\(\int (a+c x^2+b x^4)^p (A+B x^2+C x^4) \, dx\) [20]

Optimal result

Integrand size = 27, antiderivative size = 353 \[ \int \left (a+c x^2+b x^4\right )^p \left (A+B x^2+C x^4\right ) \, dx=\frac {C x \left (a+c x^2+b x^4\right )^{1+p}}{b (5+4 p)}-\frac {(a C-A b (5+4 p)) x \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{b (5+4 p)}-\frac {(c C (3+2 p)-b B (5+4 p)) x^3 \left (1+\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (1+\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {-4 a b+c^2}},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )}{3 b (5+4 p)} \] Output:

C*x*(b*x^4+c*x^2+a)^(p+1)/b/(5+4*p)-(a*C-A*b*(5+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/(5+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*C*(3+2*p)-b*B*(5+4*p))*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/(5+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 = 0.72 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.84 \[ \int \left (a+c x^2+b x^4\right )^p \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{15} x \left (\frac {c-\sqrt {-4 a b+c^2}+2 b x^2}{c-\sqrt {-4 a b+c^2}}\right )^{-p} \left (\frac {c+\sqrt {-4 a b+c^2}+2 b x^2}{c+\sqrt {-4 a b+c^2}}\right )^{-p} \left (a+c x^2+b x^4\right )^p \left (15 A \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+5 B x^2 \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )+3 C x^4 \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},-\frac {2 b x^2}{c+\sqrt {-4 a b+c^2}},\frac {2 b x^2}{-c+\sqrt {-4 a b+c^2}}\right )\right ) \] Input:

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

Output:

(x*(a + c*x^2 + b*x^4)^p*(15*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])] + 5*B*x^2*Appell 
F1[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*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*((c - Sqrt 
[-4*a*b + c^2] + 2*b*x^2)/(c - Sqrt[-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 = 0.56 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2207, 25, 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.

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

\(\Big \downarrow \) 2207

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1515

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {C x \left (a+b x^4+c x^2\right )^{p+1}}{b (4 p+5)}-\frac {x (a C-A b (4 p+5)) \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )+\frac {1}{3} x^3 \left (\frac {2 b x^2}{c-\sqrt {c^2-4 a b}}+1\right )^{-p} \left (a+b x^4+c x^2\right )^p \left (\frac {2 b x^2}{\sqrt {c^2-4 a b}+c}+1\right )^{-p} (c C (2 p+3)-b B (4 p+5)) \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 b x^2}{c-\sqrt {c^2-4 a b}},-\frac {2 b x^2}{c+\sqrt {c^2-4 a b}}\right )}{b (4 p+5)}\)

Input:

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

Output:

(C*x*(a + c*x^2 + b*x^4)^(1 + p))/(b*(5 + 4*p)) - (((a*C - A*b*(5 + 4*p))* 
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 - S 
qrt[-4*a*b + c^2]))^p*(1 + (2*b*x^2)/(c + Sqrt[-4*a*b + c^2]))^p) + ((c*C* 
(3 + 2*p) - b*B*(5 + 4*p))*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 + S 
qrt[-4*a*b + c^2]))^p))/(b*(5 + 4*p))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(16*(a + b*x**4 + c*x**2)**p*a*b**2*p**2*x + 32*(a + b*x**4 + c*x**2)**p*a 
*b**2*p*x + 15*(a + b*x**4 + c*x**2)**p*a*b**2*x + 16*(a + b*x**4 + c*x**2 
)**p*a*b*c*p**2*x + 12*(a + b*x**4 + c*x**2)**p*a*b*c*p*x + 16*(a + b*x**4 
 + c*x**2)**p*b**3*p**2*x**3 + 24*(a + b*x**4 + c*x**2)**p*b**3*p*x**3 + 5 
*(a + b*x**4 + c*x**2)**p*b**3*x**3 + 16*(a + b*x**4 + c*x**2)**p*b**2*c*p 
**2*x**5 + 8*(a + b*x**4 + c*x**2)**p*b**2*c*p**2*x + 16*(a + b*x**4 + c*x 
**2)**p*b**2*c*p*x**5 + 10*(a + b*x**4 + c*x**2)**p*b**2*c*p*x + 3*(a + b* 
x**4 + c*x**2)**p*b**2*c*x**5 + 8*(a + b*x**4 + c*x**2)**p*b*c**2*p**2*x** 
3 + 2*(a + b*x**4 + c*x**2)**p*b*c**2*p*x**3 - 4*(a + b*x**4 + c*x**2)**p* 
c**3*p**2*x - 6*(a + b*x**4 + c*x**2)**p*c**3*p*x + 4096*int((a + b*x**4 + 
 c*x**2)**p/(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**4 + 144 
*b*p**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*c*p**3*x**2 + 144*c*p**2*x**2 
+ 92*c*p*x**2 + 15*c*x**2),x)*a**2*b**2*p**6 + 17408*int((a + b*x**4 + c*x 
**2)**p/(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**4 + 144*b*p 
**2*x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*c*p**3*x**2 + 144*c*p**2*x**2 + 92 
*c*p*x**2 + 15*c*x**2),x)*a**2*b**2*p**5 + 28160*int((a + b*x**4 + c*x**2) 
**p/(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**4 + 144*b*p**2* 
x**4 + 92*b*p*x**4 + 15*b*x**4 + 64*c*p**3*x**2 + 144*c*p**2*x**2 + 92*c*p 
*x**2 + 15*c*x**2),x)*a**2*b**2*p**4 + 21376*int((a + b*x**4 + c*x**2)**p/ 
(64*a*p**3 + 144*a*p**2 + 92*a*p + 15*a + 64*b*p**3*x**4 + 144*b*p**2*x...