\(\int (a+b x^2)^p (c+d x^2) \, dx\) [516]

Optimal result

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

d*x*(b*x^2+a)^(p+1)/b/(3+2*p)+(c-a*d/(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 = 0.02 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (d \left (a+b x^2\right ) \left (1+\frac {b x^2}{a}\right )^p+(-a d+b c (3+2 p)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{b (3+2 p)} \] Input:

Integrate[(a + b*x^2)^p*(c + d*x^2),x]
 

Output:

(x*(a + b*x^2)^p*(d*(a + b*x^2)*(1 + (b*x^2)/a)^p + (-(a*d) + b*c*(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.

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

Input:

Int[(a + b*x^2)^p*(c + d*x^2),x]
 

Output:

(d*x*(a + b*x^2)^(1 + p))/(b*(3 + 2*p)) + ((c - (a*d)/(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]

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

Input:

int((b*x^2+a)^p*(d*x^2+c),x)
 

Output:

int((b*x^2+a)^p*(d*x^2+c),x)
 

Fricas [F]

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

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

Output:

integral((d*x^2 + c)*(b*x^2 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=a^{p} c x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{p} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \] Input:

integrate((b*x**2+a)**p*(d*x**2+c),x)
 

Output:

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

Maxima [F]

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

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

Output:

integrate((d*x^2 + c)*(b*x^2 + a)^p, x)
 

Giac [F]

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

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

Output:

integrate((d*x^2 + c)*(b*x^2 + a)^p, x)
 

Mupad [F(-1)]

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

int((a + b*x^2)^p*(c + d*x^2),x)
 

Output:

int((a + b*x^2)^p*(c + d*x^2), x)
 

Reduce [F]

\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right ) \, dx=\frac {2 \left (b \,x^{2}+a \right )^{p} a d p x +2 \left (b \,x^{2}+a \right )^{p} b c p x +3 \left (b \,x^{2}+a \right )^{p} b c x +2 \left (b \,x^{2}+a \right )^{p} b d p \,x^{3}+\left (b \,x^{2}+a \right )^{p} b d \,x^{3}-8 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d \,p^{3}-16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d \,p^{2}-6 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a^{2} d p +16 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a b c \,p^{4}+56 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a b c \,p^{3}+60 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a b c \,p^{2}+18 \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{4 b \,p^{2} x^{2}+8 b p \,x^{2}+4 a \,p^{2}+3 b \,x^{2}+8 a p +3 a}d x \right ) a b c p}{b \left (4 p^{2}+8 p +3\right )} \] Input:

int((b*x^2+a)^p*(d*x^2+c),x)
 

Output:

(2*(a + b*x**2)**p*a*d*p*x + 2*(a + b*x**2)**p*b*c*p*x + 3*(a + b*x**2)**p 
*b*c*x + 2*(a + b*x**2)**p*b*d*p*x**3 + (a + b*x**2)**p*b*d*x**3 - 8*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*d*p**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*d*p**2 - 6*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 
*d*p + 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*b*c*p**4 + 56*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*b*c*p**3 + 60*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*b*c*p**2 + 18*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*b*c*p)/(b*(4*p**2 + 8*p + 3))