\(\int \frac {(e x)^{-4-2 p} (a+b x^2)^p}{(c+d x)^2} \, dx\) [1917]

Optimal result

Integrand size = 26, antiderivative size = 400 \[ \int \frac {(e x)^{-4-2 p} \left (a+b x^2\right )^p}{(c+d x)^2} \, dx=-\frac {d^3 \left (b c^2+a d^2 (2+p)\right ) (e x)^{-2 p} \left (a+b x^2\right )^{1+p}}{a c^3 \left (b c^2+a d^2\right ) e^4 (1+p) \left (c^2-d^2 x^2\right )}+\frac {d (e x)^{-2 (1+p)} \left (a+b x^2\right )^{1+p}}{a c e^2 (1+p) \left (c^2-d^2 x^2\right )}-\frac {(e x)^{-3-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-3-2 p),-p,2,\frac {1}{2} (-1-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 e (3+2 p)}-\frac {d^2 (e x)^{-1-2 p} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2} (-1-2 p),-p,2,\frac {1}{2} (1-2 p),-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 e^3 (1+2 p)}+\frac {d^3 \left (2 b c^2+a d^2 (2+p)\right ) (e x)^{-2 p} \left (a+b x^2\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (b+\frac {a d^2}{c^2}\right ) x^2}{a+b x^2}\right )}{c^5 \left (b c^2+a d^2\right ) e^4 p} \] Output:

-d^3*(b*c^2+a*d^2*(2+p))*(b*x^2+a)^(p+1)/a/c^3/(a*d^2+b*c^2)/e^4/(p+1)/((e 
*x)^(2*p))/(-d^2*x^2+c^2)+d*(b*x^2+a)^(p+1)/a/c/e^2/(p+1)/((e*x)^(2*p+2))/ 
(-d^2*x^2+c^2)-(e*x)^(-3-2*p)*(b*x^2+a)^p*AppellF1(-3/2-p,2,-p,-1/2-p,d^2* 
x^2/c^2,-b*x^2/a)/c^2/e/(3+2*p)/((1+b*x^2/a)^p)-d^2*(e*x)^(-1-2*p)*(b*x^2+ 
a)^p*AppellF1(-1/2-p,2,-p,1/2-p,d^2*x^2/c^2,-b*x^2/a)/c^4/e^3/(1+2*p)/((1+ 
b*x^2/a)^p)+d^3*(2*b*c^2+a*d^2*(2+p))*(b*x^2+a)^p*hypergeom([1, -p],[1-p], 
(b+a*d^2/c^2)*x^2/(b*x^2+a))/c^5/(a*d^2+b*c^2)/e^4/p/((e*x)^(2*p))
 

Mathematica [F]

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

Integrate[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^2,x]
 

Output:

Integrate[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^2, x]
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {623, 622, 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.

\(\displaystyle \int \frac {(e x)^{-2 p-4} \left (a+b x^2\right )^p}{(c+d x)^2} \, dx\)

\(\Big \downarrow \) 623

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

\(\Big \downarrow \) 622

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

\(\Big \downarrow \) 2009

\(\displaystyle x^{2 (p+2)} (e x)^{-2 (p+2)} \left (-\frac {x^{-2 p-3} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-p-\frac {3}{2},-p,2,-p-\frac {1}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 (2 p+3)}-\frac {d^2 x^{-2 p-1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-p-\frac {1}{2},-p,2,\frac {1}{2}-p,-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 (2 p+1)}-\frac {d^3 x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{c \left (c^2-d^2 x^2\right ) \left (a d^2+b c^2\right )}+\frac {d^3 x^{-2 p} \left (a+b x^2\right )^p \left (a d^2 (p+2)+2 b c^2\right ) \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {\left (\frac {a d^2}{c^2}+b\right ) x^2}{b x^2+a}\right )}{c^5 p \left (a d^2+b c^2\right )}+\frac {d x^{-2 (p+1)} \left (a+b x^2\right )^{p+1} \left (a d^2 (p+2)+b c^2\right )}{a c^3 (p+1) \left (a d^2+b c^2\right )}\right )\)

Input:

Int[((e*x)^(-4 - 2*p)*(a + b*x^2)^p)/(c + d*x)^2,x]
 

Output:

