Integrand size = 26, antiderivative size = 466 \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\frac {3 d \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^{1+n}}{8 b c (1+n) \sqrt {1-c^2 x^2}}-\frac {i 2^{-3-n} d e^{-\frac {2 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 i (a+b \arcsin (c x))}{b}\right )}{c \sqrt {1-c^2 x^2}}+\frac {i 2^{-3-n} d e^{\frac {2 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 i (a+b \arcsin (c x))}{b}\right )}{c \sqrt {1-c^2 x^2}}-\frac {i 2^{-2 (3+n)} d e^{-\frac {4 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {4 i (a+b \arcsin (c x))}{b}\right )}{c \sqrt {1-c^2 x^2}}+\frac {i 2^{-2 (3+n)} d e^{\frac {4 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {4 i (a+b \arcsin (c x))}{b}\right )}{c \sqrt {1-c^2 x^2}} \]
3/8*d*(a+b*arcsin(c*x))^(1+n)*(-c^2*d*x^2+d)^(1/2)/b/c/(1+n)/(-c^2*x^2+1)^ (1/2)-I*2^(-3-n)*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-2*I*(a+b*arcsin(c*x))/b) *(-c^2*d*x^2+d)^(1/2)/c/exp(2*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^ 2+1)^(1/2)+I*2^(-3-n)*d*exp(2*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,2*I*(a+ b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/c/((I*(a+b*arcsin(c*x))/b)^n)/(-c^2 *x^2+1)^(1/2)-I*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-4*I*(a+b*arcsin(c*x))/b)* (-c^2*d*x^2+d)^(1/2)/(2^(6+2*n))/c/exp(4*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^ n)/(-c^2*x^2+1)^(1/2)+I*d*exp(4*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,4*I*( a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/(2^(6+2*n))/c/((I*(a+b*arcsin(c*x ))/b)^n)/(-c^2*x^2+1)^(1/2)
Time = 1.46 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.70 \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\frac {d^2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^n \left (-\frac {8 (a+b \arcsin (c x))}{b (1+n)}+8 \left (\frac {4 a+4 b \arcsin (c x)}{b+b n}-i 2^{-n} e^{-\frac {2 i a}{b}} \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {2 i (a+b \arcsin (c x))}{b}\right )+i 2^{-n} e^{\frac {2 i a}{b}} \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {2 i (a+b \arcsin (c x))}{b}\right )\right )+i 4^{-n} e^{-\frac {4 i a}{b}} \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{-n} \left (-\left (\frac {i (a+b \arcsin (c x))}{b}\right )^n \Gamma \left (1+n,-\frac {4 i (a+b \arcsin (c x))}{b}\right )+e^{\frac {8 i a}{b}} \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^n \Gamma \left (1+n,\frac {4 i (a+b \arcsin (c x))}{b}\right )\right )\right )}{64 c \sqrt {d-c^2 d x^2}} \]
(d^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^n*((-8*(a + b*ArcSin[c*x]))/(b* (1 + n)) + 8*((4*a + 4*b*ArcSin[c*x])/(b + b*n) - (I*Gamma[1 + n, ((-2*I)* (a + b*ArcSin[c*x]))/b])/(2^n*E^(((2*I)*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/ b)^n) + (I*E^(((2*I)*a)/b)*Gamma[1 + n, ((2*I)*(a + b*ArcSin[c*x]))/b])/(2 ^n*((I*(a + b*ArcSin[c*x]))/b)^n)) + (I*(-(((I*(a + b*ArcSin[c*x]))/b)^n*G amma[1 + n, ((-4*I)*(a + b*ArcSin[c*x]))/b]) + E^(((8*I)*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/b)^n*Gamma[1 + n, ((4*I)*(a + b*ArcSin[c*x]))/b]))/(4^n*E ^(((4*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^n)))/(64*c*Sqrt[d - c^2*d*x^2])
Time = 0.57 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.73, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5168, 3042, 3793, 2009}
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.
\(\displaystyle \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx\) |
\(\Big \downarrow \) 5168 |
\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \int (a+b \arcsin (c x))^n \cos ^4\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )d(a+b \arcsin (c x))}{b c \sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \int (a+b \arcsin (c x))^n \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}+\frac {\pi }{2}\right )^4d(a+b \arcsin (c x))}{b c \sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \int \left (\frac {1}{8} \cos \left (\frac {4 a}{b}-\frac {4 (a+b \arcsin (c x))}{b}\right ) (a+b \arcsin (c x))^n+\frac {1}{2} \cos \left (\frac {2 a}{b}-\frac {2 (a+b \arcsin (c x))}{b}\right ) (a+b \arcsin (c x))^n+\frac {3}{8} (a+b \arcsin (c x))^n\right )d(a+b \arcsin (c x))}{b c \sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \left (\frac {3 (a+b \arcsin (c x))^{n+1}}{8 (n+1)}-i b 2^{-n-3} e^{-\frac {2 i a}{b}} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,-\frac {2 i (a+b \arcsin (c x))}{b}\right )-i b 2^{-2 (n+3)} e^{-\frac {4 i a}{b}} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,-\frac {4 i (a+b \arcsin (c x))}{b}\right )+i b 2^{-n-3} e^{\frac {2 i a}{b}} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,\frac {2 i (a+b \arcsin (c x))}{b}\right )+i b 2^{-2 (n+3)} e^{\frac {4 i a}{b}} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,\frac {4 i (a+b \arcsin (c x))}{b}\right )\right )}{b c \sqrt {1-c^2 x^2}}\) |
(d*Sqrt[d - c^2*d*x^2]*((3*(a + b*ArcSin[c*x])^(1 + n))/(8*(1 + n)) - (I*2 ^(-3 - n)*b*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-2*I)*(a + b*ArcSin[c*x]) )/b])/(E^(((2*I)*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/b)^n) + (I*2^(-3 - n)*b *E^(((2*I)*a)/b)*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((2*I)*(a + b*ArcSin[c *x]))/b])/((I*(a + b*ArcSin[c*x]))/b)^n - (I*b*(a + b*ArcSin[c*x])^n*Gamma [1 + n, ((-4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*E^(((4*I)*a)/b)*(( (-I)*(a + b*ArcSin[c*x]))/b)^n) + (I*b*E^(((4*I)*a)/b)*(a + b*ArcSin[c*x]) ^n*Gamma[1 + n, ((4*I)*(a + b*ArcSin[c*x]))/b])/(2^(2*(3 + n))*((I*(a + b* ArcSin[c*x]))/b)^n)))/(b*c*Sqrt[1 - c^2*x^2])
3.5.89.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[(1/(b*c))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Subst[Int[ x^n*Cos[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b , c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p, 0]
\[\int \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{n}d x\]
\[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{n} \,d x } \]
Timed out. \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\text {Timed out} \]
\[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{n} \,d x } \]
Exception generated. \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\text {Exception raised: TypeError} \]
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
Timed out. \[ \int \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^n\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]