\(\int \frac {(e x)^m (a+b x^2)^3 (A+B x^2)}{(c+d x^2)^2} \, dx\) [30]

Optimal result

Integrand size = 31, antiderivative size = 264 \[ \int \frac {(e x)^m \left (a+b x^2\right )^3 \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx=\frac {b \left (3 a^2 B d^2+b^2 c (3 B c-2 A d)-3 a b d (2 B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}-\frac {b^2 (2 b B c-A b d-3 a B d) (e x)^{3+m}}{d^3 e^3 (3+m)}+\frac {b^3 B (e x)^{5+m}}{d^2 e^5 (5+m)}+\frac {(b c-a d)^3 (B c-A d) (e x)^{1+m}}{2 c d^4 e \left (c+d x^2\right )}+\frac {(b c-a d)^2 (a d (A d (1-m)+B c (1+m))+b c (A d (5+m)-B c (7+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{2 c^2 d^4 e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^m \left (a+b x^2\right )^3 \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx=\frac {x (e x)^m \left (\frac {b \left (3 a^2 B d^2+b^2 c (3 B c-2 A d)+3 a b d (-2 B c+A d)\right )}{1+m}+\frac {b^2 d (-2 b B c+A b d+3 a B d) x^2}{3+m}+\frac {b^3 B d^2 x^4}{5+m}-\frac {(b c-a d)^2 (4 b B c-3 A b d-a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (1+m)}+\frac {(b c-a d)^3 (B c-A d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c^2 (1+m)}\right )}{d^4} \] Input:

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

Output:

(x*(e*x)^m*((b*(3*a^2*B*d^2 + b^2*c*(3*B*c - 2*A*d) + 3*a*b*d*(-2*B*c + A* 
d)))/(1 + m) + (b^2*d*(-2*b*B*c + A*b*d + 3*a*B*d)*x^2)/(3 + m) + (b^3*B*d 
^2*x^4)/(5 + m) - ((b*c - a*d)^2*(4*b*B*c - 3*A*b*d - a*B*d)*Hypergeometri 
c2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m)) + ((b*c - a*d)^3* 
(B*c - A*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c^2 
*(1 + m))))/d^4
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {439, 25, 437, 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.

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 437

\(\displaystyle \frac {\int \left (-\frac {b \left (b^2 (A d (m+5)-B c (m+7)) c^2-3 a b d (A d (m+3)-B c (m+5)) c+3 a^2 d^2 (A d (m+1)-B c (m+3))\right ) (e x)^m}{d^3}+\frac {\left (-7 b^3 B c^4-b^3 B m c^4+5 A b^3 d c^3+15 a b^2 B d c^3+A b^3 d m c^3+3 a b^2 B d m c^3-9 a A b^2 d^2 c^2-9 a^2 b B d^2 c^2-3 a A b^2 d^2 m c^2-3 a^2 b B d^2 m c^2+3 a^2 A b d^3 c+a^3 B d^3 c+3 a^2 A b d^3 m c+a^3 B d^3 m c+a^3 A d^4-a^3 A d^4 m\right ) (e x)^m}{d^3 \left (d x^2+c\right )}-\frac {b^2 (3 a d (A d (m+3)-B c (m+5))-b c (A d (m+5)-B c (m+7))) (e x)^{m+2}}{d^2 e^2}-\frac {b^3 (A d (m+5)-B c (m+7)) (e x)^{m+4}}{d e^4}\right )dx}{2 c d}-\frac {\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{2 c d e \left (c+d x^2\right )}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 437

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q 
_.)*((e_) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*( 
a + b*x^2)^p*(c + d*x^2)^q*(e + f*x^2)^r, x], x] /; FreeQ[{a, b, c, d, e, f 
, g, m}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
 

rule 439

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/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p 
+ 1))   Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( 
p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 
))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G 
tQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(4*x**m*a**3*b*d**3*m**3*x + 36*x**m*a**3*b*d**3*m**2*x + 92*x**m*a* 
*3*b*d**3*m*x + 60*x**m*a**3*b*d**3*x - 6*x**m*a**2*b**2*c*d**2*m**3*x - 6 
6*x**m*a**2*b**2*c*d**2*m**2*x - 234*x**m*a**2*b**2*c*d**2*m*x - 270*x**m* 
a**2*b**2*c*d**2*x + 6*x**m*a**2*b**2*d**3*m**3*x**3 + 42*x**m*a**2*b**2*d 
**3*m**2*x**3 + 42*x**m*a**2*b**2*d**3*m*x**3 - 90*x**m*a**2*b**2*d**3*x** 
3 + 4*x**m*a*b**3*c**2*d*m**3*x + 52*x**m*a*b**3*c**2*d*m**2*x + 220*x**m* 
a*b**3*c**2*d*m*x + 300*x**m*a*b**3*c**2*d*x - 4*x**m*a*b**3*c*d**2*m**3*x 
**3 - 36*x**m*a*b**3*c*d**2*m**2*x**3 - 60*x**m*a*b**3*c*d**2*m*x**3 + 100 
*x**m*a*b**3*c*d**2*x**3 + 4*x**m*a*b**3*d**3*m**3*x**5 + 20*x**m*a*b**3*d 
**3*m**2*x**5 - 4*x**m*a*b**3*d**3*m*x**5 - 20*x**m*a*b**3*d**3*x**5 - x** 
m*b**4*c**3*m**3*x - 15*x**m*b**4*c**3*m**2*x - 71*x**m*b**4*c**3*m*x - 10 
5*x**m*b**4*c**3*x + x**m*b**4*c**2*d*m**3*x**3 + 11*x**m*b**4*c**2*d*m**2 
*x**3 + 23*x**m*b**4*c**2*d*m*x**3 - 35*x**m*b**4*c**2*d*x**3 - x**m*b**4* 
c*d**2*m**3*x**5 - 7*x**m*b**4*c*d**2*m**2*x**5 + x**m*b**4*c*d**2*m*x**5 
+ 7*x**m*b**4*c*d**2*x**5 + x**m*b**4*d**3*m**3*x**7 + 3*x**m*b**4*d**3*m* 
*2*x**7 - x**m*b**4*d**3*m*x**7 - 3*x**m*b**4*d**3*x**7 + int(x**m/(c**2*m 
 - c**2 + 2*c*d*m*x**2 - 2*c*d*x**2 + d**2*m*x**4 - d**2*x**4),x)*a**4*c*d 
**4*m**5 + 7*int(x**m/(c**2*m - c**2 + 2*c*d*m*x**2 - 2*c*d*x**2 + d**2*m* 
x**4 - d**2*x**4),x)*a**4*c*d**4*m**4 + 6*int(x**m/(c**2*m - c**2 + 2*c*d* 
m*x**2 - 2*c*d*x**2 + d**2*m*x**4 - d**2*x**4),x)*a**4*c*d**4*m**3 - 22...