\(\int (a+b x^2)^p (A+B x^2+C x^4+D x^6) \, dx\) [151]

Optimal result

Integrand size = 27, antiderivative size = 262 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {\left (15 a^2 D-3 a b C (7+2 p)+b^2 B \left (35+24 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac {(5 a D-b C (7+2 p)) x^3 \left (a+b x^2\right )^{1+p}}{b^2 (5+2 p) (7+2 p)}+\frac {D x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {\left (A b^3 \left (35+24 p+4 p^2\right )-\frac {a \left (15 a^2 D-3 a b C (7+2 p)+b^2 B \left (35+24 p+4 p^2\right )\right )}{3+2 p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b^3 (5+2 p) (7+2 p)} \] Output:

(15*a^2*D-3*a*b*C*(7+2*p)+b^2*B*(4*p^2+24*p+35))*x*(b*x^2+a)^(p+1)/b^3/(3+ 
2*p)/(5+2*p)/(7+2*p)-(5*D*a-b*C*(7+2*p))*x^3*(b*x^2+a)^(p+1)/b^2/(5+2*p)/( 
7+2*p)+D*x^5*(b*x^2+a)^(p+1)/b/(7+2*p)+(A*b^3*(4*p^2+24*p+35)-a*(15*a^2*D- 
3*a*b*C*(7+2*p)+b^2*B*(4*p^2+24*p+35))/(3+2*p))*x*(b*x^2+a)^p*hypergeom([1 
/2, -p],[3/2],-b*x^2/a)/b^3/(5+2*p)/(7+2*p)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{105} x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (105 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+35 B x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+21 C x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+15 D x^6 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-p,\frac {9}{2},-\frac {b x^2}{a}\right )\right ) \] Input:

Integrate[(a + b*x^2)^p*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(x*(a + b*x^2)^p*(105*A*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + 35 
*B*x^2*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)] + 21*C*x^4*Hypergeome 
tric2F1[5/2, -p, 7/2, -((b*x^2)/a)] + 15*D*x^6*Hypergeometric2F1[7/2, -p, 
9/2, -((b*x^2)/a)]))/(105*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2346, 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.

\(\displaystyle \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {\int \left (b x^2+a\right )^p \left (-\left ((5 a D-b C (2 p+7)) x^4\right )+b B (2 p+7) x^2+A b (2 p+7)\right )dx}{b (2 p+7)}+\frac {D x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {\int \left (b x^2+a\right )^p \left (A \left (4 p^2+24 p+35\right ) b^2+\left (15 D a^2-3 b C (2 p+7) a+b^2 B \left (4 p^2+24 p+35\right )\right ) x^2\right )dx}{b (2 p+5)}-\frac {x^3 \left (a+b x^2\right )^{p+1} (5 a D-b C (2 p+7))}{b (2 p+5)}}{b (2 p+7)}+\frac {D x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (A b^3 \left (4 p^2+24 p+35\right )-\frac {a \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{2 p+3}\right ) \int \left (b x^2+a\right )^pdx}{b}+\frac {x \left (a+b x^2\right )^{p+1} \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{b (2 p+3)}}{b (2 p+5)}-\frac {x^3 \left (a+b x^2\right )^{p+1} (5 a D-b C (2 p+7))}{b (2 p+5)}}{b (2 p+7)}+\frac {D x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {\frac {\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (A b^3 \left (4 p^2+24 p+35\right )-\frac {a \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{2 p+3}\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{b}+\frac {x \left (a+b x^2\right )^{p+1} \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{b (2 p+3)}}{b (2 p+5)}-\frac {x^3 \left (a+b x^2\right )^{p+1} (5 a D-b C (2 p+7))}{b (2 p+5)}}{b (2 p+7)}+\frac {D x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {\frac {\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right ) \left (A b^3 \left (4 p^2+24 p+35\right )-\frac {a \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{2 p+3}\right )}{b}+\frac {x \left (a+b x^2\right )^{p+1} \left (15 a^2 D-3 a b C (2 p+7)+b^2 B \left (4 p^2+24 p+35\right )\right )}{b (2 p+3)}}{b (2 p+5)}-\frac {x^3 \left (a+b x^2\right )^{p+1} (5 a D-b C (2 p+7))}{b (2 p+5)}}{b (2 p+7)}+\frac {D x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

Input:

Int[(a + b*x^2)^p*(A + B*x^2 + C*x^4 + D*x^6),x]
 

Output:

