\(\int (a+b x+c x^2)^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} (d+f x^2) \, dx\) [14]

Optimal result

Integrand size = 52, antiderivative size = 119 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=-\frac {\left (4 c^2 d+b^2 f\right ) \left (b (c d+a f)+\left (2 c^2 d+b^2 f-2 a c f\right ) x\right ) \left (a+b x+c x^2\right )^{-\frac {2 c^2 d+b^2 f-2 a c f}{4 c^2 d+b^2 f}}}{c \left (b^2-4 a c\right ) \left (2 c^2 d+b^2 f-2 a c f\right )} \] Output:

-(b^2*f+4*c^2*d)*(b*(a*f+c*d)+(-2*a*c*f+b^2*f+2*c^2*d)*x)/c/(-4*a*c+b^2)/( 
-2*a*c*f+b^2*f+2*c^2*d)/((c*x^2+b*x+a)^((-2*a*c*f+b^2*f+2*c^2*d)/(b^2*f+4* 
c^2*d)))
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=\frac {\left (4 c^2 d+b^2 f\right ) \left (b (c d+a f)+b^2 f x+2 c (c d-a f) x\right ) (a+x (b+c x))^{-\frac {2 c^2 d+b^2 f-2 a c f}{4 c^2 d+b^2 f}}}{c \left (-b^2+4 a c\right ) \left (2 c^2 d+b^2 f-2 a c f\right )} \] Input:

Integrate[(a + b*x + c*x^2)^((-6*c^2*d - 2*b^2*f + 2*a*c*f)/(4*c^2*d + b^2 
*f))*(d + f*x^2),x]
 

Output:

((4*c^2*d + b^2*f)*(b*(c*d + a*f) + b^2*f*x + 2*c*(c*d - a*f)*x))/(c*(-b^2 
 + 4*a*c)*(2*c^2*d + b^2*f - 2*a*c*f)*(a + x*(b + c*x))^((2*c^2*d + b^2*f 
- 2*a*c*f)/(4*c^2*d + b^2*f)))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2192, 25, 27, 1104}

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.

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

\(\Big \downarrow \) 2192

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1104

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

Input:

Int[(a + b*x + c*x^2)^((-6*c^2*d - 2*b^2*f + 2*a*c*f)/(4*c^2*d + b^2*f))*( 
d + f*x^2),x]
 

Output:

-((b*(c*d + a*f)*(4*c^2*d + b^2*f))/(c*(b^2 - 4*a*c)*(2*c^2*d + b^2*f - 2* 
a*c*f)*(a + b*x + c*x^2)^((2*c^2*d + b^2*f - 2*a*c*f)/(4*c^2*d + b^2*f)))) 
 - ((4*c^2*d + b^2*f)*x)/(c*(b^2 - 4*a*c)*(a + b*x + c*x^2)^((2*c^2*d + b^ 
2*f - 2*a*c*f)/(4*c^2*d + b^2*f)))
 

