\(\int (a+b x^2)^p (A+B x^2) \, dx\) [108]

Optimal result

Integrand size = 17, antiderivative size = 85 \[ \int \left (a+b x^2\right )^p \left (A+B x^2\right ) \, dx=\frac {B x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\left (A-\frac {a B}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^2\right )^p \left (A+B x^2\right ) \, dx=\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (B \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+(-a B+A b (3+2 p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{b (3+2 p)} \] Input:

Integrate[(a + b*x^2)^p*(A + B*x^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {299, 238, 237}

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.

\(\displaystyle \int \left (A+B x^2\right ) \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 299

\(\displaystyle \left (A-\frac {a B}{2 b p+3 b}\right ) \int \left (b x^2+a\right )^pdx+\frac {B x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}\)

\(\Big \downarrow \) 238

\(\displaystyle \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (A-\frac {a B}{2 b p+3 b}\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx+\frac {B x \left (a+b x^2\right )^{p+1}}{b (2 p+3)}\)

\(\Big \downarrow \) 237

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

Input:

Int[(a + b*x^2)^p*(A + B*x^2),x]
 

Output:

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

Defintions of rubi rules used

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 299

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

Maple [F]

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

Input:

int((b*x^2+a)^p*(B*x^2+A),x)
 

Output:

int((b*x^2+a)^p*(B*x^2+A),x)
 

Fricas [F]

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

integrate((b*x^2+a)^p*(B*x^2+A),x, algorithm="fricas")
 

Output:

integral((B*x^2 + A)*(b*x^2 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \left (a+b x^2\right )^p \left (A+B x^2\right ) \, dx=A a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {B a^{p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:

integrate((b*x**2+a)**p*(B*x**2+A),x)
 

Output:

A*a**p*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + B*a**p*x**3* 
hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a)/3
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((b*x^2+a)^p*(B*x^2+A),x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*(b*x^2 + a)^p, x)
 

Giac [F]

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

integrate((b*x^2+a)^p*(B*x^2+A),x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*(b*x^2 + a)^p, x)
 

Mupad [F(-1)]

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

int((A + B*x^2)*(a + b*x^2)^p,x)
 

Output:

int((A + B*x^2)*(a + b*x^2)^p, x)
 

Reduce [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2\right ) \, dx=\frac {4 \left (b \,x^{2}+a \right )^{p} a p x +3 \left (b \,x^{2}+a \right )^{p} a x +2 \left (b \,x^{2}+a \right )^{p} b p \,x^{3}+\left (b \,x^{2}+a \right )^{p} b \,x^{3}+16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} p^{4}+48 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} p^{3}+44 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} p^{2}+12 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} p}{4 p^{2}+8 p +3} \] Input:

int((b*x^2+a)^p*(B*x^2+A),x)
 

Output:

(4*(a + b*x**2)**p*a*p*x + 3*(a + b*x**2)**p*a*x + 2*(a + b*x**2)**p*b*p*x 
**3 + (a + b*x**2)**p*b*x**3 + 16*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 
3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*p**4 + 48*int((a + b* 
x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2), 
x)*a**2*p**3 + 44*int((a + b*x**2)**p/(4*a*p**2 + 8*a*p + 3*a + 4*b*p**2*x 
**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*p**2 + 12*int((a + b*x**2)**p/(4*a*p* 
*2 + 8*a*p + 3*a + 4*b*p**2*x**2 + 8*b*p*x**2 + 3*b*x**2),x)*a**2*p)/(4*p* 
*2 + 8*p + 3)