\(\int \frac {(e x)^m (A+B x^2)}{(a+b x^2)^2 (c+d x^2)^3} \, dx\) [42]

Optimal result

Integrand size = 31, antiderivative size = 452 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {d (2 A b c-3 a B c+a A d) (e x)^{1+m}}{4 a c (b c-a d)^2 e \left (c+d x^2\right )^2}+\frac {(A b-a B) (e x)^{1+m}}{2 a (b c-a d) e \left (a+b x^2\right ) \left (c+d x^2\right )^2}+\frac {d \left (A \left (4 b^2 c^2-a^2 d^2 (3-m)+a b c d (11-m)\right )-a B c (b c (11-m)+a d (1+m))\right ) (e x)^{1+m}}{8 a c^2 (b c-a d)^3 e \left (c+d x^2\right )}+\frac {b^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (1+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^4 e (1+m)}-\frac {d \left (b^2 c^2 (B c (3-m)-A d (7-m)) (5-m)-a^2 d^2 (1-m) (A d (3-m)+B c (1+m))+2 a b c d \left (B c \left (5+4 m-m^2\right )+A d \left (7-8 m+m^2\right )\right )\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{8 c^3 (b c-a d)^4 e (1+m)} \] Output:

1/4*d*(A*a*d+2*A*b*c-3*B*a*c)*(e*x)^(1+m)/a/c/(-a*d+b*c)^2/e/(d*x^2+c)^2+1 
/2*(A*b-B*a)*(e*x)^(1+m)/a/(-a*d+b*c)/e/(b*x^2+a)/(d*x^2+c)^2+1/8*d*(A*(4* 
b^2*c^2-a^2*d^2*(3-m)+a*b*c*d*(11-m))-a*B*c*(b*c*(11-m)+a*d*(1+m)))*(e*x)^ 
(1+m)/a/c^2/(-a*d+b*c)^3/e/(d*x^2+c)+1/2*b^2*(A*b*(b*c*(1-m)-a*d*(7-m))+a* 
B*(a*d*(5-m)+b*c*(1+m)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m], 
-b*x^2/a)/a^2/(-a*d+b*c)^4/e/(1+m)-1/8*d*(b^2*c^2*(B*c*(3-m)-A*d*(7-m))*(5 
-m)-a^2*d^2*(1-m)*(A*d*(3-m)+B*c*(1+m))+2*a*b*c*d*(B*c*(-m^2+4*m+5)+A*d*(m 
^2-8*m+7)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^3 
/(-a*d+b*c)^4/e/(1+m)
 

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.59 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\frac {x (e x)^m \left (\frac {b^2 (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}-\frac {b d (b B c-3 A b d+2 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c}+\frac {b^2 (-A b+a B) (-b c+a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2}-\frac {d (b c-a d) (b B c-2 A b d+a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2}+\frac {d (b c-a d)^2 (-B c+A d) \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^3}\right )}{(b c-a d)^4 (1+m)} \] Input:

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

Output:

(x*(e*x)^m*((b^2*(b*B*c - 3*A*b*d + 2*a*B*d)*Hypergeometric2F1[1, (1 + m)/ 
2, (3 + m)/2, -((b*x^2)/a)])/a - (b*d*(b*B*c - 3*A*b*d + 2*a*B*d)*Hypergeo 
metric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/c + (b^2*(-(A*b) + a*B)* 
(-(b*c) + a*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/a 
^2 - (d*(b*c - a*d)*(b*B*c - 2*A*b*d + a*B*d)*Hypergeometric2F1[2, (1 + m) 
/2, (3 + m)/2, -((d*x^2)/c)])/c^2 + (d*(b*c - a*d)^2*(-(B*c) + A*d)*Hyperg 
eometric2F1[3, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/c^3))/((b*c - a*d)^4*( 
1 + m))
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {441, 441, 27, 441, 25, 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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) (e x)^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx\)

\(\Big \downarrow \) 441

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\int \frac {2 (e x)^m \left (-b d (2 A b c-3 a B c+a A d) (3-m) x^2+A \left (-2 b^2 (1-m) c^2+8 a b d c-a^2 d^2 (3-m)\right )-a B c (2 b c+a d) (m+1)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\int \frac {(e x)^m \left (-b d (2 A b c-3 a B c+a A d) (3-m) x^2+A \left (-2 b^2 (1-m) c^2+8 a b d c-a^2 d^2 (3-m)\right )-a B c (2 b c+a d) (m+1)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^2}dx}{2 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 441

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {\frac {\int -\frac {(e x)^m \left (b d (1-m) \left (A \left (4 b^2 c^2+a b d (11-m) c-a^2 d^2 (3-m)\right )-a B c (b c (11-m)+a d (m+1))\right ) x^2+a B c \left (4 b^2 c^2+a b d (9-m) c-a^2 d^2 (1-m)\right ) (m+1)-A \left (-4 b^3 (1-m) c^3+24 a b^2 d c^2-a^2 b d^2 \left (m^2-12 m+11\right ) c+a^3 d^3 \left (m^2-4 m+3\right )\right )\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{2 c e \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {-\frac {\int \frac {(e x)^m \left (b d (1-m) \left (A \left (4 b^2 c^2+a b d (11-m) c-a^2 d^2 (3-m)\right )-a B c (b c (11-m)+a d (m+1))\right ) x^2+a B c \left (4 b^2 c^2+a b d (9-m) c-a^2 d^2 (1-m)\right ) (m+1)-A \left (-4 b^3 (1-m) c^3+24 a b^2 d c^2-a^2 b d^2 \left (m^2-12 m+11\right ) c+a^3 d^3 \left (m^2-4 m+3\right )\right )\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{2 c e \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 446

