\(\int (a+c x^2+b x^4)^p (A+B x^2+C x^4+D x^6) \, dx\) [21]

Optimal result

Integrand size = 32, antiderivative size = 484 \[ \int \left (a+c x^2+b x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=-\frac {(c D (5+2 p)-b C (7+4 p)) x \left (a+c x^2+b x^4\right )^{1+p}}{b^2 (5+4 p) (7+4 p)}+\frac {D x^3 \left (a+c x^2+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\left (A b^2 \left (35+48 p+16 p^2\right )+a (c D (5+2 p)-b C (7+4 p))\right ) x \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{b^2 (5+4 p) (7+4 p)}+\frac {\left (c^2 D \left (15+16 p+4 p^2\right )+b^2 B \left (35+48 p+16 p^2\right )-b \left (3 a D (5+4 p)+c C \left (21+26 p+8 p^2\right )\right )\right ) x^3 \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{3 b^2 (5+4 p) (7+4 p)} \] Output:

-(c*D*(5+2*p)-b*C*(7+4*p))*x*(b*x^4+c*x^2+a)^(p+1)/b^2/(5+4*p)/(7+4*p)+D*x 
^3*(b*x^4+c*x^2+a)^(p+1)/b/(7+4*p)+(A*b^2*(16*p^2+48*p+35)+a*(c*D*(5+2*p)- 
b*C*(7+4*p)))*x*(b*x^4+c*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,-2*b*x^2/(c-(-4*a 
*b+c^2)^(1/2)),-2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))/b^2/(5+4*p)/(7+4*p)/((1+2* 
b*x^2/(c-(-4*a*b+c^2)^(1/2)))^p)/((1+2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))^p)+1/ 
3*(c^2*D*(4*p^2+16*p+15)+b^2*B*(16*p^2+48*p+35)-b*(3*a*D*(5+4*p)+c*C*(8*p^ 
2+26*p+21)))*x^3*(b*x^4+c*x^2+a)^p*AppellF1(3/2,-p,-p,5/2,-2*b*x^2/(c-(-4* 
a*b+c^2)^(1/2)),-2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))/b^2/(5+4*p)/(7+4*p)/((1+2 
*b*x^2/(c-(-4*a*b+c^2)^(1/2)))^p)/((1+2*b*x^2/(c+(-4*a*b+c^2)^(1/2)))^p)
 

Mathematica [A] (warning: unable to verify)

Time = 1.03 (sec) , antiderivative size = 363, normalized size of antiderivative = 0.75 \[ \int \left (a+c x^2+b x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{105} x \left (\frac {c-\sqrt {-4 a b+c^2}+2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (\frac {c+\sqrt {-4 a b+c^2}+2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \left (105 A \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+35 B x^2 \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+21 C x^4 \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+15 D x^6 \operatorname {AppellF1}\left (\frac {7}{2},-p,-p,\frac {9}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )\right ) \] Input:

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

Output:

(x*(a + c*x^2 + b*x^4)^p*(105*A*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c + 
 Sqrt[-4*a*b + c^2]), (2*b*x^2)/(-c + Sqrt[-4*a*b + c^2])] + 35*B*x^2*Appe 
llF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (2*b*x^2)/(-c 
+ Sqrt[-4*a*b + c^2])] + 21*C*x^4*AppellF1[5/2, -p, -p, 7/2, (-2*b*x^2)/(c 
 + Sqrt[-4*a*b + c^2]), (2*b*x^2)/(-c + Sqrt[-4*a*b + c^2])] + 15*D*x^6*Ap 
pellF1[7/2, -p, -p, 9/2, (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2]), (2*b*x^2)/(- 
c + Sqrt[-4*a*b + c^2])]))/(105*((c - Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c - S 
qrt[-4*a*b + c^2]))^p*((c + Sqrt[-4*a*b + c^2] + 2*b*x^2)/(c + Sqrt[-4*a*b 
 + c^2]))^p)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 466, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2207, 2207, 1515, 2009}

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.

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

\(\Big \downarrow \) 2207

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

\(\Big \downarrow \) 2207

\(\displaystyle \frac {\frac {\int \left (A \left (16 p^2+48 p+35\right ) b^2-a C (4 p+7) b+\left (B \left (16 p^2+48 p+35\right ) b^2-\left (3 a D (4 p+5)+c C \left (8 p^2+26 p+21\right )\right ) b+c^2 D \left (4 p^2+16 p+15\right )\right ) x^2+a c D (2 p+5)\right ) \left (b x^4+c x^2+a\right )^pdx}{b (4 p+5)}-\frac {x \left (a+b x^4+c x^2\right )^{p+1} (c D (2 p+5)-b C (4 p+7))}{b (4 p+5)}}{b (4 p+7)}+\frac {D x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 1515

\(\displaystyle \frac {\frac {\int \left (\left (B \left (16 p^2+48 p+35\right ) b^2-\left (3 a D (4 p+5)+c C \left (8 p^2+26 p+21\right )\right ) b+c^2 D \left (4 p^2+16 p+15\right )\right ) x^2 \left (b x^4+c x^2+a\right )^p+a c D (2 p+5) \left (\frac {b (4 p+7) (A b (4 p+5)-a C)}{a c D (2 p+5)}+1\right ) \left (b x^4+c x^2+a\right )^p\right )dx}{b (4 p+5)}-\frac {x \left (a+b x^4+c x^2\right )^{p+1} (c D (2 p+5)-b C (4 p+7))}{b (4 p+5)}}{b (4 p+7)}+\frac {D x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {x \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right ) \left (-a b C (4 p+7)+a c D (2 p+5)+A b^2 \left (16 p^2+48 p+35\right )\right )+\frac {1}{3} x^3 \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right ) \left (-b \left (3 a D (4 p+5)+c C \left (8 p^2+26 p+21\right )\right )+b^2 B \left (16 p^2+48 p+35\right )+c^2 D \left (4 p^2+16 p+15\right )\right )}{b (4 p+5)}-\frac {x \left (a+b x^4+c x^2\right )^{p+1} (c D (2 p+5)-b C (4 p+7))}{b (4 p+5)}}{b (4 p+7)}+\frac {D x^3 \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+7)}\)

