\(\int \frac {(c x)^m (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [508]

Optimal result

Integrand size = 24, antiderivative size = 113 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B (c x)^{1+m}}{b c m \sqrt {a+b x^2}}-\frac {\left (\frac {B}{b m}-\frac {A}{a+a m}\right ) (c x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{c \sqrt {a+b x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (c x)^m \sqrt {1+\frac {b x^2}{a}} \left (A (3+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+B (1+m) x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )}{a (1+m) (3+m) \sqrt {a+b x^2}} \] Input:

Integrate[((c*x)^m*(A + B*x^2))/(a + b*x^2)^(3/2),x]
 

Output:

(x*(c*x)^m*Sqrt[1 + (b*x^2)/a]*(A*(3 + m)*Hypergeometric2F1[3/2, (1 + m)/2 
, (3 + m)/2, -((b*x^2)/a)] + B*(1 + m)*x^2*Hypergeometric2F1[3/2, (3 + m)/ 
2, (5 + m)/2, -((b*x^2)/a)]))/(a*(1 + m)*(3 + m)*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {362, 279, 278}

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.

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {(c x)^{m+1} (A b-a B)}{a b c \sqrt {a+b x^2}}-\frac {\sqrt {\frac {b x^2}{a}+1} (c x)^{m+1} (A b m-a B (m+1)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a b c (m+1) \sqrt {a+b x^2}}\)

Input:

Int[((c*x)^m*(A + B*x^2))/(a + b*x^2)^(3/2),x]
 

Output:

((A*b - a*B)*(c*x)^(1 + m))/(a*b*c*Sqrt[a + b*x^2]) - ((A*b*m - a*B*(1 + m 
))*(c*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(a*b*c*(1 + m)*Sqrt[a + b*x^2])
 

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

Maple [F]

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

Input:

int((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x)
 

Output:

int((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x)
 

Fricas [F]

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

integrate((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

integral((B*x^2 + A)*sqrt(b*x^2 + a)*(c*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), 
x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int \frac {(c x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A c^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B c^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

integrate((c*x)**m*(B*x**2+A)/(b*x**2+a)**(3/2),x)
 

Output:

A*c**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((3/2, m/2 + 1/2), (m/2 + 3/2,), b 
*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(m/2 + 3/2)) + B*c**m*x**(m + 3) 
*gamma(m/2 + 3/2)*hyper((3/2, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(3/2)*gamma(m/2 + 5/2))
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*(c*x)^m/(b*x^2 + a)^(3/2), x)
 

Giac [F]

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

integrate((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*(c*x)^m/(b*x^2 + a)^(3/2), x)
 

Mupad [F(-1)]

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

int(((A + B*x^2)*(c*x)^m)/(a + b*x^2)^(3/2),x)
 

Output:

int(((A + B*x^2)*(c*x)^m)/(a + b*x^2)^(3/2), x)
 

Reduce [F]

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

int((c*x)^m*(B*x^2+A)/(b*x^2+a)^(3/2),x)
 

Output:

c**m*int(x**m/sqrt(a + b*x**2),x)