(D*x^5*(a + b*x^2)^(1 + p))/(b*(7 + 2*p)) + (-(((5*a*D - b*C*(7 + 2*p))*x^ 
3*(a + b*x^2)^(1 + p))/(b*(5 + 2*p))) + (((15*a^2*D - 3*a*b*C*(7 + 2*p) + 
b^2*B*(35 + 24*p + 4*p^2))*x*(a + b*x^2)^(1 + p))/(b*(3 + 2*p)) + ((A*b^3* 
(35 + 24*p + 4*p^2) - (a*(15*a^2*D - 3*a*b*C*(7 + 2*p) + b^2*B*(35 + 24*p 
+ 4*p^2)))/(3 + 2*p))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b 
*x^2)/a)])/(b*(1 + (b*x^2)/a)^p))/(b*(5 + 2*p)))/(b*(7 + 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]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [F]

\[\int \left (b \,x^{2}+a \right )^{p} \left (D x^{6}+C \,x^{4}+x^{2} B +A \right )d x\]

Input:

int((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
 

Output:

int((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
 

Fricas [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x, algorithm="fricas")
 

Output:

integral((D*x^6 + 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 = 18.44 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.42 \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=A a^{p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {B a^{p} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {C a^{p} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {D a^{p} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - p \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \] Input:

integrate((b*x**2+a)**p*(D*x**6+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 + D*a**p*x**7*hyper((7/2, -p 
), (9/2,), b*x**2*exp_polar(I*pi)/a)/7
 

Maxima [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x, algorithm="maxima")
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^p, x)
 

Giac [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int {\left (b\,x^2+a\right )}^p\,\left (A+B\,x^2+C\,x^4+x^6\,D\right ) \,d x \] Input:

int((a + b*x^2)^p*(A + B*x^2 + C*x^4 + x^6*D),x)
 

Output:

int((a + b*x^2)^p*(A + B*x^2 + C*x^4 + x^6*D), x)
 

Reduce [F]

\[ \int \left (a+b x^2\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\text {too large to display} \] Input:

int((b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
 

Output:

(30*(a + b*x**2)**p*a**3*d*p*x - 12*(a + b*x**2)**p*a**2*b*c*p**2*x - 42*( 
a + b*x**2)**p*a**2*b*c*p*x - 20*(a + b*x**2)**p*a**2*b*d*p**2*x**3 - 10*( 
a + b*x**2)**p*a**2*b*d*p*x**3 + 16*(a + b*x**2)**p*a*b**3*p**3*x + 108*(a 
 + b*x**2)**p*a*b**3*p**2*x + 212*(a + b*x**2)**p*a*b**3*p*x + 105*(a + b* 
x**2)**p*a*b**3*x + 8*(a + b*x**2)**p*a*b**2*c*p**3*x**3 + 32*(a + b*x**2) 
**p*a*b**2*c*p**2*x**3 + 14*(a + b*x**2)**p*a*b**2*c*p*x**3 + 8*(a + b*x** 
2)**p*a*b**2*d*p**3*x**5 + 16*(a + b*x**2)**p*a*b**2*d*p**2*x**5 + 6*(a + 
b*x**2)**p*a*b**2*d*p*x**5 + 8*(a + b*x**2)**p*b**4*p**3*x**3 + 52*(a + b* 
x**2)**p*b**4*p**2*x**3 + 94*(a + b*x**2)**p*b**4*p*x**3 + 35*(a + b*x**2) 
**p*b**4*x**3 + 8*(a + b*x**2)**p*b**3*c*p**3*x**5 + 44*(a + b*x**2)**p*b* 
*3*c*p**2*x**5 + 62*(a + b*x**2)**p*b**3*c*p*x**5 + 21*(a + b*x**2)**p*b** 
3*c*x**5 + 8*(a + b*x**2)**p*b**3*d*p**3*x**7 + 36*(a + b*x**2)**p*b**3*d* 
p**2*x**7 + 46*(a + b*x**2)**p*b**3*d*p*x**7 + 15*(a + b*x**2)**p*b**3*d*x 
**7 - 480*int((a + b*x**2)**p/(16*a*p**4 + 128*a*p**3 + 344*a*p**2 + 352*a 
*p + 105*a + 16*b*p**4*x**2 + 128*b*p**3*x**2 + 344*b*p**2*x**2 + 352*b*p* 
x**2 + 105*b*x**2),x)*a**4*d*p**5 - 3840*int((a + b*x**2)**p/(16*a*p**4 + 
128*a*p**3 + 344*a*p**2 + 352*a*p + 105*a + 16*b*p**4*x**2 + 128*b*p**3*x* 
*2 + 344*b*p**2*x**2 + 352*b*p*x**2 + 105*b*x**2),x)*a**4*d*p**4 - 10320*i 
nt((a + b*x**2)**p/(16*a*p**4 + 128*a*p**3 + 344*a*p**2 + 352*a*p + 105*a 
+ 16*b*p**4*x**2 + 128*b*p**3*x**2 + 344*b*p**2*x**2 + 352*b*p*x**2 + 1...