\(\int (a+b x^2)^p (A+B x^2+C x^4) \, dx\) [126]

Optimal result

Integrand size = 22, antiderivative size = 165 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=-\frac {(3 a C-b B (5+2 p)) x \left (a+b x^2\right )^{1+p}}{b^2 (3+2 p) (5+2 p)}+\frac {C x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {\left (A b^2 \left (15+16 p+4 p^2\right )+a (3 a C-b B (5+2 p))\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b^2 (3+2 p) (5+2 p)} \] Output:

-(3*C*a-b*B*(5+2*p))*x*(b*x^2+a)^(p+1)/b^2/(3+2*p)/(5+2*p)+C*x^3*(b*x^2+a) 
^(p+1)/b/(5+2*p)+(A*b^2*(4*p^2+16*p+15)+a*(3*C*a-b*B*(5+2*p)))*x*(b*x^2+a) 
^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/b^2/(3+2*p)/(5+2*p)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.61 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{15} x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (15 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+5 B x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+3 C x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

Integrate[(a + b*x^2)^p*(A + B*x^2 + C*x^4),x]
 

Output:

(x*(a + b*x^2)^p*(15*A*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + 5*B 
*x^2*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)] + 3*C*x^4*Hypergeometri 
c2F1[5/2, -p, 7/2, -((b*x^2)/a)]))/(15*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1473, 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.

\(\displaystyle \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\int \left (b x^2+a\right )^p \left (A b (2 p+5)-(3 a C-b B (2 p+5)) x^2\right )dx}{b (2 p+5)}+\frac {C x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\left (\frac {a (3 a C-b B (2 p+5))}{b (2 p+3)}+A b (2 p+5)\right ) \int \left (b x^2+a\right )^pdx-\frac {x \left (a+b x^2\right )^{p+1} (3 a C-b B (2 p+5))}{b (2 p+3)}}{b (2 p+5)}+\frac {C x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (\frac {a (3 a C-b B (2 p+5))}{b (2 p+3)}+A b (2 p+5)\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx-\frac {x \left (a+b x^2\right )^{p+1} (3 a C-b B (2 p+5))}{b (2 p+3)}}{b (2 p+5)}+\frac {C x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \left (\frac {a (3 a C-b B (2 p+5))}{b (2 p+3)}+A b (2 p+5)\right )-\frac {x \left (a+b x^2\right )^{p+1} (3 a C-b B (2 p+5))}{b (2 p+3)}}{b (2 p+5)}+\frac {C x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)}\)

Input:

Int[(a + b*x^2)^p*(A + B*x^2 + C*x^4),x]
 

Output:

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

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

Maple [F]

\[\int \left (b \,x^{2}+a \right )^{p} \left (C \,x^{4}+x^{2} B +A \right )d x\]

Input:

int((b*x^2+a)^p*(C*x^4+B*x^2+A),x)
 

Output:

int((b*x^2+a)^p*(C*x^4+B*x^2+A),x)
 

Fricas [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(C*x^4+B*x^2+A),x, algorithm="fricas")
 

Output:

integral((C*x^4 + B*x^2 + A)*(b*x^2 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.50 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=A a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {B a^{p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {C a^{p} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} \] Input:

integrate((b*x**2+a)**p*(C*x**4+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 + C*a**p*x**5*hyper(( 
5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5
 

Maxima [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(C*x^4+B*x^2+A),x, algorithm="maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(b*x^2 + a)^p, x)
 

Giac [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(C*x^4+B*x^2+A),x, algorithm="giac")
 

Output:

integrate((C*x^4 + B*x^2 + A)*(b*x^2 + a)^p, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx=\int {\left (b\,x^2+a\right )}^p\,\left (C\,x^4+B\,x^2+A\right ) \,d x \] Input:

int((a + b*x^2)^p*(A + B*x^2 + C*x^4),x)
 

Output:

int((a + b*x^2)^p*(A + B*x^2 + C*x^4), x)
 

Reduce [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4\right ) \, dx =\text {Too large to display} \] Input:

int((b*x^2+a)^p*(C*x^4+B*x^2+A),x)
 

Output:

( - 6*(a + b*x**2)**p*a**2*c*p*x + 8*(a + b*x**2)**p*a*b**2*p**2*x + 26*(a 
 + b*x**2)**p*a*b**2*p*x + 15*(a + b*x**2)**p*a*b**2*x + 4*(a + b*x**2)**p 
*a*b*c*p**2*x**3 + 2*(a + b*x**2)**p*a*b*c*p*x**3 + 4*(a + b*x**2)**p*b**3 
*p**2*x**3 + 12*(a + b*x**2)**p*b**3*p*x**3 + 5*(a + b*x**2)**p*b**3*x**3 
+ 4*(a + b*x**2)**p*b**2*c*p**2*x**5 + 8*(a + b*x**2)**p*b**2*c*p*x**5 + 3 
*(a + b*x**2)**p*b**2*c*x**5 + 48*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p** 
2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x* 
*2),x)*a**3*c*p**4 + 216*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a* 
p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a* 
*3*c*p**3 + 276*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a 
+ 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*c*p**2 
 + 90*int((a + b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3 
*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*c*p + 64*int((a 
+ b*x**2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b* 
p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**2*b**2*p**6 + 512*int((a + b*x* 
*2)**p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x 
**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**2*b**2*p**5 + 1536*int((a + b*x**2)** 
p/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 
 46*b*p*x**2 + 15*b*x**2),x)*a**2*b**2*p**4 + 2128*int((a + b*x**2)**p/(8* 
a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 4...