Integrand size = 34, antiderivative size = 529 \[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=-\frac {\left (3 c \left (35 C d^3-d e (9+2 q) (5 B d-A e (7+2 q))\right )+e (9+2 q) \left (a e (7+2 q) (3 C d-B e (5+2 q))-b \left (15 C d^2-e (7+2 q) (3 B d-A e (5+2 q))\right )\right )\right ) x \left (d+e x^2\right )^{1+q}}{e^4 (3+2 q) (5+2 q) (7+2 q) (9+2 q)}-\frac {\left (e (9+2 q) (5 b C d-b B e (7+2 q)-a C e (7+2 q))-c \left (35 C d^2-e (9+2 q) (5 B d-A e (7+2 q))\right )\right ) x^3 \left (d+e x^2\right )^{1+q}}{e^3 (5+2 q) (7+2 q) (9+2 q)}-\frac {(7 c C d-B c e (9+2 q)-b C e (9+2 q)) x^5 \left (d+e x^2\right )^{1+q}}{e^2 (7+2 q) (9+2 q)}+\frac {c C x^7 \left (d+e x^2\right )^{1+q}}{e (9+2 q)}+\frac {\left (a A e^4 \left (315+286 q+84 q^2+8 q^3\right )-\frac {d \left (e (9+2 q) \left (15 b C d^2-b e (7+2 q) (3 B d-A e (5+2 q))-a e (7+2 q) (3 C d-B e (5+2 q))\right )-3 c d \left (35 C d^2-e (9+2 q) (5 B d-A e (7+2 q))\right )\right )}{3+2 q}\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{e^4 (5+2 q) (7+2 q) (9+2 q)} \] Output:
-(3*c*(35*C*d^3-d*e*(9+2*q)*(5*B*d-A*e*(7+2*q)))+e*(9+2*q)*(a*e*(7+2*q)*(3 *C*d-B*e*(5+2*q))-b*(15*C*d^2-e*(7+2*q)*(3*B*d-A*e*(5+2*q)))))*x*(e*x^2+d) ^(1+q)/e^4/(3+2*q)/(5+2*q)/(7+2*q)/(9+2*q)-(e*(9+2*q)*(5*C*b*d-b*B*e*(7+2* q)-a*C*e*(7+2*q))-c*(35*C*d^2-e*(9+2*q)*(5*B*d-A*e*(7+2*q))))*x^3*(e*x^2+d )^(1+q)/e^3/(5+2*q)/(7+2*q)/(9+2*q)-(7*C*c*d-B*c*e*(9+2*q)-b*C*e*(9+2*q))* x^5*(e*x^2+d)^(1+q)/e^2/(7+2*q)/(9+2*q)+c*C*x^7*(e*x^2+d)^(1+q)/e/(9+2*q)+ (a*A*e^4*(8*q^3+84*q^2+286*q+315)-d*(e*(9+2*q)*(15*b*C*d^2-b*e*(7+2*q)*(3* B*d-A*e*(5+2*q))-a*e*(7+2*q)*(3*C*d-B*e*(5+2*q)))-3*c*d*(35*C*d^2-e*(9+2*q )*(5*B*d-A*e*(7+2*q))))/(3+2*q))*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],- e*x^2/d)/e^4/(5+2*q)/(7+2*q)/(9+2*q)/((1+e*x^2/d)^q)
Time = 0.67 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.33 \[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\frac {1}{315} x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \left (315 a A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )+105 (A b+a B) x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )+63 (b B+A c+a C) x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-q,\frac {7}{2},-\frac {e x^2}{d}\right )+45 (B c+b C) x^6 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-q,\frac {9}{2},-\frac {e x^2}{d}\right )+35 c C x^8 \operatorname {Hypergeometric2F1}\left (\frac {9}{2},-q,\frac {11}{2},-\frac {e x^2}{d}\right )\right ) \] Input:
Integrate[(d + e*x^2)^q*(a + b*x^2 + c*x^4)*(A + B*x^2 + C*x^4),x]
Output:
(x*(d + e*x^2)^q*(315*a*A*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)] + 105*(A*b + a*B)*x^2*Hypergeometric2F1[3/2, -q, 5/2, -((e*x^2)/d)] + 63*(b* B + A*c + a*C)*x^4*Hypergeometric2F1[5/2, -q, 7/2, -((e*x^2)/d)] + 45*(B*c + b*C)*x^6*Hypergeometric2F1[7/2, -q, 9/2, -((e*x^2)/d)] + 35*c*C*x^8*Hyp ergeometric2F1[9/2, -q, 11/2, -((e*x^2)/d)]))/(315*(1 + (e*x^2)/d)^q)
Time = 0.44 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.51, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2256, 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.
