3.5.88 \(\int x (d-c^2 d x^2)^{3/2} (a+b \arcsin (c x))^n \, dx\) [488]

3.5.88.1 Optimal result

Integrand size = 27, antiderivative size = 595 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=-\frac {d e^{-\frac {i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {i (a+b \arcsin (c x))}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {d e^{\frac {i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {i (a+b \arcsin (c x))}{b}\right )}{16 c^2 \sqrt {1-c^2 x^2}}-\frac {3^{-n} d e^{-\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {3 i (a+b \arcsin (c x))}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {3^{-n} d e^{\frac {3 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {3 i (a+b \arcsin (c x))}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {5^{-1-n} d e^{-\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,-\frac {5 i (a+b \arcsin (c x))}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}}-\frac {5^{-1-n} d e^{\frac {5 i a}{b}} \sqrt {d-c^2 d x^2} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (1+n,\frac {5 i (a+b \arcsin (c x))}{b}\right )}{32 c^2 \sqrt {1-c^2 x^2}} \]

output

-1/16*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+ 
d)^(1/2)/c^2/exp(I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/ 
16*d*exp(I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,I*(a+b*arcsin(c*x))/b)*(-c^2 
*d*x^2+d)^(1/2)/c^2/((I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)-1/32*d* 
(a+b*arcsin(c*x))^n*GAMMA(1+n,-3*I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/ 
2)/(3^n)/c^2/exp(3*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^(1/2)- 
1/32*d*exp(3*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n,3*I*(a+b*arcsin(c*x))/b) 
*(-c^2*d*x^2+d)^(1/2)/(3^n)/c^2/((I*(a+b*arcsin(c*x))/b)^n)/(-c^2*x^2+1)^( 
1/2)-1/32*5^(-1-n)*d*(a+b*arcsin(c*x))^n*GAMMA(1+n,-5*I*(a+b*arcsin(c*x))/ 
b)*(-c^2*d*x^2+d)^(1/2)/c^2/exp(5*I*a/b)/((-I*(a+b*arcsin(c*x))/b)^n)/(-c^ 
2*x^2+1)^(1/2)-1/32*5^(-1-n)*d*exp(5*I*a/b)*(a+b*arcsin(c*x))^n*GAMMA(1+n, 
5*I*(a+b*arcsin(c*x))/b)*(-c^2*d*x^2+d)^(1/2)/c^2/((I*(a+b*arcsin(c*x))/b) 
^n)/(-c^2*x^2+1)^(1/2)
 

3.5.88.2 Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 464, normalized size of antiderivative = 0.78 \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=-\frac {15^{-1-n} d^2 e^{-\frac {5 i a}{b}} \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^n \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{-3 n} \left (2\ 15^{1+n} e^{\frac {4 i a}{b}} \left (\frac {i (a+b \arcsin (c x))}{b}\right )^n \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{2 n} \Gamma \left (1+n,-\frac {i (a+b \arcsin (c x))}{b}\right )+\left (-\frac {i (a+b \arcsin (c x))}{b}\right )^n \left (2\ 15^{1+n} e^{\frac {6 i a}{b}} \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{2 n} \Gamma \left (1+n,\frac {i (a+b \arcsin (c x))}{b}\right )+3 \left (5^{1+n} e^{\frac {2 i a}{b}} \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{2 n} \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^n \Gamma \left (1+n,-\frac {3 i (a+b \arcsin (c x))}{b}\right )+5^{1+n} e^{\frac {8 i a}{b}} \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{2 n} \Gamma \left (1+n,\frac {3 i (a+b \arcsin (c x))}{b}\right )+3^n \left (\left (-\frac {i (a+b \arcsin (c x))}{b}\right )^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{3 n} \Gamma \left (1+n,-\frac {5 i (a+b \arcsin (c x))}{b}\right )+e^{\frac {10 i a}{b}} \left (\frac {(a+b \arcsin (c x))^2}{b^2}\right )^{2 n} \Gamma \left (1+n,\frac {5 i (a+b \arcsin (c x))}{b}\right )\right )\right )\right )\right )}{32 c^2 \sqrt {d-c^2 d x^2}} \]

