\(\int \frac {(e x)^m (A+B x^2)}{(a+b x^2) (c+d x^2)^2} \, dx\) [34]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {(B c-A d) (e x)^{1+m}}{2 c (b c-a d) e \left (c+d x^2\right )}+\frac {b (A b-a B) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (b c-a d)^2 e (1+m)}+\frac {(b c (B c (1-m)-A d (3-m))+a d (A d (1-m)+B c (1+m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^2 e (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.72 \[ \int \frac {(e x)^m \left (A+B x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )^2} \, dx=\frac {x (e x)^m \left (b (A b-a B) c^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a (-A b+a B) c d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+a (b c-a d) (B c-A d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )}{a c^2 (b c-a d)^2 (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.08, 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.

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

\(\Big \downarrow \) 441

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

\(\Big \downarrow \) 446

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

e**m*int(x**m/(c**2 + 2*c*d*x**2 + d**2*x**4),x)