\(\int \frac {x^m (a+b \arcsin (c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [150]

Optimal result

Integrand size = 27, antiderivative size = 272 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {x^{1+m} (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}-\frac {m x^{1+m} \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )}{d (1+m) \sqrt {d-c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{d (2+m) \sqrt {d-c^2 d x^2}}+\frac {b c m x^{2+m} \sqrt {1-c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )}{d \left (2+3 m+m^2\right ) \sqrt {d-c^2 d x^2}} \] Output:

x^(1+m)*(a+b*arcsin(c*x))/d/(-c^2*d*x^2+d)^(1/2)-m*x^(1+m)*(-c^2*x^2+1)^(1 
/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],c^2*x^2)/d/(1 
+m)/(-c^2*d*x^2+d)^(1/2)-b*c*x^(2+m)*(-c^2*x^2+1)^(1/2)*hypergeom([1, 1+1/ 
2*m],[2+1/2*m],c^2*x^2)/d/(2+m)/(-c^2*d*x^2+d)^(1/2)+b*c*m*x^(2+m)*(-c^2*x 
^2+1)^(1/2)*hypergeom([1, 1+1/2*m, 1+1/2*m],[2+1/2*m, 3/2+1/2*m],c^2*x^2)/ 
d/(m^2+3*m+2)/(-c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.76 \[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {x^{1+m} \left (-m (2+m) \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},c^2 x^2\right )+(1+m) \left ((2+m) (a+b \arcsin (c x))-b c x \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},c^2 x^2\right )\right )+b c m x \sqrt {1-c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};c^2 x^2\right )\right )}{d (1+m) (2+m) \sqrt {d-c^2 d x^2}} \] Input:

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

Output:

(x^(1 + m)*(-(m*(2 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Hypergeometr 
ic2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2]) + (1 + m)*((2 + m)*(a + b*ArcSi 
n[c*x]) - b*c*x*Sqrt[1 - c^2*x^2]*Hypergeometric2F1[1, 1 + m/2, 2 + m/2, c 
^2*x^2]) + b*c*m*x*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/ 
2}, {3/2 + m/2, 2 + m/2}, c^2*x^2]))/(d*(1 + m)*(2 + m)*Sqrt[d - c^2*d*x^2 
])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5208, 278, 5220}

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.

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

\(\Big \downarrow \) 5208

\(\displaystyle -\frac {m \int \frac {x^m (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}}dx}{d}-\frac {b c \sqrt {1-c^2 x^2} \int \frac {x^{m+1}}{1-c^2 x^2}dx}{d \sqrt {d-c^2 d x^2}}+\frac {x^{m+1} (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {m \int \frac {x^m (a+b \arcsin (c x))}{\sqrt {d-c^2 d x^2}}dx}{d}+\frac {x^{m+1} (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} x^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{d (m+2) \sqrt {d-c^2 d x^2}}\)

\(\Big \downarrow \) 5220

\(\displaystyle -\frac {m \left (\frac {\sqrt {1-c^2 x^2} x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},c^2 x^2\right ) (a+b \arcsin (c x))}{(m+1) \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;c^2 x^2\right )}{\left (m^2+3 m+2\right ) \sqrt {d-c^2 d x^2}}\right )}{d}+\frac {x^{m+1} (a+b \arcsin (c x))}{d \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} x^{m+2} \operatorname {Hypergeometric2F1}\left (1,\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{d (m+2) \sqrt {d-c^2 d x^2}}\)

Input:

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

Output:

(x^(1 + m)*(a + b*ArcSin[c*x]))/(d*Sqrt[d - c^2*d*x^2]) - (b*c*x^(2 + m)*S 
qrt[1 - c^2*x^2]*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, c^2*x^2])/(d*( 
2 + m)*Sqrt[d - c^2*d*x^2]) - (m*((x^(1 + m)*Sqrt[1 - c^2*x^2]*(a + b*ArcS 
in[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/((1 + m)*S 
qrt[d - c^2*d*x^2]) - (b*c*x^(2 + m)*Sqrt[1 - c^2*x^2]*HypergeometricPFQ[{ 
1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/((2 + 3*m + m^2)*Sqr 
t[d - c^2*d*x^2])))/d
 

Defintions of rubi rules used

rule 278

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])
 

rule 5208

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])
 

rule 5220

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)*x^m/(c^4*d^2*x^4 - 2*c^2 
*d^2*x^2 + d^2), x)
 

Sympy [F]

\[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^{m} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral(x**m*(a + b*asin(c*x))/(-d*(c*x - 1)*(c*x + 1))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^m (a+b \arcsin (c x))}{\left (d-c^2 d x^2\right )^{3/2}} \, dx=\frac {-\left (\int \frac {x^{m}}{\sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) a -\left (\int \frac {x^{m} \mathit {asin} \left (c x \right )}{\sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}-\sqrt {-c^{2} x^{2}+1}}d x \right ) b}{\sqrt {d}\, d} \] Input:

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

Output:

( - (int(x**m/(sqrt( - c**2*x**2 + 1)*c**2*x**2 - sqrt( - c**2*x**2 + 1)), 
x)*a + int((x**m*asin(c*x))/(sqrt( - c**2*x**2 + 1)*c**2*x**2 - sqrt( - c* 
*2*x**2 + 1)),x)*b))/(sqrt(d)*d)