} (a+b \arcsin (c x))}{6 d^3 \sqrt {1-c^2 x^2}}-\frac {b c (3-m) x^{2+m} (a+b \arcsin (c x))}{4 d^3 \sqrt {1-c^2 x^2}}+\frac {x^{1+m} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \arcsin (c x))^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac {b c (1-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{6 d^3 (2+m)}+\frac {b c (3-m) (1+m) x^{2+m} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {b^2 c^2 (1-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}+\frac {b^2 c^2 (3-m) x^{3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{4 d^3 (3+m)}+\frac {b^2 c^2 x^{3+m} \operatorname {Hypergeometric2F1}\left (2,\frac {3+m}{2},\frac {5+m}{2},c^2 x^2\right )}{6 d^3 (3+m)}-\frac {b^2 c^2 (1-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{6 d^3 \left (6+5 m+m^2\right )}-\frac {b^2 c^2 (3-m) (1+m) x^{3+m} \, _3F_2\left (1,\frac {3}{2}+\frac {m}{2},\frac {3}{2}+\frac {m}{2};2+\frac {m}{2},\frac {5}{2}+\frac {m}{2};c^2 x^2\right )}{4 d^3 \left (6+5 m+m^2\right )}+\frac {(1-m) (3-m) \text {Int}\left (\frac {x^m (a+b \arcsin (c x))^2}{d-c^2 d x^2},x\right )}{8 d^2} \]
-1/6*b*c*x^(2+m)*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1)^(3/2)+1/4*x^(1+m)*(a+b *arcsin(c*x))^2/d^3/(-c^2*x^2+1)^2+1/8*(3-m)*x^(1+m)*(a+b*arcsin(c*x))^2/d ^3/(-c^2*x^2+1)+1/6*b*c*(1-m)*(1+m)*x^(2+m)*(a+b*arcsin(c*x))*hypergeom([1 /2, 1+1/2*m],[2+1/2*m],c^2*x^2)/d^3/(2+m)+1/4*b*c*(3-m)*(1+m)*x^(2+m)*(a+b *arcsin(c*x))*hypergeom([1/2, 1+1/2*m],[2+1/2*m],c^2*x^2)/d^3/(2+m)+1/6*b^ 2*c^2*(1-m)*x^(3+m)*hypergeom([1, 3/2+1/2*m],[5/2+1/2*m],c^2*x^2)/d^3/(3+m )+1/4*b^2*c^2*(3-m)*x^(3+m)*hypergeom([1, 3/2+1/2*m],[5/2+1/2*m],c^2*x^2)/ d^3/(3+m)+1/6*b^2*c^2*x^(3+m)*hypergeom([2, 3/2+1/2*m],[5/2+1/2*m],c^2*x^2 )/d^3/(3+m)-1/6*b^2*c^2*(1-m)*(1+m)*x^(3+m)*hypergeom([1, 3/2+1/2*m, 3/2+1 /2*m],[2+1/2*m, 5/2+1/2*m],c^2*x^2)/d^3/(m^2+5*m+6)-1/4*b^2*c^2*(3-m)*(1+m )*x^(3+m)*hypergeom([1, 3/2+1/2*m, 3/2+1/2*m],[2+1/2*m, 5/2+1/2*m],c^2*x^2 )/d^3/(m^2+5*m+6)-1/6*b*c*(1-m)*x^(2+m)*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1) ^(1/2)-1/4*b*c*(3-m)*x^(2+m)*(a+b*arcsin(c*x))/d^3/(-c^2*x^2+1)^(1/2)+1/8* (1-m)*(3-m)*Unintegrable(x^m*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d),x)/d^2
Not integrable
Time = 9.81 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx \]
Not integrable
Time = 1.75 (sec) , antiderivative size = 27, 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 = {5208, 27, 5208, 278, 5208, 278, 5220, 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.
