3.2.53 \(\int \frac {x^m (a+b \arcsin (c x))}{(d-c^2 d x^2)^{3/2}} \, dx\) [153]

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

3.2.53.2 Mathematica [A] (verified)

Time = 0.17 (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 
])
 

3.2.53.3 Rubi [A] (verified)

Time = 0.54 (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
 

3.2.53.3.1 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]
 

3.2.53.4 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)
 

3.2.53.5 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)
 

3.2.53.6 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)
 

3.2.53.7 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)
 

3.2.53.8 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
 

3.2.53.9 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)