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {-\frac {\int \left (\frac {4 b^2 c^2 (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (m+1))) (e x)^m}{(b c-a d) \left (b x^2+a\right )}+\frac {a d \left (-b^2 (B c (3-m)-A d (7-m)) (5-m) c^2-2 a b d \left (B c \left (-m^2+4 m+5\right )+A d \left (m^2-8 m+7\right )\right ) c+a^2 d^2 (1-m) (A d (3-m)+B c (m+1))\right ) (e x)^m}{(b c-a d) \left (d x^2+c\right )}\right )dx}{2 c (b c-a d)}-\frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{2 c e \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right )^2 (b c-a d)}-\frac {\frac {-\frac {\frac {4 b^2 c^2 (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (7-m))+a B (a d (5-m)+b c (m+1)))}{a e (m+1) (b c-a d)}-\frac {a d (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right ) \left (-a^2 d^2 (1-m) (A d (3-m)+B c (m+1))+2 a b c d \left (A d \left (m^2-8 m+7\right )+B c \left (-m^2+4 m+5\right )\right )+b^2 c^2 (5-m) (B c (3-m)-A d (7-m))\right )}{c e (m+1) (b c-a d)}}{2 c (b c-a d)}-\frac {d (e x)^{m+1} \left (A \left (-a^2 d^2 (3-m)+a b c d (11-m)+4 b^2 c^2\right )-a B c (a d (m+1)+b c (11-m))\right )}{2 c e \left (c+d x^2\right ) (b c-a d)}}{2 c (b c-a d)}-\frac {d (e x)^{m+1} (a A d-3 a B c+2 A b c)}{2 c e \left (c+d x^2\right )^2 (b c-a d)}}{2 a (b c-a d)}\)

Input:

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

Output:

((A*b - a*B)*(e*x)^(1 + m))/(2*a*(b*c - a*d)*e*(a + b*x^2)*(c + d*x^2)^2) 
- (-1/2*(d*(2*A*b*c - 3*a*B*c + a*A*d)*(e*x)^(1 + m))/(c*(b*c - a*d)*e*(c 
+ d*x^2)^2) + (-1/2*(d*(A*(4*b^2*c^2 - a^2*d^2*(3 - m) + a*b*c*d*(11 - m)) 
 - a*B*c*(b*c*(11 - m) + a*d*(1 + m)))*(e*x)^(1 + m))/(c*(b*c - a*d)*e*(c 
+ d*x^2)) - ((4*b^2*c^2*(A*b*(b*c*(1 - m) - a*d*(7 - m)) + a*B*(a*d*(5 - m 
) + b*c*(1 + m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, 
 -((b*x^2)/a)])/(a*(b*c - a*d)*e*(1 + m)) - (a*d*(b^2*c^2*(B*c*(3 - m) - A 
*d*(7 - m))*(5 - m) - a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) + 2*a*b* 
c*d*(B*c*(5 + 4*m - m^2) + A*d*(7 - 8*m + m^2)))*(e*x)^(1 + m)*Hypergeomet 
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)*e*(1 + m)))/ 
(2*c*(b*c - a*d)))/(2*c*(b*c - a*d)))/(2*a*(b*c - a*d))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (x^{2} B +A \right )}{\left (b \,x^{2}+a \right )^{2} \left (x^{2} d +c \right )^{3}}d x\]

Input:

int((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x)
 

Output:

int((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x)
 

Fricas [F]

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

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")
 

Output:

integral((B*x^2 + A)*(e*x)^m/(b^2*d^3*x^10 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^8 
 + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d^3)*x^6 + a^2*c^3 + (b^2*c^3 + 6*a*b* 
c^2*d + 3*a^2*c*d^2)*x^4 + (2*a*b*c^3 + 3*a^2*c^2*d)*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate((e*x)**m*(B*x**2+A)/(b*x**2+a)**2/(d*x**2+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*(e*x)^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)
 

Giac [F]

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

integrate((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*(e*x)^m/((b*x^2 + a)^2*(d*x^2 + c)^3), x)
 

Mupad [F(-1)]

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

int(((A + B*x^2)*(e*x)^m)/((a + b*x^2)^2*(c + d*x^2)^3),x)
 

Output:

int(((A + B*x^2)*(e*x)^m)/((a + b*x^2)^2*(c + d*x^2)^3), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx=e^{m} \left (\int \frac {x^{m}}{b \,d^{3} x^{8}+a \,d^{3} x^{6}+3 b c \,d^{2} x^{6}+3 a c \,d^{2} x^{4}+3 b \,c^{2} d \,x^{4}+3 a \,c^{2} d \,x^{2}+b \,c^{3} x^{2}+a \,c^{3}}d x \right ) \] Input:

int((e*x)^m*(B*x^2+A)/(b*x^2+a)^2/(d*x^2+c)^3,x)
 

Output:

e**m*int(x**m/(a*c**3 + 3*a*c**2*d*x**2 + 3*a*c*d**2*x**4 + a*d**3*x**6 + 
b*c**3*x**2 + 3*b*c**2*d*x**4 + 3*b*c*d**2*x**6 + b*d**3*x**8),x)