\(\int \frac {(d+f x^2)^q}{(a+b x+c x^2)^2} \, dx\) [13]

Optimal result

Integrand size = 22, antiderivative size = 1032 \[ \int \frac {\left (d+f x^2\right )^q}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Output:

-(b*(b^2*f+c*(-3*a*f+c*d))+c*(-2*a*c*f+b^2*f+2*c^2*d)*x)*(f*x^2+d)^(1+q)/( 
-4*a*c+b^2)/(b^2*d*f+(-a*f+c*d)^2)/(c*x^2+b*x+a)+2*c*(2*c^3*d^2-4*a*c^2*d* 
f*(1-q)+a*b*(b+(-4*a*c+b^2)^(1/2))*f^2*q+c*f*(2*a^2*f*(1-2*q)+b^2*d*(2-q)+ 
b*(-4*a*c+b^2)^(1/2)*d*q))*x*(f*x^2+d)^q*AppellF1(1/2,1,-q,3/2,4*c^2*x^2/( 
b-(-4*a*c+b^2)^(1/2))^2,-f*x^2/d)/(-4*a*c+b^2)/(b^2-4*a*c-b*(-4*a*c+b^2)^( 
1/2))/(b^2*d*f+(-a*f+c*d)^2)/((1+f*x^2/d)^q)+2*c*(2*c^3*d^2-4*a*c^2*d*f*(1 
-q)+a*b*(b-(-4*a*c+b^2)^(1/2))*f^2*q+c*f*(2*a^2*f*(1-2*q)+b^2*d*(2-q)-b*(- 
4*a*c+b^2)^(1/2)*d*q))*x*(f*x^2+d)^q*AppellF1(1/2,1,-q,3/2,4*c^2*x^2/(b+(- 
4*a*c+b^2)^(1/2))^2,-f*x^2/d)/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2 
)/(b^2*d*f+(-a*f+c*d)^2)/((1+f*x^2/d)^q)+f*(-2*a*c*f+b^2*f+2*c^2*d)*(1+2*q 
)*x*(f*x^2+d)^q*hypergeom([1/2, -q],[3/2],-f*x^2/d)/(-4*a*c+b^2)/(b^2*d*f+ 
(-a*f+c*d)^2)/((1+f*x^2/d)^q)+2*c^2*(2*c^3*d^2-4*a*c^2*d*f*(1-q)+a*b*(b+(- 
4*a*c+b^2)^(1/2))*f^2*q+c*f*(2*a^2*f*(1-2*q)+b^2*d*(2-q)+b*(-4*a*c+b^2)^(1 
/2)*d*q))*(f*x^2+d)^(1+q)*hypergeom([1, 1+q],[2+q],4*c^2*(f*x^2+d)/(4*c^2* 
d+(b-(-4*a*c+b^2)^(1/2))^2*f))/(-4*a*c+b^2)^(3/2)/(4*c^2*d+(b-(-4*a*c+b^2) 
^(1/2))^2*f)/(b^2*d*f+(-a*f+c*d)^2)/(1+q)+2*c^2*(b*f*(a*f+c*d)*q-(2*c^3*d^ 
2+c*f*(2*a^2*f*(1-2*q)+b^2*d*(2-q))-4*a*c^2*d*f*(1-q)+a*b^2*f^2*q)/(-4*a*c 
+b^2)^(1/2))*(f*x^2+d)^(1+q)*hypergeom([1, 1+q],[2+q],4*c^2*(f*x^2+d)/(4*c 
^2*d+(b+(-4*a*c+b^2)^(1/2))^2*f))/(-4*a*c+b^2)/(4*c^2*d+(b+(-4*a*c+b^2)^(1 
/2))^2*f)/(b^2*d*f+(-a*f+c*d)^2)/(1+q)
 

Mathematica [F]

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

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

Output:

Integrate[(d + f*x^2)^q/(a + b*x + c*x^2)^2, 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.