\(\displaystyle \int \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \left (d+e x^2\right )^q \, dx\) |
\(\Big \downarrow \) 2256 |
\(\displaystyle \int \left (x^4 \left (d+e x^2\right )^q (a C+A c+b B)+x^2 (a B+A b) \left (d+e x^2\right )^q+a A \left (d+e x^2\right )^q+x^6 (b C+B c) \left (d+e x^2\right )^q+c C x^8 \left (d+e x^2\right )^q\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} x^5 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-q,\frac {7}{2},-\frac {e x^2}{d}\right ) (a C+A c+b B)+\frac {1}{3} x^3 (a B+A b) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-q,\frac {5}{2},-\frac {e x^2}{d}\right )+a A x \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )+\frac {1}{7} x^7 (b C+B c) \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-q,\frac {9}{2},-\frac {e x^2}{d}\right )+\frac {1}{9} c C x^9 \left (d+e x^2\right )^q \left (\frac {e x^2}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {9}{2},-q,\frac {11}{2},-\frac {e x^2}{d}\right )\) |
Input:
Int[(d + e*x^2)^q*(a + b*x^2 + c*x^4)*(A + B*x^2 + C*x^4),x]
Output:
(a*A*x*(d + e*x^2)^q*Hypergeometric2F1[1/2, -q, 3/2, -((e*x^2)/d)])/(1 + ( e*x^2)/d)^q + ((A*b + a*B)*x^3*(d + e*x^2)^q*Hypergeometric2F1[3/2, -q, 5/ 2, -((e*x^2)/d)])/(3*(1 + (e*x^2)/d)^q) + ((b*B + A*c + a*C)*x^5*(d + e*x^ 2)^q*Hypergeometric2F1[5/2, -q, 7/2, -((e*x^2)/d)])/(5*(1 + (e*x^2)/d)^q) + ((B*c + b*C)*x^7*(d + e*x^2)^q*Hypergeometric2F1[7/2, -q, 9/2, -((e*x^2) /d)])/(7*(1 + (e*x^2)/d)^q) + (c*C*x^9*(d + e*x^2)^q*Hypergeometric2F1[9/2 , -q, 11/2, -((e*x^2)/d)])/(9*(1 + (e*x^2)/d)^q)
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4 )^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
\[\int \left (e \,x^{2}+d \right )^{q} \left (c \,x^{4}+b \,x^{2}+a \right ) \left (C \,x^{4}+B \,x^{2}+A \right )d x\]
Input:
int((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x)
Output:
int((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x)
\[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x, algorithm="fricas ")
Output:
integral((C*c*x^8 + (C*b + B*c)*x^6 + (C*a + B*b + A*c)*x^4 + (B*a + A*b)* x^2 + A*a)*(e*x^2 + d)^q, x)
Result contains complex when optimal does not.
Time = 60.59 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.51 \[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=A a d^{q} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - q \\ \frac {3}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )} + \frac {A b d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{3} + \frac {A c d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5} + \frac {B a d^{q} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - q \\ \frac {5}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{3} + \frac {B b d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5} + \frac {B c d^{q} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - q \\ \frac {9}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{7} + \frac {C a d^{q} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - q \\ \frac {7}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{5} + \frac {C b d^{q} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - q \\ \frac {9}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{7} + \frac {C c d^{q} x^{9} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{2}, - q \\ \frac {11}{2} \end {matrix}\middle | {\frac {e x^{2} e^{i \pi }}{d}} \right )}}{9} \] Input:
integrate((e*x**2+d)**q*(c*x**4+b*x**2+a)*(C*x**4+B*x**2+A),x)
Output:
A*a*d**q*x*hyper((1/2, -q), (3/2,), e*x**2*exp_polar(I*pi)/d) + A*b*d**q*x **3*hyper((3/2, -q), (5/2,), e*x**2*exp_polar(I*pi)/d)/3 + A*c*d**q*x**5*h yper((5/2, -q), (7/2,), e*x**2*exp_polar(I*pi)/d)/5 + B*a*d**q*x**3*hyper( (3/2, -q), (5/2,), e*x**2*exp_polar(I*pi)/d)/3 + B*b*d**q*x**5*hyper((5/2, -q), (7/2,), e*x**2*exp_polar(I*pi)/d)/5 + B*c*d**q*x**7*hyper((7/2, -q), (9/2,), e*x**2*exp_polar(I*pi)/d)/7 + C*a*d**q*x**5*hyper((5/2, -q), (7/2 ,), e*x**2*exp_polar(I*pi)/d)/5 + C*b*d**q*x**7*hyper((7/2, -q), (9/2,), e *x**2*exp_polar(I*pi)/d)/7 + C*c*d**q*x**9*hyper((9/2, -q), (11/2,), e*x** 2*exp_polar(I*pi)/d)/9
\[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x, algorithm="maxima ")
Output:
integrate((C*x^4 + B*x^2 + A)*(c*x^4 + b*x^2 + a)*(e*x^2 + d)^q, x)
\[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\int { {\left (C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )} {\left (e x^{2} + d\right )}^{q} \,d x } \] Input:
integrate((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x, algorithm="giac")
Output:
integrate((C*x^4 + B*x^2 + A)*(c*x^4 + b*x^2 + a)*(e*x^2 + d)^q, x)
Timed out. \[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\int {\left (e\,x^2+d\right )}^q\,\left (C\,x^4+B\,x^2+A\right )\,\left (c\,x^4+b\,x^2+a\right ) \,d x \] Input:
int((d + e*x^2)^q*(A + B*x^2 + C*x^4)*(a + b*x^2 + c*x^4),x)
Output:
int((d + e*x^2)^q*(A + B*x^2 + C*x^4)*(a + b*x^2 + c*x^4), x)
\[ \int \left (d+e x^2\right )^q \left (a+b x^2+c x^4\right ) \left (A+B x^2+C x^4\right ) \, dx=\text {too large to display} \] Input:
int((e*x^2+d)^q*(c*x^4+b*x^2+a)*(C*x^4+B*x^2+A),x)
Output:
(16*(d + e*x**2)**q*a**2*e**4*q**4*x + 192*(d + e*x**2)**q*a**2*e**4*q**3* x + 824*(d + e*x**2)**q*a**2*e**4*q**2*x + 1488*(d + e*x**2)**q*a**2*e**4* q*x + 945*(d + e*x**2)**q*a**2*e**4*x + 32*(d + e*x**2)**q*a*b*d*e**3*q**4 *x + 336*(d + e*x**2)**q*a*b*d*e**3*q**3*x + 1144*(d + e*x**2)**q*a*b*d*e* *3*q**2*x + 1260*(d + e*x**2)**q*a*b*d*e**3*q*x + 32*(d + e*x**2)**q*a*b*e **4*q**4*x**3 + 352*(d + e*x**2)**q*a*b*e**4*q**3*x**3 + 1312*(d + e*x**2) **q*a*b*e**4*q**2*x**3 + 1832*(d + e*x**2)**q*a*b*e**4*q*x**3 + 630*(d + e *x**2)**q*a*b*e**4*x**3 - 48*(d + e*x**2)**q*a*c*d**2*e**2*q**3*x - 384*(d + e*x**2)**q*a*c*d**2*e**2*q**2*x - 756*(d + e*x**2)**q*a*c*d**2*e**2*q*x + 32*(d + e*x**2)**q*a*c*d*e**3*q**4*x**3 + 272*(d + e*x**2)**q*a*c*d*e** 3*q**3*x**3 + 632*(d + e*x**2)**q*a*c*d*e**3*q**2*x**3 + 252*(d + e*x**2)* *q*a*c*d*e**3*q*x**3 + 32*(d + e*x**2)**q*a*c*e**4*q**4*x**5 + 320*(d + e* x**2)**q*a*c*e**4*q**3*x**5 + 1040*(d + e*x**2)**q*a*c*e**4*q**2*x**5 + 12 00*(d + e*x**2)**q*a*c*e**4*q*x**5 + 378*(d + e*x**2)**q*a*c*e**4*x**5 - 2 4*(d + e*x**2)**q*b**2*d**2*e**2*q**3*x - 192*(d + e*x**2)**q*b**2*d**2*e* *2*q**2*x - 378*(d + e*x**2)**q*b**2*d**2*e**2*q*x + 16*(d + e*x**2)**q*b* *2*d*e**3*q**4*x**3 + 136*(d + e*x**2)**q*b**2*d*e**3*q**3*x**3 + 316*(d + e*x**2)**q*b**2*d*e**3*q**2*x**3 + 126*(d + e*x**2)**q*b**2*d*e**3*q*x**3 + 16*(d + e*x**2)**q*b**2*e**4*q**4*x**5 + 160*(d + e*x**2)**q*b**2*e**4* q**3*x**5 + 520*(d + e*x**2)**q*b**2*e**4*q**2*x**5 + 600*(d + e*x**2)*...