input

Integrate[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^n,x]
 

output

-1/32*(15^(-1 - n)*d^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^n*(2*15^(1 + 
n)*E^(((4*I)*a)/b)*((I*(a + b*ArcSin[c*x]))/b)^n*((a + b*ArcSin[c*x])^2/b^ 
2)^(2*n)*Gamma[1 + n, ((-I)*(a + b*ArcSin[c*x]))/b] + (((-I)*(a + b*ArcSin 
[c*x]))/b)^n*(2*15^(1 + n)*E^(((6*I)*a)/b)*((a + b*ArcSin[c*x])^2/b^2)^(2* 
n)*Gamma[1 + n, (I*(a + b*ArcSin[c*x]))/b] + 3*(5^(1 + n)*E^(((2*I)*a)/b)* 
((I*(a + b*ArcSin[c*x]))/b)^(2*n)*((a + b*ArcSin[c*x])^2/b^2)^n*Gamma[1 + 
n, ((-3*I)*(a + b*ArcSin[c*x]))/b] + 5^(1 + n)*E^(((8*I)*a)/b)*((a + b*Arc 
Sin[c*x])^2/b^2)^(2*n)*Gamma[1 + n, ((3*I)*(a + b*ArcSin[c*x]))/b] + 3^n*( 
(((-I)*(a + b*ArcSin[c*x]))/b)^n*((I*(a + b*ArcSin[c*x]))/b)^(3*n)*Gamma[1 
 + n, ((-5*I)*(a + b*ArcSin[c*x]))/b] + E^(((10*I)*a)/b)*((a + b*ArcSin[c* 
x])^2/b^2)^(2*n)*Gamma[1 + n, ((5*I)*(a + b*ArcSin[c*x]))/b])))))/(c^2*E^( 
((5*I)*a)/b)*Sqrt[d - c^2*d*x^2]*((a + b*ArcSin[c*x])^2/b^2)^(3*n))
 

3.5.88.3 Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 440, normalized size of antiderivative = 0.74, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5224, 25, 4906, 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.

\(\displaystyle \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx\)

\(\Big \downarrow \) 5224

\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \int -(a+b \arcsin (c x))^n \cos ^4\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )d(a+b \arcsin (c x))}{b c^2 \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {d \sqrt {d-c^2 d x^2} \int (a+b \arcsin (c x))^n \cos ^4\left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )d(a+b \arcsin (c x))}{b c^2 \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 4906

\(\displaystyle -\frac {d \sqrt {d-c^2 d x^2} \int \left (\frac {1}{16} \sin \left (\frac {5 a}{b}-\frac {5 (a+b \arcsin (c x))}{b}\right ) (a+b \arcsin (c x))^n+\frac {3}{16} \sin \left (\frac {3 a}{b}-\frac {3 (a+b \arcsin (c x))}{b}\right ) (a+b \arcsin (c x))^n+\frac {1}{8} \sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right ) (a+b \arcsin (c x))^n\right )d(a+b \arcsin (c x))}{b c^2 \sqrt {1-c^2 x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d \sqrt {d-c^2 d x^2} \left (-\frac {1}{16} b e^{-\frac {i a}{b}} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,-\frac {i (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} b 3^{-n} e^{-\frac {3 i a}{b}} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,-\frac {3 i (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} b 5^{-n-1} e^{-\frac {5 i a}{b}} (a+b \arcsin (c x))^n \left (-\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,-\frac {5 i (a+b \arcsin (c x))}{b}\right )-\frac {1}{16} b e^{\frac {i a}{b}} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,\frac {i (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} b 3^{-n} e^{\frac {3 i a}{b}} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,\frac {3 i (a+b \arcsin (c x))}{b}\right )-\frac {1}{32} b 5^{-n-1} e^{\frac {5 i a}{b}} (a+b \arcsin (c x))^n \left (\frac {i (a+b \arcsin (c x))}{b}\right )^{-n} \Gamma \left (n+1,\frac {5 i (a+b \arcsin (c x))}{b}\right )\right )}{b c^2 \sqrt {1-c^2 x^2}}\)

