Integrand size = 32, antiderivative size = 484 \[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=-\frac {(b D (5+2 p)-c C (7+4 p)) x \left (a+b x^2+c x^4\right )^{1+p}}{c^2 (5+4 p) (7+4 p)}+\frac {D x^3 \left (a+b x^2+c x^4\right )^{1+p}}{c (7+4 p)}+\frac {\left (A c^2 \left (35+48 p+16 p^2\right )+a (b D (5+2 p)-c C (7+4 p))\right ) x \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{c^2 (5+4 p) (7+4 p)}-\frac {\left (3 a c D (5+4 p)-b^2 D \left (15+16 p+4 p^2\right )+b c C \left (21+26 p+8 p^2\right )-B c^2 \left (35+48 p+16 p^2\right )\right ) x^3 \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 c^2 (5+4 p) (7+4 p)} \] Output:
-(b*D*(5+2*p)-c*C*(7+4*p))*x*(c*x^4+b*x^2+a)^(p+1)/c^2/(5+4*p)/(7+4*p)+D*x ^3*(c*x^4+b*x^2+a)^(p+1)/c/(7+4*p)+(A*c^2*(16*p^2+48*p+35)+a*(b*D*(5+2*p)- c*C*(7+4*p)))*x*(c*x^4+b*x^2+a)^p*AppellF1(1/2,-p,-p,3/2,-2*c*x^2/(b-(-4*a *c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))/c^2/(5+4*p)/(7+4*p)/((1+2* c*x^2/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^p)-1/ 3*(3*a*c*D*(5+4*p)-b^2*D*(4*p^2+16*p+15)+b*c*C*(8*p^2+26*p+21)-B*c^2*(16*p ^2+48*p+35))*x^3*(c*x^4+b*x^2+a)^p*AppellF1(3/2,-p,-p,5/2,-2*c*x^2/(b-(-4* a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))/c^2/(5+4*p)/(7+4*p)/((1+2 *c*x^2/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^p)
Time = 1.03 (sec) , antiderivative size = 363, normalized size of antiderivative = 0.75 \[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{105} x \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \left (105 A \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+35 B x^2 \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+21 C x^4 \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+15 D x^6 \operatorname {AppellF1}\left (\frac {7}{2},-p,-p,\frac {9}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right ) \] Input:
Integrate[(a + b*x^2 + c*x^4)^p*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
(x*(a + b*x^2 + c*x^4)^p*(105*A*AppellF1[1/2, -p, -p, 3/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 35*B*x^2*Appell F1[3/2, -p, -p, 5/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + S qrt[b^2 - 4*a*c])] + 21*C*x^4*AppellF1[5/2, -p, -p, 7/2, (-2*c*x^2)/(b + S qrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + 15*D*x^6*AppellF1 [7/2, -p, -p, 9/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqr t[b^2 - 4*a*c])]))/(105*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 0.88 (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.
\(\displaystyle \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx\) |
\(\Big \downarrow \) 2207 |
\(\displaystyle \frac {\int \left (c x^4+b x^2+a\right )^p \left (-\left ((b D (2 p+5)-c C (4 p+7)) x^4\right )-(3 a D-B c (4 p+7)) x^2+A c (4 p+7)\right )dx}{c (4 p+7)}+\frac {D x^3 \left (a+b x^2+c x^4\right )^{p+1}}{c (4 p+7)}\) |
\(\Big \downarrow \) 2207 |
\(\displaystyle \frac {\frac {\int \left (A \left (16 p^2+48 p+35\right ) c^2-a C (4 p+7) c-\left (-D \left (4 p^2+16 p+15\right ) b^2+c C \left (8 p^2+26 p+21\right ) b+3 a c D (4 p+5)-B c^2 \left (16 p^2+48 p+35\right )\right ) x^2+a b D (2 p+5)\right ) \left (c x^4+b x^2+a\right )^pdx}{c (4 p+5)}-\frac {x \left (a+b x^2+c x^4\right )^{p+1} (b D (2 p+5)-c C (4 p+7))}{c (4 p+5)}}{c (4 p+7)}+\frac {D x^3 \left (a+b x^2+c x^4\right )^{p+1}}{c (4 p+7)}\) |
\(\Big \downarrow \) 1515 |
\(\displaystyle \frac {\frac {\int \left (\left (D \left (4 p^2+16 p+15\right ) b^2-c C \left (8 p^2+26 p+21\right ) b-3 a c D (4 p+5)+B c^2 \left (16 p^2+48 p+35\right )\right ) x^2 \left (c x^4+b x^2+a\right )^p+a b D (2 p+5) \left (\frac {c (4 p+7) (A c (4 p+5)-a C)}{a b D (2 p+5)}+1\right ) \left (c x^4+b x^2+a\right )^p\right )dx}{c (4 p+5)}-\frac {x \left (a+b x^2+c x^4\right )^{p+1} (b D (2 p+5)-c C (4 p+7))}{c (4 p+5)}}{c (4 p+7)}+\frac {D x^3 \left (a+b x^2+c x^4\right )^{p+1}}{c (4 p+7)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {x \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right ) \left (a b D (2 p+5)-a c C (4 p+7)+A c^2 \left (16 p^2+48 p+35\right )\right )-\frac {1}{3} x^3 \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right ) \left (3 a c D (4 p+5)+b^2 (-D) \left (4 p^2+16 p+15\right )+b c C \left (8 p^2+26 p+21\right )-B c^2 \left (16 p^2+48 p+35\right )\right )}{c (4 p+5)}-\frac {x \left (a+b x^2+c x^4\right )^{p+1} (b D (2 p+5)-c C (4 p+7))}{c (4 p+5)}}{c (4 p+7)}+\frac {D x^3 \left (a+b x^2+c x^4\right )^{p+1}}{c (4 p+7)}\) |