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

\(\Big \downarrow \) 1306

\(\displaystyle \frac {\int -\frac {\left (f x^2+d\right )^q \left (2 d^2 c^3-6 a d f c^2-f \left (f b^2+2 c^2 d-2 a c f\right ) (2 q+1) x^2 c+2 f \left (2 f a^2+b^2 d\right ) c-a b^2 f^2-b f \left (f (2 q+1) b^2+2 c^2 d (q+1)-2 a c f (3 q+1)\right ) x\right )}{c x^2+b x+a}dx}{\left (b^2-4 a c\right ) \left ((c d-a f)^2+b^2 d f\right )}-\frac {\left (d+f x^2\right )^{q+1} \left (c x \left (-2 a c f+b^2 f+2 c^2 d\right )+b \left (c (c d-3 a f)+b^2 f\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left ((c d-a f)^2+b^2 d f\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \frac {\left (f x^2+d\right )^q \left (2 d^2 c^3-6 a d f c^2-f \left (f b^2+2 c^2 d-2 a c f\right ) (2 q+1) x^2 c+2 f \left (2 f a^2+b^2 d\right ) c-a b^2 f^2-b f \left (f (2 q+1) b^2+2 c^2 d (q+1)-2 a c f (3 q+1)\right ) x\right )}{c x^2+b x+a}dx}{\left (b^2-4 a c\right ) \left ((c d-a f)^2+b^2 d f\right )}-\frac {\left (d+f x^2\right )^{q+1} \left (c x \left (-2 a c f+b^2 f+2 c^2 d\right )+b \left (c (c d-3 a f)+b^2 f\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left ((c d-a f)^2+b^2 d f\right )}\)

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 25

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

rule 1306

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x 
_Symbol] :> Simp[(b^3*f + b*c*(c*d - 3*a*f) + c*(2*c^2*d + b^2*f - c*(2*a*f 
))*x)*(a + b*x + c*x^2)^(p + 1)*((d + f*x^2)^(q + 1)/((b^2 - 4*a*c)*(b^2*d* 
f + (c*d - a*f)^2)*(p + 1))), x] - Simp[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - 
a*f)^2)*(p + 1))   Int[(a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^q*Simp[2*c*(b^ 
2*d*f + (c*d - a*f)^2)*(p + 1) - (2*c^2*d + b^2*f - c*(2*a*f))*(a*f*(p + 1) 
 - c*d*(p + 2)) + (2*f*(b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + 
 b^2*f - c*(2*a*f))*(b*f*(p + 1)))*x + c*f*(2*c^2*d + b^2*f - c*(2*a*f))*(2 
*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x] && NeQ[b^2 - 
4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ 
[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

Maple [F]

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

Input:

int((f*x^2+d)^q/(c*x^2+b*x+a)^2,x)
 

Output:

int((f*x^2+d)^q/(c*x^2+b*x+a)^2,x)
 

Fricas [F]

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

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

Output:

integral((f*x^2 + d)^q/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 
+ a^2), x)
 

Sympy [F(-1)]

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

integrate((f*x**2+d)**q/(c*x**2+b*x+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((f*x^2 + d)^q/(c*x^2 + b*x + a)^2, x)
 

Giac [F]

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

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

Output:

integrate((f*x^2 + d)^q/(c*x^2 + b*x + a)^2, x)
 

Mupad [F(-1)]

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

int((d + f*x^2)^q/(a + b*x + c*x^2)^2,x)
 

Output:

int((d + f*x^2)^q/(a + b*x + c*x^2)^2, x)
 

Reduce [F]

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

int((f*x^2+d)^q/(c*x^2+b*x+a)^2,x)
 

Output:

int((d + f*x**2)**q/(a**2 + 2*a*b*x + 2*a*c*x**2 + b**2*x**2 + 2*b*c*x**3 
+ c**2*x**4),x)