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 1104

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

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.12

method	result	size
gosper	\(-\frac {\left (b^{2} f +4 c^{2} d \right ) \left (c \,x^{2}+b x +a \right )^{\frac {2 a c f -2 b^{2} f -6 c^{2} d}{b^{2} f +4 c^{2} d}+1} \left (-2 a c f x +b^{2} f x +2 c^{2} d x +a b f +d b c \right )}{c \left (8 a^{2} c^{2} f -6 a \,b^{2} c f -8 a \,c^{3} d +b^{4} f +2 b^{2} c^{2} d \right )}\)	\(133\)
orering	\(-\frac {\left (-2 a \,b^{2} c \,f^{2} x -8 a \,c^{3} d f x +b^{4} f^{2} x +6 b^{2} c^{2} d f x +8 c^{4} d^{2} x +a \,b^{3} f^{2}+4 a b \,c^{2} d f +b^{3} c d f +4 b \,c^{3} d^{2}\right ) \left (c \,x^{2}+b x +a \right ) \left (c \,x^{2}+b x +a \right )^{\frac {2 a c f -2 b^{2} f -6 c^{2} d}{b^{2} f +4 c^{2} d}}}{c \left (8 a^{2} c^{2} f -6 a \,b^{2} c f -8 a \,c^{3} d +b^{4} f +2 b^{2} c^{2} d \right )}\)	\(183\)
risch	\(-\frac {\left (b^{2} f +4 c^{2} d \right ) \left (-2 a \,c^{2} f \,x^{3}+b^{2} c f \,x^{3}+2 c^{3} d \,x^{3}-a b c f \,x^{2}+b^{3} f \,x^{2}+3 b \,c^{2} d \,x^{2}-2 a^{2} c f x +2 a \,b^{2} f x +2 a d x \,c^{2}+b^{2} c x d +f \,a^{2} b +a b c d \right ) \left (c \,x^{2}+b x +a \right )^{\frac {2 a c f -2 b^{2} f -6 c^{2} d}{b^{2} f +4 c^{2} d}}}{c \left (4 a c -b^{2}\right ) \left (2 a c f -b^{2} f -2 c^{2} d \right )}\)	\(192\)
norman	\(\frac {\left (b^{2} f +4 c^{2} d \right ) x^{3} {\mathrm e}^{\frac {\left (2 a c f -2 b^{2} f -6 c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{b^{2} f +4 c^{2} d}}}{4 a c -b^{2}}+\frac {\left (2 a^{2} b^{2} c \,f^{2}+8 a^{2} c^{3} d f -2 a \,b^{4} f^{2}-10 a \,b^{2} c^{2} d f -8 a \,c^{4} d^{2}-b^{4} c d f -4 c^{3} b^{2} d^{2}\right ) x \,{\mathrm e}^{\frac {\left (2 a c f -2 b^{2} f -6 c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{b^{2} f +4 c^{2} d}}}{c \left (8 a^{2} c^{2} f -6 a \,b^{2} c f -8 a \,c^{3} d +b^{4} f +2 b^{2} c^{2} d \right )}+\frac {b \left (a \,b^{2} c \,f^{2}+4 a \,c^{3} d f -b^{4} f^{2}-7 b^{2} c^{2} d f -12 c^{4} d^{2}\right ) x^{2} {\mathrm e}^{\frac {\left (2 a c f -2 b^{2} f -6 c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{b^{2} f +4 c^{2} d}}}{c \left (8 a^{2} c^{2} f -6 a \,b^{2} c f -8 a \,c^{3} d +b^{4} f +2 b^{2} c^{2} d \right )}-\frac {\left (a \,b^{2} f^{2}+4 a \,c^{2} d f +b^{2} c d f +4 c^{3} d^{2}\right ) a b \,{\mathrm e}^{\frac {\left (2 a c f -2 b^{2} f -6 c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{b^{2} f +4 c^{2} d}}}{c \left (8 a^{2} c^{2} f -6 a \,b^{2} c f -8 a \,c^{3} d +b^{4} f +2 b^{2} c^{2} d \right )}\)	\(500\)
parallelrisch	\(\text {Expression too large to display}\)	\(1211\)

Input:

int((c*x^2+b*x+a)^((2*a*c*f-2*b^2*f-6*c^2*d)/(b^2*f+4*c^2*d))*(f*x^2+d),x, 
method=_RETURNVERBOSE)
 

Output:

-1/c*(b^2*f+4*c^2*d)*(c*x^2+b*x+a)^(1+2*(a*c*f-b^2*f-3*c^2*d)/(b^2*f+4*c^2 
*d))/(8*a^2*c^2*f-6*a*b^2*c*f-8*a*c^3*d+b^4*f+2*b^2*c^2*d)*(-2*a*c*f*x+b^2 
*f*x+2*c^2*d*x+a*b*f+b*c*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (120) = 240\).

Time = 0.08 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.52 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=-\frac {4 \, a b c^{3} d^{2} + a^{2} b^{3} f^{2} + {\left (8 \, c^{5} d^{2} + 2 \, {\left (3 \, b^{2} c^{3} - 4 \, a c^{4}\right )} d f + {\left (b^{4} c - 2 \, a b^{2} c^{2}\right )} f^{2}\right )} x^{3} + {\left (a b^{3} c + 4 \, a^{2} b c^{2}\right )} d f + {\left (12 \, b c^{4} d^{2} + {\left (7 \, b^{3} c^{2} - 4 \, a b c^{3}\right )} d f + {\left (b^{5} - a b^{3} c\right )} f^{2}\right )} x^{2} + {\left (4 \, {\left (b^{2} c^{3} + 2 \, a c^{4}\right )} d^{2} + {\left (b^{4} c + 10 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} d f + 2 \, {\left (a b^{4} - a^{2} b^{2} c\right )} f^{2}\right )} x}{{\left (2 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d + {\left (b^{4} c - 6 \, a b^{2} c^{2} + 8 \, a^{2} c^{3}\right )} f\right )} {\left (c x^{2} + b x + a\right )}^{\frac {2 \, {\left (3 \, c^{2} d + {\left (b^{2} - a c\right )} f\right )}}{4 \, c^{2} d + b^{2} f}}} \] Input:

integrate((c*x^2+b*x+a)^((2*a*c*f-2*b^2*f-6*c^2*d)/(b^2*f+4*c^2*d))*(f*x^2 
+d),x, algorithm="fricas")
 

Output:

