3.1.48 \(\int \frac {(e x)^m (a+b x^2)^p (A+B x^2)}{(c+d x^2)^2} \, dx\) [48]

3.1.48.1 Optimal result

Integrand size = 31, antiderivative size = 295 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{2 c (b c-a d) e \left (c+d x^2\right )}-\frac {(a d (A d (1-m)+B c (1+m))-b c (A d (1-m-2 p)+B c (1+m+2 p))) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{2 c^2 d (b c-a d) e (1+m)}-\frac {b (B c-A d) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 c d (b c-a d) e (1+m)} \]

output

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

3.1.48.2 Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.43 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx=\frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (B c \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-B c+A d) \operatorname {AppellF1}\left (\frac {1+m}{2},-p,2,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{c^2 d (1+m)} \]

input

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

output

(x*(e*x)^m*(a + b*x^2)^p*(B*c*AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*x 
^2)/a), -((d*x^2)/c)] + (-(B*c) + A*d)*AppellF1[(1 + m)/2, -p, 2, (3 + m)/ 
2, -((b*x^2)/a), -((d*x^2)/c)]))/(c^2*d*(1 + m)*(1 + (b*x^2)/a)^p)
 

3.1.48.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {441, 446, 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.

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

\(\Big \downarrow \) 441

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

\(\Big \downarrow \) 446

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

\(\Big \downarrow \) 2009

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

input

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

output

((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(2*c*(b*c - a*d)*e*(c + d* 
x^2)) + (-(((a*d*(A*d*(1 - m) + B*c*(1 + m)) - b*c*(A*d*(1 - m - 2*p) + B* 
c*(1 + m + 2*p)))*(e*x)^(1 + m)*(a + b*x^2)^p*AppellF1[(1 + m)/2, -p, 1, ( 
3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(c*d*e*(1 + m)*(1 + (b*x^2)/a)^p)) 
- (b*(B*c - A*d)*(1 + m + 2*p)*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2 
F1[(1 + m)/2, -p, (3 + m)/2, -((b*x^2)/a)])/(d*e*(1 + m)*(1 + (b*x^2)/a)^p 
))/(2*c*(b*c - a*d))
 

3.1.48.3.1 Defintions of rubi rules used

rule 441

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

rule 446

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/( 
(c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^2)^ 
p*((e + f*x^2)/(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
]
 

rule 2009

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

3.1.48.4 Maple [F]

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

input

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

output

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

3.1.48.5 Fricas [F]

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

input

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

output

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

3.1.48.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.1.48.7 Maxima [F]

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

input

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

output

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

3.1.48.8 Giac [F]

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

input

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

output

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

3.1.48.9 Mupad [F(-1)]

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

input

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

output

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