\(\int \frac {(d+e x^2)^q}{(a+c x^4)^3} \, dx\) [479]

Optimal result

Integrand size = 19, antiderivative size = 553 \[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\frac {c x \left (d-e x^2\right ) \left (d+e x^2\right )^{1+q}}{8 a \left (c d^2+a e^2\right ) \left (a+c x^4\right )^2}+\frac {c x \left (d+e x^2\right )^{1+q} \left (d \left (7 c d^2+a e^2 (11-4 q)\right )-e \left (a e^2 (11-2 q)+c d^2 (7+2 q)\right ) x^2\right )}{32 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^4\right )}+\frac {\left (21 c^2 d^4+8 \sqrt {-a} \sqrt {c} d e \left (2 c d^2+a e^2 (3-q)\right ) q+2 a c d^2 e^2 \left (21-6 q-2 q^2\right )+a^2 e^4 \left (21-20 q+4 q^2\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},-\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{64 a^3 \left (c d^2+a e^2\right )^2}+\frac {\left (21 c^2 d^4-8 \sqrt {-a} \sqrt {c} d e \left (2 c d^2+a e^2 (3-q)\right ) q+2 a c d^2 e^2 \left (21-6 q-2 q^2\right )+a^2 e^4 \left (21-20 q+4 q^2\right )\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {AppellF1}\left (\frac {1}{2},-q,1,\frac {3}{2},-\frac {e x^2}{d},\frac {\sqrt {c} x^2}{\sqrt {-a}}\right )}{64 a^3 \left (c d^2+a e^2\right )^2}+\frac {e^2 (1+2 q) \left (a e^2 (11-2 q)+c d^2 (7+2 q)\right ) x \left (d+e x^2\right )^q \left (1+\frac {e x^2}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-q,\frac {3}{2},-\frac {e x^2}{d}\right )}{32 a^2 \left (c d^2+a e^2\right )^2} \] Output:

1/8*c*x*(-e*x^2+d)*(e*x^2+d)^(1+q)/a/(a*e^2+c*d^2)/(c*x^4+a)^2+1/32*c*x*(e 
*x^2+d)^(1+q)*(d*(7*c*d^2+a*e^2*(11-4*q))-e*(a*e^2*(11-2*q)+c*d^2*(7+2*q)) 
*x^2)/a^2/(a*e^2+c*d^2)^2/(c*x^4+a)+1/64*(21*c^2*d^4+8*(-a)^(1/2)*c^(1/2)* 
d*e*(2*c*d^2+a*e^2*(3-q))*q+2*a*c*d^2*e^2*(-2*q^2-6*q+21)+a^2*e^4*(4*q^2-2 
0*q+21))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,-c^(1/2)*x^2/(-a)^(1/2),-e*x^ 
2/d)/a^3/(a*e^2+c*d^2)^2/((1+e*x^2/d)^q)+1/64*(21*c^2*d^4-8*(-a)^(1/2)*c^( 
1/2)*d*e*(2*c*d^2+a*e^2*(3-q))*q+2*a*c*d^2*e^2*(-2*q^2-6*q+21)+a^2*e^4*(4* 
q^2-20*q+21))*x*(e*x^2+d)^q*AppellF1(1/2,1,-q,3/2,c^(1/2)*x^2/(-a)^(1/2),- 
e*x^2/d)/a^3/(a*e^2+c*d^2)^2/((1+e*x^2/d)^q)+1/32*e^2*(1+2*q)*(a*e^2*(11-2 
*q)+c*d^2*(7+2*q))*x*(e*x^2+d)^q*hypergeom([1/2, -q],[3/2],-e*x^2/d)/a^2/( 
a*e^2+c*d^2)^2/((1+e*x^2/d)^q)
 

Mathematica [F]

\[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx \] Input:

Integrate[(d + e*x^2)^q/(a + c*x^4)^3,x]
 

Output:

Integrate[(d + e*x^2)^q/(a + c*x^4)^3, x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 1571

\(\displaystyle \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3}dx\)

Input:

Int[(d + e*x^2)^q/(a + c*x^4)^3,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 1571

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> U 
nintegrable[(d + e*x^2)^q*(a + c*x^4)^p, x] /; FreeQ[{a, c, d, e, p, q}, x]
 

Maple [F]

\[\int \frac {\left (e \,x^{2}+d \right )^{q}}{\left (c \,x^{4}+a \right )^{3}}d x\]

Input:

int((e*x^2+d)^q/(c*x^4+a)^3,x)
 

Output:

int((e*x^2+d)^q/(c*x^4+a)^3,x)
 

Fricas [F]

\[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{3}} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a)^3,x, algorithm="fricas")
 

Output:

integral((e*x^2 + d)^q/(c^3*x^12 + 3*a*c^2*x^8 + 3*a^2*c*x^4 + a^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**q/(c*x**4+a)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{3}} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a)^3,x, algorithm="maxima")
 

Output:

integrate((e*x^2 + d)^q/(c*x^4 + a)^3, x)
 

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{q}}{{\left (c x^{4} + a\right )}^{3}} \,d x } \] Input:

integrate((e*x^2+d)^q/(c*x^4+a)^3,x, algorithm="giac")
 

Output:

integrate((e*x^2 + d)^q/(c*x^4 + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int \frac {{\left (e\,x^2+d\right )}^q}{{\left (c\,x^4+a\right )}^3} \,d x \] Input:

int((d + e*x^2)^q/(a + c*x^4)^3,x)
 

Output:

int((d + e*x^2)^q/(a + c*x^4)^3, x)
 

Reduce [F]

\[ \int \frac {\left (d+e x^2\right )^q}{\left (a+c x^4\right )^3} \, dx=\int \frac {\left (e \,x^{2}+d \right )^{q}}{c^{3} x^{12}+3 a \,c^{2} x^{8}+3 a^{2} c \,x^{4}+a^{3}}d x \] Input:

int((e*x^2+d)^q/(c*x^4+a)^3,x)
 

Output:

int((d + e*x**2)**q/(a**3 + 3*a**2*c*x**4 + 3*a*c**2*x**8 + c**3*x**12),x)