\(\int (e x)^{-2 p} (a+b x^2)^p (c+d x^2)^2 \, dx\) [1615]

Optimal result

Integrand size = 26, antiderivative size = 196 \[ \int (e x)^{-2 p} \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\frac {d (10 b c-a d (3-2 p)) (e x)^{1-2 p} \left (a+b x^2\right )^{1+p}}{15 b^2 e}+\frac {d^2 (e x)^{3-2 p} \left (a+b x^2\right )^{1+p}}{5 b e^3}+\frac {\left (15 b^2 c^2-10 a b c d (1-2 p)+a^2 d^2 \left (3-8 p+4 p^2\right )\right ) (e x)^{1-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 p),-p,\frac {1}{2} (3-2 p),-\frac {b x^2}{a}\right )}{15 b^2 e (1-2 p)} \] Output:

1/15*d*(10*b*c-a*d*(3-2*p))*(e*x)^(1-2*p)*(b*x^2+a)^(p+1)/b^2/e+1/5*d^2*(e 
*x)^(3-2*p)*(b*x^2+a)^(p+1)/b/e^3+1/15*(15*b^2*c^2-10*a*b*c*d*(1-2*p)+a^2* 
d^2*(4*p^2-8*p+3))*(e*x)^(1-2*p)*(b*x^2+a)^p*hypergeom([-p, 1/2-p],[3/2-p] 
,-b*x^2/a)/b^2/e/(1-2*p)/((1+b*x^2/a)^p)
 

Mathematica [A] (verified)

Time = 6.16 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.91 \[ \int (e x)^{-2 p} \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=-\frac {x (e x)^{-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c^2 \left (15-16 p+4 p^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,-p,\frac {3}{2}-p,-\frac {b x^2}{a}\right )+d (-1+2 p) x^2 \left (2 c (-5+2 p) \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,-p,\frac {5}{2}-p,-\frac {b x^2}{a}\right )+d (-3+2 p) x^2 \operatorname {Hypergeometric2F1}\left (\frac {5}{2}-p,-p,\frac {7}{2}-p,-\frac {b x^2}{a}\right )\right )\right )}{(-5+2 p) (-3+2 p) (-1+2 p)} \] Input:

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

Output:

-((x*(a + b*x^2)^p*(c^2*(15 - 16*p + 4*p^2)*Hypergeometric2F1[1/2 - p, -p, 
 3/2 - p, -((b*x^2)/a)] + d*(-1 + 2*p)*x^2*(2*c*(-5 + 2*p)*Hypergeometric2 
F1[3/2 - p, -p, 5/2 - p, -((b*x^2)/a)] + d*(-3 + 2*p)*x^2*Hypergeometric2F 
1[5/2 - p, -p, 7/2 - p, -((b*x^2)/a)])))/((-5 + 2*p)*(-3 + 2*p)*(-1 + 2*p) 
*(e*x)^(2*p)*(1 + (b*x^2)/a)^p))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {367, 363, 279, 278}

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.

\(\displaystyle \int \left (c+d x^2\right )^2 (e x)^{-2 p} \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 367

\(\displaystyle \frac {\int (e x)^{-2 p} \left (b x^2+a\right )^p \left (5 b c^2+d (10 b c-a d (3-2 p)) x^2\right )dx}{5 b}+\frac {d^2 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {\frac {\left (15 b^2 c^2-a d (1-2 p) (10 b c-a d (3-2 p))\right ) \int (e x)^{-2 p} \left (b x^2+a\right )^pdx}{3 b}+\frac {d (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} (10 b c-a d (3-2 p))}{3 b e}}{5 b}+\frac {d^2 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 b^2 c^2-a d (1-2 p) (10 b c-a d (3-2 p))\right ) \int (e x)^{-2 p} \left (\frac {b x^2}{a}+1\right )^pdx}{3 b}+\frac {d (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} (10 b c-a d (3-2 p))}{3 b e}}{5 b}+\frac {d^2 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\frac {(e x)^{1-2 p} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (15 b^2 c^2-a d (1-2 p) (10 b c-a d (3-2 p))\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1-2 p),-p,\frac {1}{2} (3-2 p),-\frac {b x^2}{a}\right )}{3 b e (1-2 p)}+\frac {d (e x)^{1-2 p} \left (a+b x^2\right )^{p+1} (10 b c-a d (3-2 p))}{3 b e}}{5 b}+\frac {d^2 (e x)^{3-2 p} \left (a+b x^2\right )^{p+1}}{5 b e^3}\)

Input:

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

Output:

(d^2*(e*x)^(3 - 2*p)*(a + b*x^2)^(1 + p))/(5*b*e^3) + ((d*(10*b*c - a*d*(3 
 - 2*p))*(e*x)^(1 - 2*p)*(a + b*x^2)^(1 + p))/(3*b*e) + ((15*b^2*c^2 - a*d 
*(10*b*c - a*d*(3 - 2*p))*(1 - 2*p))*(e*x)^(1 - 2*p)*(a + b*x^2)^p*Hyperge 
ometric2F1[(1 - 2*p)/2, -p, (3 - 2*p)/2, -((b*x^2)/a)])/(3*b*e*(1 - 2*p)*( 
1 + (b*x^2)/a)^p))/(5*b)
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 367

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, 
x_Symbol] :> Simp[d^2*(e*x)^(m + 3)*((a + b*x^2)^(p + 1)/(b*e^3*(m + 2*p + 
5))), x] + Simp[1/(b*(m + 2*p + 5))   Int[(e*x)^m*(a + b*x^2)^p*Simp[b*c^2* 
(m + 2*p + 5) - d*(a*d*(m + 3) - 2*b*c*(m + 2*p + 5))*x^2, x], x], x] /; Fr 
eeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + 2*p + 5, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*(b*x^2 + a)^p/(e*x)^(2*p), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 81.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.77 \[ \int (e x)^{-2 p} \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\frac {a^{p} c d e^{- 2 p} x^{3 - 2 p} \Gamma \left (\frac {3}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {3}{2} - p \\ \frac {5}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {5}{2} - p\right )} + \frac {a^{p} d^{2} e^{- 2 p} x^{5 - 2 p} \Gamma \left (\frac {5}{2} - p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {5}{2} - p \\ \frac {7}{2} - p \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {7}{2} - p\right )} + \sqrt {b} b^{p - \frac {1}{2}} c^{2} e^{- 2 p} x {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - p \\ \frac {1}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )} \] Input:

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

Output:

a**p*c*d*x**(3 - 2*p)*gamma(3/2 - p)*hyper((-p, 3/2 - p), (5/2 - p,), b*x* 
*2*exp_polar(I*pi)/a)/(e**(2*p)*gamma(5/2 - p)) + a**p*d**2*x**(5 - 2*p)*g 
amma(5/2 - p)*hyper((-p, 5/2 - p), (7/2 - p,), b*x**2*exp_polar(I*pi)/a)/( 
2*e**(2*p)*gamma(7/2 - p)) + sqrt(b)*b**(p - 1/2)*c**2*x*hyper((-1/2, -p), 
 (1/2,), a*exp_polar(I*pi)/(b*x**2))/e**(2*p)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^{-2 p} \left (a+b x^2\right )^p \left (c+d x^2\right )^2 \, dx=\frac {4 \left (b \,x^{2}+a \right )^{p} a^{2} d^{2} p^{2} x -6 \left (b \,x^{2}+a \right )^{p} a^{2} d^{2} p x +20 \left (b \,x^{2}+a \right )^{p} a b c d p x +2 \left (b \,x^{2}+a \right )^{p} a b \,d^{2} p \,x^{3}+15 \left (b \,x^{2}+a \right )^{p} b^{2} c^{2} x +10 \left (b \,x^{2}+a \right )^{p} b^{2} c d \,x^{3}+3 \left (b \,x^{2}+a \right )^{p} b^{2} d^{2} x^{5}+8 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{2} p^{3}-16 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{2} p^{2}+6 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{3} d^{2} p +40 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c d \,p^{2}-20 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a^{2} b c d p +30 x^{2 p} \left (\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2 p} a +x^{2 p} b \,x^{2}}d x \right ) a \,b^{2} c^{2} p}{15 x^{2 p} e^{2 p} b^{2}} \] Input:

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

Output:

(4*(a + b*x**2)**p*a**2*d**2*p**2*x - 6*(a + b*x**2)**p*a**2*d**2*p*x + 20 
*(a + b*x**2)**p*a*b*c*d*p*x + 2*(a + b*x**2)**p*a*b*d**2*p*x**3 + 15*(a + 
 b*x**2)**p*b**2*c**2*x + 10*(a + b*x**2)**p*b**2*c*d*x**3 + 3*(a + b*x**2 
)**p*b**2*d**2*x**5 + 8*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a + x**(2*p 
)*b*x**2),x)*a**3*d**2*p**3 - 16*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a 
+ x**(2*p)*b*x**2),x)*a**3*d**2*p**2 + 6*x**(2*p)*int((a + b*x**2)**p/(x** 
(2*p)*a + x**(2*p)*b*x**2),x)*a**3*d**2*p + 40*x**(2*p)*int((a + b*x**2)** 
p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**2*b*c*d*p**2 - 20*x**(2*p)*int((a + 
 b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a**2*b*c*d*p + 30*x**(2*p)*i 
nt((a + b*x**2)**p/(x**(2*p)*a + x**(2*p)*b*x**2),x)*a*b**2*c**2*p)/(15*x* 
*(2*p)*e**(2*p)*b**2)