Input:

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

Output:

(D*x^3*(a + c*x^2 + b*x^4)^(1 + p))/(b*(7 + 4*p)) + (-(((c*D*(5 + 2*p) - b 
*C*(7 + 4*p))*x*(a + c*x^2 + b*x^4)^(1 + p))/(b*(5 + 4*p))) + (((a*c*D*(5 
+ 2*p) - a*b*C*(7 + 4*p) + A*b^2*(35 + 48*p + 16*p^2))*x*(a + c*x^2 + b*x^ 
4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*b*x^2)/(c - Sqrt[-4*a*b + c^2]), (-2*b 
*x^2)/(c + Sqrt[-4*a*b + c^2])])/((1 + (2*b*x^2)/(c - Sqrt[-4*a*b + c^2])) 
^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p) + ((c^2*D*(15 + 16*p + 4*p^ 
2) + b^2*B*(35 + 48*p + 16*p^2) - b*(3*a*D*(5 + 4*p) + c*C*(21 + 26*p + 8* 
p^2)))*x^3*(a + c*x^2 + b*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*b*x^2)/(c 
- Sqrt[-4*a*b + c^2]), (-2*b*x^2)/(c + Sqrt[-4*a*b + c^2])])/(3*(1 + (2*b* 
x^2)/(c - Sqrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^ 
p))/(b*(5 + 4*p)))/(b*(7 + 4*p))
 

Defintions of rubi rules used

rule 1515

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2207

Int[(Px_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{n = 
 Expon[Px, x^2], e = Coeff[Px, x^2, Expon[Px, x^2]]}, Simp[e*x^(2*n - 3)*(( 
a + b*x^2 + c*x^4)^(p + 1)/(c*(2*n + 4*p + 1))), x] + Simp[1/(c*(2*n + 4*p 
+ 1))   Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*n + 4*p + 1)*Px - a*e*(2 
*n - 3)*x^(2*n - 4) - b*e*(2*n + 2*p - 1)*x^(2*n - 2) - c*e*(2*n + 4*p + 1) 
*x^(2*n), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Px, x^2] && Expon[ 
Px, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate((b*x^4+c*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^4 + c*x^2 + a)^p, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((b*x^4+c*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^4 + c*x^2 + a)^p, x)
 

Giac [F]

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

integrate((b*x^4+c*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^4 + c*x^2 + a)^p, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(64*(a + b*x**4 + c*x**2)**p*a*b**3*p**3*x + 240*(a + b*x**4 + c*x**2)**p* 
a*b**3*p**2*x + 284*(a + b*x**4 + c*x**2)**p*a*b**3*p*x + 105*(a + b*x**4 
+ c*x**2)**p*a*b**3*x + 64*(a + b*x**4 + c*x**2)**p*a*b**2*c*p**3*x + 160* 
(a + b*x**4 + c*x**2)**p*a*b**2*c*p**2*x + 84*(a + b*x**4 + c*x**2)**p*a*b 
**2*c*p*x + 64*(a + b*x**4 + c*x**2)**p*a*b**2*d*p**3*x**3 + 96*(a + b*x** 
4 + c*x**2)**p*a*b**2*d*p**2*x**3 + 20*(a + b*x**4 + c*x**2)**p*a*b**2*d*p 
*x**3 - 32*(a + b*x**4 + c*x**2)**p*a*b*c*d*p**3*x - 128*(a + b*x**4 + c*x 
**2)**p*a*b*c*d*p**2*x - 90*(a + b*x**4 + c*x**2)**p*a*b*c*d*p*x + 64*(a + 
 b*x**4 + c*x**2)**p*b**4*p**3*x**3 + 208*(a + b*x**4 + c*x**2)**p*b**4*p* 
*2*x**3 + 188*(a + b*x**4 + c*x**2)**p*b**4*p*x**3 + 35*(a + b*x**4 + c*x* 
*2)**p*b**4*x**3 + 64*(a + b*x**4 + c*x**2)**p*b**3*c*p**3*x**5 + 32*(a + 
b*x**4 + c*x**2)**p*b**3*c*p**3*x + 176*(a + b*x**4 + c*x**2)**p*b**3*c*p* 
*2*x**5 + 96*(a + b*x**4 + c*x**2)**p*b**3*c*p**2*x + 124*(a + b*x**4 + c* 
x**2)**p*b**3*c*p*x**5 + 70*(a + b*x**4 + c*x**2)**p*b**3*c*p*x + 21*(a + 
b*x**4 + c*x**2)**p*b**3*c*x**5 + 64*(a + b*x**4 + c*x**2)**p*b**3*d*p**3* 
x**7 + 144*(a + b*x**4 + c*x**2)**p*b**3*d*p**2*x**7 + 92*(a + b*x**4 + c* 
x**2)**p*b**3*d*p*x**7 + 15*(a + b*x**4 + c*x**2)**p*b**3*d*x**7 + 32*(a + 
 b*x**4 + c*x**2)**p*b**2*c**2*p**3*x**3 + 64*(a + b*x**4 + c*x**2)**p*b** 
2*c**2*p**2*x**3 + 14*(a + b*x**4 + c*x**2)**p*b**2*c**2*p*x**3 + 32*(a + 
b*x**4 + c*x**2)**p*b**2*c*d*p**3*x**5 + 32*(a + b*x**4 + c*x**2)**p*b*...