\(\int (a+b x^2)^p (c+d x^2)^3 \, dx\) [550]

Optimal result

Integrand size = 19, antiderivative size = 282 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\frac {3 d \left (5 a^2 d^2-3 a b c d (7+2 p)+b^2 c^2 \left (35+24 p+4 p^2\right )\right ) x \left (a+b x^2\right )^{1+p}}{b^3 (3+2 p) (5+2 p) (7+2 p)}-\frac {d^2 (5 a d-3 b c (7+2 p)) x^3 \left (a+b x^2\right )^{1+p}}{b^2 (5+2 p) (7+2 p)}+\frac {d^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {\left (b^3 c^3 \left (35+24 p+4 p^2\right )-\frac {3 a d \left (5 a^2 d^2-3 a b c d (7+2 p)+b^2 c^2 \left (35+24 p+4 p^2\right )\right )}{3+2 p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b^3 (5+2 p) (7+2 p)} \] Output:

3*d*(5*a^2*d^2-3*a*b*c*d*(7+2*p)+b^2*c^2*(4*p^2+24*p+35))*x*(b*x^2+a)^(p+1 
)/b^3/(3+2*p)/(5+2*p)/(7+2*p)-d^2*(5*a*d-3*b*c*(7+2*p))*x^3*(b*x^2+a)^(p+1 
)/b^2/(5+2*p)/(7+2*p)+d^3*x^5*(b*x^2+a)^(p+1)/b/(7+2*p)+(b^3*c^3*(4*p^2+24 
*p+35)-3*a*d*(5*a^2*d^2-3*a*b*c*d*(7+2*p)+b^2*c^2*(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 = 5.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\frac {1}{35} x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (35 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )+d x^2 \left (35 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+d x^2 \left (21 c \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )+5 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-p,\frac {9}{2},-\frac {b x^2}{a}\right )\right )\right )\right ) \] Input:

Integrate[(a + b*x^2)^p*(c + d*x^2)^3,x]
 

Output:

(x*(a + b*x^2)^p*(35*c^3*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)] + d 
*x^2*(35*c^2*Hypergeometric2F1[3/2, -p, 5/2, -((b*x^2)/a)] + d*x^2*(21*c*H 
ypergeometric2F1[5/2, -p, 7/2, -((b*x^2)/a)] + 5*d*x^2*Hypergeometric2F1[7 
/2, -p, 9/2, -((b*x^2)/a)]))))/(35*(1 + (b*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {318, 25, 403, 25, 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.

\(\displaystyle \int \left (c+d x^2\right )^3 \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 318

\(\displaystyle \frac {\int -\left (b x^2+a\right )^p \left (d x^2+c\right ) \left (d (5 a d-b c (2 p+11)) x^2+c (a d-b c (2 p+7))\right )dx}{b (2 p+7)}+\frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\int \left (b x^2+a\right )^p \left (d x^2+c\right ) \left (d (5 a d-b c (2 p+11)) x^2+c (a d-b c (2 p+7))\right )dx}{b (2 p+7)}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\frac {\int -\left (b x^2+a\right )^p \left (d \left (b^2 \left (4 p^2+28 p+57\right ) c^2-8 a b d (p+6) c+15 a^2 d^2\right ) x^2+c \left (b^2 \left (4 p^2+24 p+35\right ) c^2-4 a b d (p+4) c+5 a^2 d^2\right )\right )dx}{b (2 p+5)}+\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b (2 p+5)}}{b (2 p+7)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b (2 p+5)}-\frac {\int \left (b x^2+a\right )^p \left (d \left (b^2 \left (4 p^2+28 p+57\right ) c^2-8 a b d (p+6) c+15 a^2 d^2\right ) x^2+c \left (b^2 \left (4 p^2+24 p+35\right ) c^2-4 a b d (p+4) c+5 a^2 d^2\right )\right )dx}{b (2 p+5)}}{b (2 p+7)}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b (2 p+3)}-\frac {\left (15 a^3 d^3-9 a^2 b c d^2 (2 p+7)+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \int \left (b x^2+a\right )^pdx}{b (2 p+3)}}{b (2 p+5)}}{b (2 p+7)}\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b (2 p+3)}-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 a^3 d^3-9 a^2 b c d^2 (2 p+7)+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \int \left (\frac {b x^2}{a}+1\right )^pdx}{b (2 p+3)}}{b (2 p+5)}}{b (2 p+7)}\)

\(\Big \downarrow \) 237

\(\displaystyle \frac {d x \left (c+d x^2\right )^2 \left (a+b x^2\right )^{p+1}}{b (2 p+7)}-\frac {\frac {d x \left (c+d x^2\right ) \left (a+b x^2\right )^{p+1} (5 a d-b c (2 p+11))}{b (2 p+5)}-\frac {\frac {d x \left (a+b x^2\right )^{p+1} \left (15 a^2 d^2-8 a b c d (p+6)+b^2 c^2 \left (4 p^2+28 p+57\right )\right )}{b (2 p+3)}-\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 a^3 d^3-9 a^2 b c d^2 (2 p+7)+3 a b^2 c^2 d \left (4 p^2+24 p+35\right )-b^3 c^3 \left (8 p^3+60 p^2+142 p+105\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^2}{a}\right )}{b (2 p+3)}}{b (2 p+5)}}{b (2 p+7)}\)

Input:

Int[(a + b*x^2)^p*(c + d*x^2)^3,x]
 

Output:

(d*x*(a + b*x^2)^(1 + p)*(c + d*x^2)^2)/(b*(7 + 2*p)) - ((d*(5*a*d - b*c*( 
11 + 2*p))*x*(a + b*x^2)^(1 + p)*(c + d*x^2))/(b*(5 + 2*p)) - ((d*(15*a^2* 
d^2 - 8*a*b*c*d*(6 + p) + b^2*c^2*(57 + 28*p + 4*p^2))*x*(a + b*x^2)^(1 + 
p))/(b*(3 + 2*p)) - ((15*a^3*d^3 - 9*a^2*b*c*d^2*(7 + 2*p) + 3*a*b^2*c^2*d 
*(35 + 24*p + 4*p^2) - b^3*c^3*(105 + 142*p + 60*p^2 + 8*p^3))*x*(a + b*x^ 
2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(b*(3 + 2*p)*(1 + (b*x 
^2)/a)^p))/(b*(5 + 2*p)))/(b*(7 + 2*p))
 

Defintions of rubi rules used

rule 25

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

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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[a, b, c, 
d, 2, p, q, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

Maple [F]

\[\int \left (b \,x^{2}+a \right )^{p} \left (x^{2} d +c \right )^{3}d x\]

Input:

int((b*x^2+a)^p*(d*x^2+c)^3,x)
 

Output:

int((b*x^2+a)^p*(d*x^2+c)^3,x)
 

Fricas [F]

\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*(b*x^2 + a)^p, x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 18.72 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.43 \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=a^{p} c^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + a^{p} c^{2} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {3 a^{p} c d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + \frac {a^{p} d^{3} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - p \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} \] Input:

integrate((b*x**2+a)**p*(d*x**2+c)**3,x)
 

Output:

a**p*c**3*x*hyper((1/2, -p), (3/2,), b*x**2*exp_polar(I*pi)/a) + a**p*c**2 
*d*x**3*hyper((3/2, -p), (5/2,), b*x**2*exp_polar(I*pi)/a) + 3*a**p*c*d**2 
*x**5*hyper((5/2, -p), (7/2,), b*x**2*exp_polar(I*pi)/a)/5 + a**p*d**3*x** 
7*hyper((7/2, -p), (9/2,), b*x**2*exp_polar(I*pi)/a)/7
 

Maxima [F]

\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^3*(b*x^2 + a)^p, x)
 

Giac [F]

\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\int { {\left (d x^{2} + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \,d x } \] Input:

integrate((b*x^2+a)^p*(d*x^2+c)^3,x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^3*(b*x^2 + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\int {\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^3 \,d x \] Input:

int((a + b*x^2)^p*(c + d*x^2)^3,x)
 

Output:

int((a + b*x^2)^p*(c + d*x^2)^3, x)
 

Reduce [F]

\[ \int \left (a+b x^2\right )^p \left (c+d x^2\right )^3 \, dx=\text {too large to display} \] Input:

int((b*x^2+a)^p*(d*x^2+c)^3,x)
 

Output:

(30*(a + b*x**2)**p*a**3*d**3*p*x - 36*(a + b*x**2)**p*a**2*b*c*d**2*p**2* 
x - 126*(a + b*x**2)**p*a**2*b*c*d**2*p*x - 20*(a + b*x**2)**p*a**2*b*d**3 
*p**2*x**3 - 10*(a + b*x**2)**p*a**2*b*d**3*p*x**3 + 24*(a + b*x**2)**p*a* 
b**2*c**2*d*p**3*x + 144*(a + b*x**2)**p*a*b**2*c**2*d*p**2*x + 210*(a + b 
*x**2)**p*a*b**2*c**2*d*p*x + 24*(a + b*x**2)**p*a*b**2*c*d**2*p**3*x**3 + 
 96*(a + b*x**2)**p*a*b**2*c*d**2*p**2*x**3 + 42*(a + b*x**2)**p*a*b**2*c* 
d**2*p*x**3 + 8*(a + b*x**2)**p*a*b**2*d**3*p**3*x**5 + 16*(a + b*x**2)**p 
*a*b**2*d**3*p**2*x**5 + 6*(a + b*x**2)**p*a*b**2*d**3*p*x**5 + 8*(a + b*x 
**2)**p*b**3*c**3*p**3*x + 60*(a + b*x**2)**p*b**3*c**3*p**2*x + 142*(a + 
b*x**2)**p*b**3*c**3*p*x + 105*(a + b*x**2)**p*b**3*c**3*x + 24*(a + b*x** 
2)**p*b**3*c**2*d*p**3*x**3 + 156*(a + b*x**2)**p*b**3*c**2*d*p**2*x**3 + 
282*(a + b*x**2)**p*b**3*c**2*d*p*x**3 + 105*(a + b*x**2)**p*b**3*c**2*d*x 
**3 + 24*(a + b*x**2)**p*b**3*c*d**2*p**3*x**5 + 132*(a + b*x**2)**p*b**3* 
c*d**2*p**2*x**5 + 186*(a + b*x**2)**p*b**3*c*d**2*p*x**5 + 63*(a + b*x**2 
)**p*b**3*c*d**2*x**5 + 8*(a + b*x**2)**p*b**3*d**3*p**3*x**7 + 36*(a + b* 
x**2)**p*b**3*d**3*p**2*x**7 + 46*(a + b*x**2)**p*b**3*d**3*p*x**7 + 15*(a 
 + b*x**2)**p*b**3*d**3*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 + 3 
44*b*p**2*x**2 + 352*b*p*x**2 + 105*b*x**2),x)*a**4*d**3*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 +...