3.1.41 \(\int \frac {(d x)^m (A+B x+C x^2)}{(a+b x^2+c x^4)^2} \, dx\) [41]

3.1.41.1 Optimal result

Integrand size = 30, antiderivative size = 685 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {B (d x)^{2+m} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) d^2 \left (a+b x^2+c x^4\right )}+\frac {(d x)^{1+m} \left (A \left (b^2-2 a c\right )-a b C+c (A b-2 a C) x^2\right )}{2 a \left (b^2-4 a c\right ) d \left (a+b x^2+c x^4\right )}+\frac {c \left (2 a C \left (2 b-\sqrt {b^2-4 a c} (1-m)\right )+A \left (b^2 (1-m)+b \sqrt {b^2-4 a c} (1-m)-4 a c (3-m)\right )\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}-\frac {c \left (2 a C \left (2 b+\sqrt {b^2-4 a c} (1-m)\right )+A \left (b^2 (1-m)-b \sqrt {b^2-4 a c} (1-m)-4 a c (3-m)\right )\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}-\frac {B c \left (4 a c (2-m)+b \left (b+\sqrt {b^2-4 a c}\right ) m\right ) (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) d^2 (2+m)}+\frac {B c \left (4 a c (2-m)+b \left (b-\sqrt {b^2-4 a c}\right ) m\right ) (d x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d^2 (2+m)} \]

output

1/2*B*(d*x)^(2+m)*(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/d^2/(c*x^4+b*x^2+a)+1 
/2*(d*x)^(1+m)*(A*(-2*a*c+b^2)-a*b*C+c*(A*b-2*C*a)*x^2)/a/(-4*a*c+b^2)/d/( 
c*x^4+b*x^2+a)+1/2*B*c*(d*x)^(2+m)*hypergeom([1, 1+1/2*m],[2+1/2*m],-2*c*x 
^2/(b+(-4*a*c+b^2)^(1/2)))*(4*a*c*(2-m)+b*m*(b-(-4*a*c+b^2)^(1/2)))/a/(-4* 
a*c+b^2)^(3/2)/d^2/(2+m)/(b+(-4*a*c+b^2)^(1/2))-1/2*B*c*(d*x)^(2+m)*hyperg 
eom([1, 1+1/2*m],[2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))*(4*a*c*(2-m)+b 
*m*(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/d^2/(2+m)/(b-(-4*a*c+b^2)^ 
(1/2))-1/2*c*(d*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2*c*x^2/(b+ 
(-4*a*c+b^2)^(1/2)))*(2*a*C*(2*b+(1-m)*(-4*a*c+b^2)^(1/2))+A*(b^2*(1-m)-4* 
a*c*(3-m)-b*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/d/(1+m)/(b+(-4 
*a*c+b^2)^(1/2))+1/2*c*(d*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2 
*c*x^2/(b-(-4*a*c+b^2)^(1/2)))*(2*a*C*(2*b-(1-m)*(-4*a*c+b^2)^(1/2))+A*(b^ 
2*(1-m)-4*a*c*(3-m)+b*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/d/(1 
+m)/(b-(-4*a*c+b^2)^(1/2))
 

3.1.41.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 2.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.35 \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {x (d x)^m \left (A \left (6+5 m+m^2\right ) \operatorname {AppellF1}\left (\frac {1+m}{2},2,2,\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x \left (B (3+m) \operatorname {AppellF1}\left (\frac {2+m}{2},2,2,\frac {4+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )+C (2+m) x \operatorname {AppellF1}\left (\frac {3+m}{2},2,2,\frac {5+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{a^2 (1+m) (2+m) (3+m)} \]

input

Integrate[((d*x)^m*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2,x]
 

output

(x*(d*x)^m*(A*(6 + 5*m + m^2)*AppellF1[(1 + m)/2, 2, 2, (3 + m)/2, (-2*c*x 
^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + (1 + m) 
*x*(B*(3 + m)*AppellF1[(2 + m)/2, 2, 2, (4 + m)/2, (-2*c*x^2)/(b + Sqrt[b^ 
2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + C*(2 + m)*x*AppellF1[(3 
 + m)/2, 2, 2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(- 
b + Sqrt[b^2 - 4*a*c])])))/(a^2*(1 + m)*(2 + m)*(3 + m))
 

3.1.41.3 Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 687, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2193, 27, 1441, 1600, 25, 1608, 27, 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.

\(\displaystyle \int \frac {(d x)^m \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 2193

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1441

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

\(\Big \downarrow \) 1600

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1608

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 278

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

input

Int[((d*x)^m*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2,x]
 

output

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

3.1.41.3.1 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 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 1441

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 
1)/(2*a*d*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c)) 
  Int[(d*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[b^2*(m + 2*p + 3) - 2*a*c*(m 
 + 4*p + 5) + b*c*(m + 4*p + 7)*x^2, x], x], x] /; FreeQ[{a, b, c, d, m}, x 
] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] || 
IntegerQ[m])
 

rule 1600

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(a + b*x^2 + c*x^4)^(p + 1) 
*((d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2)/(2*a*f*(p + 1)*(b^2 - 4*a 
*c))), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(f*x)^m*(a + b*x^2 + c 
*x^4)^(p + 1)*Simp[d*(b^2*(m + 2*(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) 
- a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x], x] /; 
FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Int 
egerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1608

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (b_.)*(x_)^2 + (c_.) 
*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - 
b*e)/(2*q))   Int[(f*x)^m/(b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d 
- b*e)/(2*q))   Int[(f*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, 
 d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0]
 

rule 2193

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_S 
ymbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[Pq, x, 2*k]*x^(2*k), 
{k, 0, q/2 + 1}]*(d*x)^m*(a + b*x^2 + c*x^4)^p, x] + Simp[1/d   Int[Sum[Coe 
ff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q + 1)/2}]*(d*x)^(m + 1)*(a + b*x^2 + c 
*x^4)^p, x], x]] /; FreeQ[{a, b, c, d, m, p}, x] && PolyQ[Pq, x] &&  !PolyQ 
[Pq, x^2]
 

3.1.41.4 Maple [F]

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

input

int((d*x)^m*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x)
 

output

int((d*x)^m*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x)
 

3.1.41.5 Fricas [F]

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

input

integrate((d*x)^m*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")
 

output

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

3.1.41.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]

input

integrate((d*x)**m*(C*x**2+B*x+A)/(c*x**4+b*x**2+a)**2,x)
 

output

Timed out
 

3.1.41.7 Maxima [F]

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

input

integrate((d*x)^m*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")
 

output

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

3.1.41.8 Giac [F]

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

input

integrate((d*x)^m*(C*x^2+B*x+A)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")
 

output

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

3.1.41.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d x)^m \left (A+B x+C x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {{\left (d\,x\right )}^m\,\left (C\,x^2+B\,x+A\right )}{{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \]

input

int(((d*x)^m*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2,x)
 

output

int(((d*x)^m*(A + B*x + C*x^2))/(a + b*x^2 + c*x^4)^2, x)