3.1.14 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^2}{(a+b x^2)^3} \, dx\) [14]

3.1.14.1 Optimal result

Integrand size = 31, antiderivative size = 292 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx=\frac {d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m)) (e x)^{1+m}}{8 a^2 b^3 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^2}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) (e x)^{1+m} \left (c (A b (3-m)+a B (1+m))-d (A b (1+m)-a B (5+m)) x^2\right )}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {(a d (b c (1+m)-a d (3+m)) (A b (1+m)-a B (5+m))-b c (A b (3-m)+a B (1+m)) (a d (1+m)+b (c-c m))) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{8 a^3 b^3 e (1+m)} \]

output

1/8*d*(b*c*(1+m)-a*d*(3+m))*(A*b*(1+m)-a*B*(5+m))*(e*x)^(1+m)/a^2/b^3/e/(1 
+m)+1/4*(A*b-B*a)*(e*x)^(1+m)*(d*x^2+c)^2/a/b/e/(b*x^2+a)^2+1/8*(-a*d+b*c) 
*(e*x)^(1+m)*(c*(A*b*(3-m)+a*B*(1+m))-d*(A*b*(1+m)-a*B*(5+m))*x^2)/a^2/b^2 
/e/(b*x^2+a)-1/8*(a*d*(b*c*(1+m)-a*d*(3+m))*(A*b*(1+m)-a*B*(5+m))-b*c*(A*b 
*(3-m)+a*B*(1+m))*(a*d*(1+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^3/b^3/e/(1+m)
 

3.1.14.2 Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.57 \[ \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^2}{\left (a+b x^2\right )^3} \, dx=\frac {x (e x)^m \left (B d^2+\frac {d (2 b B c+A b d-3 a B d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a}+\frac {(b c-a d) (b B c+2 A b d-3 a B d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2}+\frac {(A b-a B) (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^3}\right )}{b^3 (1+m)} \]

input

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

output

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

3.1.14.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {439, 25, 439, 25, 363, 278}

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.

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 439

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 278

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

input

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

output

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

3.1.14.3.1 Defintions of rubi rules used

rule 25

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

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 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])
 

3.1.14.4 Maple [F]

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

input

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

output

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

3.1.14.5 Fricas [F]

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

input

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

3.1.14.6 Sympy [F]

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

input

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

output

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

3.1.14.7 Maxima [F]

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

input

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

output

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

3.1.14.8 Giac [F]

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

input

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

output

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

3.1.14.9 Mupad [F(-1)]

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

input

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

output

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