(x^(2*(2 + p))*((d*(b*c^2 + a*d^2*(2 + p))*(a + b*x^2)^(1 + p))/(a*c^3*(b* 
c^2 + a*d^2)*(1 + p)*x^(2*(1 + p))) - (d^3*(a + b*x^2)^(1 + p))/(c*(b*c^2 
+ a*d^2)*x^(2*(1 + p))*(c^2 - d^2*x^2)) - (x^(-3 - 2*p)*(a + b*x^2)^p*Appe 
llF1[-3/2 - p, -p, 2, -1/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^2*(3 + 2* 
p)*(1 + (b*x^2)/a)^p) - (d^2*x^(-1 - 2*p)*(a + b*x^2)^p*AppellF1[-1/2 - p, 
 -p, 2, 1/2 - p, -((b*x^2)/a), (d^2*x^2)/c^2])/(c^4*(1 + 2*p)*(1 + (b*x^2) 
/a)^p) + (d^3*(2*b*c^2 + a*d^2*(2 + p))*(a + b*x^2)^p*Hypergeometric2F1[1, 
 -p, 1 - p, ((b + (a*d^2)/c^2)*x^2)/(a + b*x^2)])/(c^5*(b*c^2 + a*d^2)*p*x 
^(2*p))))/(e*x)^(2*(2 + p))
 

Defintions of rubi rules used

rule 622

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[x^m*(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 
 - d^2*x^2)))^(-n), x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[n, -1]
 

rule 623

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (e x \right )^{-4-2 p} \left (b \,x^{2}+a \right )^{p}}{\left (d x +c \right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((e*x)**(-4-2*p)*(b*x**2+a)**p/(d*x+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^(-4-2*p)*(b*x^2+a)^p/(d*x+c)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 20*(a + b*x**2)**p*a**2*c**2*d*p**2 - 30*(a + b*x**2)**p*a**2*c**2*d*p 
 - 10*(a + b*x**2)**p*a**2*c**2*d - 8*(a + b*x**2)**p*a**2*c*d**2*p**3*x - 
 32*(a + b*x**2)**p*a**2*c*d**2*p**2*x - 38*(a + b*x**2)**p*a**2*c*d**2*p* 
x - 12*(a + b*x**2)**p*a**2*c*d**2*x + 8*(a + b*x**2)**p*a**2*d**3*p**3*x* 
*2 + 40*(a + b*x**2)**p*a**2*d**3*p**2*x**2 + 66*(a + b*x**2)**p*a**2*d**3 
*p*x**2 + 36*(a + b*x**2)**p*a**2*d**3*x**2 + 6*(a + b*x**2)**p*a*b*c**3*p 
**2*x + 27*(a + b*x**2)**p*a*b*c**3*p*x + 12*(a + b*x**2)**p*a*b*c**3*x - 
26*(a + b*x**2)**p*a*b*c**2*d*p**2*x**2 - 53*(a + b*x**2)**p*a*b*c**2*d*p* 
x**2 - 36*(a + b*x**2)**p*a*b*c**2*d*x**2 + 4*(a + b*x**2)**p*a*b*c*d**2*p 
**2*x**3 + 14*(a + b*x**2)**p*a*b*c*d**2*p*x**3 + 12*(a + b*x**2)**p*a*b*c 
*d**2*x**3 + 4*(a + b*x**2)**p*a*b*d**3*p**2*x**4 + 14*(a + b*x**2)**p*a*b 
*d**3*p*x**4 + 12*(a + b*x**2)**p*a*b*d**3*x**4 + 2*(a + b*x**2)**p*b**2*c 
**3*p*x**3 - 7*(a + b*x**2)**p*b**2*c**3*x**3 + 2*(a + b*x**2)**p*b**2*c** 
2*d*p*x**4 - 7*(a + b*x**2)**p*b**2*c**2*d*x**4 + 48*x**(2*p)*int((a + b*x 
**2)**p/(x**(2*p)*a*c**2*p*x + x**(2*p)*a*c**2*x + 2*x**(2*p)*a*c*d*p*x**2 
 + 2*x**(2*p)*a*c*d*x**2 + x**(2*p)*a*d**2*p*x**3 + x**(2*p)*a*d**2*x**3 + 
 x**(2*p)*b*c**2*p*x**3 + x**(2*p)*b*c**2*x**3 + 2*x**(2*p)*b*c*d*p*x**4 + 
 2*x**(2*p)*b*c*d*x**4 + x**(2*p)*b*d**2*p*x**5 + x**(2*p)*b*d**2*x**5),x) 
*a**3*c*d**4*p**5*x**3 + 288*x**(2*p)*int((a + b*x**2)**p/(x**(2*p)*a*c**2 
*p*x + x**(2*p)*a*c**2*x + 2*x**(2*p)*a*c*d*p*x**2 + 2*x**(2*p)*a*c*d*x...