\(\displaystyle \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 5208 |
\(\displaystyle -\frac {b c \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{2 d^3}+\frac {(3-m) \int \frac {x^m (a+b \arcsin (c x))^2}{d^2 \left (1-c^2 x^2\right )^2}dx}{4 d}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b c \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{2 d^3}+\frac {(3-m) \int \frac {x^m (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^2}dx}{4 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 5208 |
\(\displaystyle -\frac {b c \left (\frac {1}{3} (1-m) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx-\frac {1}{3} b c \int \frac {x^{m+2}}{\left (1-c^2 x^2\right )^2}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}\right )}{2 d^3}+\frac {(3-m) \left (-b c \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx+\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {b c \left (\frac {1}{3} (1-m) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (2,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{3 (m+3)}\right )}{2 d^3}+\frac {(3-m) \left (-b c \int \frac {x^{m+1} (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{3/2}}dx+\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 5208 |
\(\displaystyle -\frac {b c \left (\frac {1}{3} (1-m) \left (-(m+1) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx-b c \int \frac {x^{m+2}}{1-c^2 x^2}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (2,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{3 (m+3)}\right )}{2 d^3}+\frac {(3-m) \left (-b c \left (-(m+1) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx-b c \int \frac {x^{m+2}}{1-c^2 x^2}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}\right )+\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {b c \left (\frac {1}{3} (1-m) \left (-(m+1) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (2,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{3 (m+3)}\right )}{2 d^3}+\frac {(3-m) \left (-b c \left (-(m+1) \int \frac {x^{m+1} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}dx+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 5220 |
\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx-b c \left (-(m+1) \left (\frac {x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{m^2+5 m+6}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}-\frac {b c \left (\frac {1}{3} (1-m) \left (-(m+1) \left (\frac {x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{m^2+5 m+6}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (2,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{3 (m+3)}\right )}{2 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
\(\Big \downarrow \) 5234 |
\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {x^m (a+b \arcsin (c x))^2}{1-c^2 x^2}dx-b c \left (-(m+1) \left (\frac {x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{m^2+5 m+6}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {x^{m+1} (a+b \arcsin (c x))^2}{2 \left (1-c^2 x^2\right )}\right )}{4 d^3}-\frac {b c \left (\frac {1}{3} (1-m) \left (-(m+1) \left (\frac {x^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{m+2}-\frac {b c x^{m+3} \, _3F_2\left (1,\frac {m}{2}+\frac {3}{2},\frac {m}{2}+\frac {3}{2};\frac {m}{2}+2,\frac {m}{2}+\frac {5}{2};c^2 x^2\right )}{m^2+5 m+6}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (1,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{m+3}\right )+\frac {x^{m+2} (a+b \arcsin (c x))}{3 \left (1-c^2 x^2\right )^{3/2}}-\frac {b c x^{m+3} \operatorname {Hypergeometric2F1}\left (2,\frac {m+3}{2},\frac {m+5}{2},c^2 x^2\right )}{3 (m+3)}\right )}{2 d^3}+\frac {x^{m+1} (a+b \arcsin (c x))^2}{4 d^3 \left (1-c^2 x^2\right )^2}\) |
3.3.81.3.1 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[((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 + 1)*((a + b*ArcSin[c*x])^n/(2*d*f*(p + 1))), x] + (Simp[(m + 2*p + 3)/(2*d*(p + 1)) Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Simp[b*c *(n/(2*f*(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] && LtQ[p, -1] && !G tQ[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_. )*(x_)^2], x_Symbol] :> Simp[((f*x)^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 - c^2* x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*S imp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m /2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && !IntegerQ[m]
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 = 0.32 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
\[\int \frac {x^{m} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (-c^{2} d \,x^{2}+d \right )^{3}}d x\]
Not integrable
Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*x^m/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)
Not integrable
Time = 117.24 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.41 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=- \frac {\int \frac {a^{2} x^{m}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b^{2} x^{m} \operatorname {asin}^{2}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {2 a b x^{m} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]
-(Integral(a**2*x**m/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Int egral(b**2*x**m*asin(c*x)**2/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(2*a*b*x**m*asin(c*x)/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x))/d**3
Not integrable
Time = 0.90 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
Not integrable
Time = 0.88 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]
Not integrable
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {x^m (a+b \arcsin (c x))^2}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {x^m\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]