input

Int[x*(d - c^2*d*x^2)^(3/2)*(a + b*ArcSin[c*x])^n,x]
 

output

(d*Sqrt[d - c^2*d*x^2]*(-1/16*(b*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-I)* 
(a + b*ArcSin[c*x]))/b])/(E^((I*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - 
(b*E^((I*a)/b)*(a + b*ArcSin[c*x])^n*Gamma[1 + n, (I*(a + b*ArcSin[c*x]))/ 
b])/(16*((I*(a + b*ArcSin[c*x]))/b)^n) - (b*(a + b*ArcSin[c*x])^n*Gamma[1 
+ n, ((-3*I)*(a + b*ArcSin[c*x]))/b])/(32*3^n*E^(((3*I)*a)/b)*(((-I)*(a + 
b*ArcSin[c*x]))/b)^n) - (b*E^(((3*I)*a)/b)*(a + b*ArcSin[c*x])^n*Gamma[1 + 
 n, ((3*I)*(a + b*ArcSin[c*x]))/b])/(32*3^n*((I*(a + b*ArcSin[c*x]))/b)^n) 
 - (5^(-1 - n)*b*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((-5*I)*(a + b*ArcSin[ 
c*x]))/b])/(32*E^(((5*I)*a)/b)*(((-I)*(a + b*ArcSin[c*x]))/b)^n) - (5^(-1 
- n)*b*E^(((5*I)*a)/b)*(a + b*ArcSin[c*x])^n*Gamma[1 + n, ((5*I)*(a + b*Ar 
cSin[c*x]))/b])/(32*((I*(a + b*ArcSin[c*x]))/b)^n)))/(b*c^2*Sqrt[1 - c^2*x 
^2])
 

3.5.88.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x 
]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG 
tQ[p, 0]
 

rule 5224

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 
2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x 
^2)^p]   Subst[Int[x^n*Sin[-a/b + x/b]^m*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 + 2, 0] && IGtQ[m, 0]
 

3.5.88.4 Maple [F]

\[\int x \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{n}d x\]

input

int(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x)
 

output

int(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x)
 

3.5.88.5 Fricas [F]

\[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{n} x \,d x } \]

input

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x, algorithm="fricas" 
)
 

output

integral(-(c^2*d*x^3 - d*x)*sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)^n, x)
 

3.5.88.6 Sympy [F(-1)]

Timed out. \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\text {Timed out} \]

input

integrate(x*(-c**2*d*x**2+d)**(3/2)*(a+b*asin(c*x))**n,x)
 

output

Timed out
 

3.5.88.7 Maxima [F]

\[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int { {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \arcsin \left (c x\right ) + a\right )}^{n} x \,d x } \]

input

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x, algorithm="maxima" 
)
 

output

integrate((-c^2*d*x^2 + d)^(3/2)*(b*arcsin(c*x) + a)^n*x, x)
 

3.5.88.8 Giac [F(-2)]

Exception generated. \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\text {Exception raised: TypeError} \]

input

integrate(x*(-c^2*d*x^2+d)^(3/2)*(a+b*arcsin(c*x))^n,x, algorithm="giac")
 

output

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
 

3.5.88.9 Mupad [F(-1)]

Timed out. \[ \int x \left (d-c^2 d x^2\right )^{3/2} (a+b \arcsin (c x))^n \, dx=\int x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^n\,{\left (d-c^2\,d\,x^2\right )}^{3/2} \,d x \]

input

int(x*(a + b*asin(c*x))^n*(d - c^2*d*x^2)^(3/2),x)
 

output

int(x*(a + b*asin(c*x))^n*(d - c^2*d*x^2)^(3/2), x)