Input:
Int[(a + b*x^2 + c*x^4)^p*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
(D*x^3*(a + b*x^2 + c*x^4)^(1 + p))/(c*(7 + 4*p)) + (-(((b*D*(5 + 2*p) - c *C*(7 + 4*p))*x*(a + b*x^2 + c*x^4)^(1 + p))/(c*(5 + 4*p))) + (((a*b*D*(5 + 2*p) - a*c*C*(7 + 4*p) + A*c^2*(35 + 48*p + 16*p^2))*x*(a + b*x^2 + c*x^ 4)^p*AppellF1[1/2, -p, -p, 3/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c* x^2)/(b + Sqrt[b^2 - 4*a*c])])/((1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p* (1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p) - ((3*a*c*D*(5 + 4*p) - b^2*D*( 15 + 16*p + 4*p^2) + b*c*C*(21 + 26*p + 8*p^2) - B*c^2*(35 + 48*p + 16*p^2 ))*x^3*(a + b*x^2 + c*x^4)^p*AppellF1[3/2, -p, -p, 5/2, (-2*c*x^2)/(b - Sq rt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*(1 + (2*c*x^2)/( b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p))/(c*( 5 + 4*p)))/(c*(7 + 4*p))
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]
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]
\[\int \left (c \,x^{4}+b \,x^{2}+a \right )^{p} \left (D x^{6}+C \,x^{4}+B \,x^{2}+A \right )d x\]
Input:
int((c*x^4+b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
Output:
int((c*x^4+b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
\[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+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)*(c*x^4 + b*x^2 + a)^p, x)
Timed out. \[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\text {Timed out} \] Input:
integrate((c*x**4+b*x**2+a)**p*(D*x**6+C*x**4+B*x**2+A),x)
Output:
Timed out
\[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+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)*(c*x^4 + b*x^2 + a)^p, x)
\[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int { {\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (c x^{4} + b x^{2} + a\right )}^{p} \,d x } \] Input:
integrate((c*x^4+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)*(c*x^4 + b*x^2 + a)^p, x)
Timed out. \[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int {\left (c\,x^4+b\,x^2+a\right )}^p\,\left (A+B\,x^2+C\,x^4+x^6\,D\right ) \,d x \] Input:
int((a + b*x^2 + c*x^4)^p*(A + B*x^2 + C*x^4 + x^6*D),x)
Output:
int((a + b*x^2 + c*x^4)^p*(A + B*x^2 + C*x^4 + x^6*D), x)
\[ \int \left (a+b x^2+c x^4\right )^p \left (A+B x^2+C x^4+D x^6\right ) \, dx=\text {too large to display} \] Input:
int((c*x^4+b*x^2+a)^p*(D*x^6+C*x^4+B*x^2+A),x)
Output:
( - 32*(a + b*x**2 + c*x**4)**p*a*b*c*d*p**3*x - 128*(a + b*x**2 + c*x**4) **p*a*b*c*d*p**2*x - 90*(a + b*x**2 + c*x**4)**p*a*b*c*d*p*x + 128*(a + b* x**2 + c*x**4)**p*a*c**3*p**3*x + 400*(a + b*x**2 + c*x**4)**p*a*c**3*p**2 *x + 368*(a + b*x**2 + c*x**4)**p*a*c**3*p*x + 105*(a + b*x**2 + c*x**4)** p*a*c**3*x + 64*(a + b*x**2 + c*x**4)**p*a*c**2*d*p**3*x**3 + 96*(a + b*x* *2 + c*x**4)**p*a*c**2*d*p**2*x**3 + 20*(a + b*x**2 + c*x**4)**p*a*c**2*d* p*x**3 + 8*(a + b*x**2 + c*x**4)**p*b**3*d*p**3*x + 32*(a + b*x**2 + c*x** 4)**p*b**3*d*p**2*x + 30*(a + b*x**2 + c*x**4)**p*b**3*d*p*x + 16*(a + b*x **2 + c*x**4)**p*b**2*c**2*p**3*x + 44*(a + b*x**2 + c*x**4)**p*b**2*c**2* p**2*x + 28*(a + b*x**2 + c*x**4)**p*b**2*c**2*p*x - 16*(a + b*x**2 + c*x* *4)**p*b**2*c*d*p**3*x**3 - 44*(a + b*x**2 + c*x**4)**p*b**2*c*d*p**2*x**3 - 10*(a + b*x**2 + c*x**4)**p*b**2*c*d*p*x**3 + 96*(a + b*x**2 + c*x**4)* *p*b*c**3*p**3*x**3 + 272*(a + b*x**2 + c*x**4)**p*b*c**3*p**2*x**3 + 202* (a + b*x**2 + c*x**4)**p*b*c**3*p*x**3 + 35*(a + b*x**2 + c*x**4)**p*b*c** 3*x**3 + 32*(a + b*x**2 + c*x**4)**p*b*c**2*d*p**3*x**5 + 32*(a + b*x**2 + c*x**4)**p*b*c**2*d*p**2*x**5 + 6*(a + b*x**2 + c*x**4)**p*b*c**2*d*p*x** 5 + 64*(a + b*x**2 + c*x**4)**p*c**4*p**3*x**5 + 176*(a + b*x**2 + c*x**4) **p*c**4*p**2*x**5 + 124*(a + b*x**2 + c*x**4)**p*c**4*p*x**5 + 21*(a + b* x**2 + c*x**4)**p*c**4*x**5 + 64*(a + b*x**2 + c*x**4)**p*c**3*d*p**3*x**7 + 144*(a + b*x**2 + c*x**4)**p*c**3*d*p**2*x**7 + 92*(a + b*x**2 + c*x...