\(\int \frac {(d x)^m}{(a+b x^2+c x^4)^2} \, dx\) [1087]

Optimal result

Integrand size = 20, antiderivative size = 315 \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^2\right )}{2 a \left (b^2-4 a c\right ) d \left (a+b x^2+c x^4\right )}+\frac {c \left (b^2 (1-m)+b \sqrt {b^2-4 a c} (1-m)-4 a c (3-m)\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}-\frac {c \left (b^2 (1-m)-b \sqrt {b^2-4 a c} (1-m)-4 a c (3-m)\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m)} \] Output:

1/2*(d*x)^(1+m)*(b*c*x^2-2*a*c+b^2)/a/(-4*a*c+b^2)/d/(c*x^4+b*x^2+a)+1/2*c 
*(b^2*(1-m)+b*(-4*a*c+b^2)^(1/2)*(1-m)-4*a*c*(3-m))*(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)))/a/(-4*a*c+b^2) 
^(3/2)/(b-(-4*a*c+b^2)^(1/2))/d/(1+m)-1/2*c*(b^2*(1-m)-b*(-4*a*c+b^2)^(1/2 
)*(1-m)-4*a*c*(3-m))*(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)))/a/(-4*a*c+b^2)^(3/2)/(b+(-4*a*c+b^2)^(1/2))/d 
/(1+m)
 

Mathematica [C] (verified)

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

Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.25 \[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {x (d x)^m \operatorname {AppellF1}\left (\frac {1+m}{2},2,2,\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )}{a^2 (1+m)} \] Input:

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

Output:

(x*(d*x)^m*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])])/(a^2*(1 + m))
 

Rubi [A] (verified)

Time = 1.04 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1441, 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.

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

\(\Big \downarrow \) 1441

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1608

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

((d*x)^(1 + 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*(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 - S 
qrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])*d*(1 + m)) 
- (c*(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)))/(2*a*(b^ 
2 - 4*a*c))
 

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 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((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)
 

Sympy [F]

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

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

Output:

Integral((d*x)**m/(a + b*x**2 + c*x**4)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d x)^m}{\left (a+b x^2+c x^4\right )^2} \, dx=d^{m} \left (\int \frac {x^{m}}{c^{2} x^{8}+2 b c \,x^{6}+2 a c \,x^{4}+b^{2} x^{4}+2 a b \,x^{2}+a^{2}}d x \right ) \] Input:

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

Output:

d**m*int(x**m/(a**2 + 2*a*b*x**2 + 2*a*c*x**4 + b**2*x**4 + 2*b*c*x**6 + c 
**2*x**8),x)