\(\int (c x)^m \sqrt {a+b x^2} (A+B x^2) \, dx\) [506]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int (c x)^m \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {B (c x)^{1+m} \left (a+b x^2\right )^{3/2}}{b c (4+m)}+\frac {\left (\frac {A}{1+m}-\frac {a B}{b (4+m)}\right ) (c x)^{1+m} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{c \sqrt {1+\frac {b x^2}{a}}} \] Output:

B*(c*x)^(1+m)*(b*x^2+a)^(3/2)/b/c/(4+m)+(A/(1+m)-a*B/b/(4+m))*(c*x)^(1+m)* 
(b*x^2+a)^(1/2)*hypergeom([-1/2, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/c/(1+b*x 
^2/a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int (c x)^m \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {x (c x)^m \sqrt {a+b x^2} \left (A (3+m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+B (1+m) x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )}{(1+m) (3+m) \sqrt {1+\frac {b x^2}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {363, 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.

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

\(\Big \downarrow \) 363

\(\displaystyle \left (A-\frac {a B (m+1)}{b (m+4)}\right ) \int (c x)^m \sqrt {b x^2+a}dx+\frac {B \left (a+b x^2\right )^{3/2} (c x)^{m+1}}{b c (m+4)}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\sqrt {a+b x^2} \left (A-\frac {a B (m+1)}{b (m+4)}\right ) \int (c x)^m \sqrt {\frac {b x^2}{a}+1}dx}{\sqrt {\frac {b x^2}{a}+1}}+\frac {B \left (a+b x^2\right )^{3/2} (c x)^{m+1}}{b c (m+4)}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\sqrt {a+b x^2} (c x)^{m+1} \left (A-\frac {a B (m+1)}{b (m+4)}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{c (m+1) \sqrt {\frac {b x^2}{a}+1}}+\frac {B \left (a+b x^2\right )^{3/2} (c x)^{m+1}}{b c (m+4)}\)

Input:

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

Output:

(B*(c*x)^(1 + m)*(a + b*x^2)^(3/2))/(b*c*(4 + m)) + ((A - (a*B*(1 + m))/(b 
*(4 + m)))*(c*x)^(1 + m)*Sqrt[a + b*x^2]*Hypergeometric2F1[-1/2, (1 + m)/2 
, (3 + m)/2, -((b*x^2)/a)])/(c*(1 + m)*Sqrt[1 + (b*x^2)/a])
 

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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

Maple [F]

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

Input:

int((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x)
 

Output:

int((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x)
 

Fricas [F]

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

integrate((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.57 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int (c x)^m \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx=\frac {A \sqrt {a} c^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B \sqrt {a} c^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

integrate((c*x)**m*(b*x**2+a)**(1/2)*(B*x**2+A),x)
 

Output:

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

Maxima [F]

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

integrate((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((A + B*x^2)*(c*x)^m*(a + b*x^2)^(1/2),x)
 

Output:

int((A + B*x^2)*(c*x)^m*(a + b*x^2)^(1/2), x)
 

Reduce [F]

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

int((c*x)^m*(b*x^2+a)^(1/2)*(B*x^2+A),x)
 

Output:

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