3.1.13 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^2}{(a+b x^2)^2} \, dx\) [13]

3.1.13.1 Optimal result

Integrand size = 31, antiderivative size = 247 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=-\frac {d (A b (2 b c (1+m)-a d (3+m))-a B (2 b c (3+m)-a d (5+m))) (e x)^{1+m}}{2 a b^3 e (1+m)}-\frac {d^2 (A b (3+m)-a B (5+m)) (e x)^{3+m}}{2 a b^2 e^3 (3+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{2 a b e \left (a+b x^2\right )}+\frac {(b c-a d) (a B (b c (1+m)-a d (5+m))+A b (a d (3+m)+b (c-c m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 b^3 e (1+m)} \]

output

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

3.1.13.2 Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.63 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^2} \, dx=\frac {x (e x)^m \left (\frac {d (2 b B c+A b d-2 a B d)}{1+m}+\frac {b B d^2 x^2}{3+m}+\frac {(b c-a d) (b B c+2 A b d-3 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a (1+m)}+\frac {(A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 (1+m)}\right )}{b^3} \]

input

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

output

(x*(e*x)^m*((d*(2*b*B*c + A*b*d - 2*a*B*d))/(1 + m) + (b*B*d^2*x^2)/(3 + m 
) + ((b*c - a*d)*(b*B*c + 2*A*b*d - 3*a*B*d)*Hypergeometric2F1[1, (1 + m)/ 
2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m)) + ((A*b - a*B)*(b*c - a*d)^2*Hype 
rgeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a^2*(1 + m))))/b^3
 

3.1.13.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 437

\(\displaystyle \frac {\int \left (-\frac {d (A b (2 b c (m+1)-a d (m+3))-a B (2 b c (m+3)-a d (m+5))) (e x)^m}{b^2}+\frac {\left (5 B d^2 a^3+B d^2 m a^3-3 A b d^2 a^2-6 b B c d a^2-A b d^2 m a^2-2 b B c d m a^2+b^2 B c^2 a+2 A b^2 c d a+b^2 B c^2 m a+2 A b^2 c d m a+A b^3 c^2-A b^3 c^2 m\right ) (e x)^m}{b^2 \left (b x^2+a\right )}-\frac {d^2 (A b (m+3)-a B (m+5)) (e x)^{m+2}}{b e^2}\right )dx}{2 a b}+\frac {\left (c+d x^2\right )^2 (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )}\)

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.13.3.1 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]
 

3.1.13.4 Maple [F]

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

input

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

output

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

3.1.13.5 Fricas [F]

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

input

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

output

integral((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)*(e*x)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
 

3.1.13.6 Sympy [F]

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

input

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

output

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

3.1.13.7 Maxima [F]

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

input

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

output

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

3.1.13.8 Giac [F]

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

input

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

output

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

3.1.13.9 Mupad [F(-1)]

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

input

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

output

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