-(4*a*b*c^3*d^2 + a^2*b^3*f^2 + (8*c^5*d^2 + 2*(3*b^2*c^3 - 4*a*c^4)*d*f + 
 (b^4*c - 2*a*b^2*c^2)*f^2)*x^3 + (a*b^3*c + 4*a^2*b*c^2)*d*f + (12*b*c^4* 
d^2 + (7*b^3*c^2 - 4*a*b*c^3)*d*f + (b^5 - a*b^3*c)*f^2)*x^2 + (4*(b^2*c^3 
 + 2*a*c^4)*d^2 + (b^4*c + 10*a*b^2*c^2 - 8*a^2*c^3)*d*f + 2*(a*b^4 - a^2* 
b^2*c)*f^2)*x)/((2*(b^2*c^3 - 4*a*c^4)*d + (b^4*c - 6*a*b^2*c^2 + 8*a^2*c^ 
3)*f)*(c*x^2 + b*x + a)^(2*(3*c^2*d + (b^2 - a*c)*f)/(4*c^2*d + b^2*f)))
 

Sympy [F(-1)]

Timed out. \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**((2*a*c*f-2*b**2*f-6*c**2*d)/(b**2*f+4*c**2*d))* 
(f*x**2+d),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (120) = 240\).

Time = 0.14 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.99 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=-\frac {{\left ({\left (8 \, c^{5} d^{2} + 6 \, b^{2} c^{3} d f + b^{4} c f^{2} - 2 \, {\left (4 \, c^{4} d f + b^{2} c^{2} f^{2}\right )} a\right )} x^{3} + {\left (4 \, b c^{2} d f + b^{3} f^{2}\right )} a^{2} + {\left (12 \, b c^{4} d^{2} + 7 \, b^{3} c^{2} d f + b^{5} f^{2} - {\left (4 \, b c^{3} d f + b^{3} c f^{2}\right )} a\right )} x^{2} + {\left (4 \, b c^{3} d^{2} + b^{3} c d f\right )} a + {\left (4 \, b^{2} c^{3} d^{2} + b^{4} c d f - 2 \, {\left (4 \, c^{3} d f + b^{2} c f^{2}\right )} a^{2} + 2 \, {\left (4 \, c^{4} d^{2} + 5 \, b^{2} c^{2} d f + b^{4} f^{2}\right )} a\right )} x\right )} e^{\left (-\frac {6 \, c^{2} d \log \left (c x^{2} + b x + a\right )}{4 \, c^{2} d + b^{2} f} - \frac {2 \, b^{2} f \log \left (c x^{2} + b x + a\right )}{4 \, c^{2} d + b^{2} f} + \frac {2 \, a c f \log \left (c x^{2} + b x + a\right )}{4 \, c^{2} d + b^{2} f}\right )}}{2 \, b^{2} c^{3} d + b^{4} c f + 8 \, a^{2} c^{3} f - 2 \, {\left (4 \, c^{4} d + 3 \, b^{2} c^{2} f\right )} a} \] Input:

integrate((c*x^2+b*x+a)^((2*a*c*f-2*b^2*f-6*c^2*d)/(b^2*f+4*c^2*d))*(f*x^2 
+d),x, algorithm="maxima")
 

Output:

-((8*c^5*d^2 + 6*b^2*c^3*d*f + b^4*c*f^2 - 2*(4*c^4*d*f + b^2*c^2*f^2)*a)* 
x^3 + (4*b*c^2*d*f + b^3*f^2)*a^2 + (12*b*c^4*d^2 + 7*b^3*c^2*d*f + b^5*f^ 
2 - (4*b*c^3*d*f + b^3*c*f^2)*a)*x^2 + (4*b*c^3*d^2 + b^3*c*d*f)*a + (4*b^ 
2*c^3*d^2 + b^4*c*d*f - 2*(4*c^3*d*f + b^2*c*f^2)*a^2 + 2*(4*c^4*d^2 + 5*b 
^2*c^2*d*f + b^4*f^2)*a)*x)*e^(-6*c^2*d*log(c*x^2 + b*x + a)/(4*c^2*d + b^ 
2*f) - 2*b^2*f*log(c*x^2 + b*x + a)/(4*c^2*d + b^2*f) + 2*a*c*f*log(c*x^2 
+ b*x + a)/(4*c^2*d + b^2*f))/(2*b^2*c^3*d + b^4*c*f + 8*a^2*c^3*f - 2*(4* 
c^4*d + 3*b^2*c^2*f)*a)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1690 vs. \(2 (120) = 240\).

Time = 0.53 (sec) , antiderivative size = 1690, normalized size of antiderivative = 14.20 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=\text {Too large to display} \] Input:

integrate((c*x^2+b*x+a)^((2*a*c*f-2*b^2*f-6*c^2*d)/(b^2*f+4*c^2*d))*(f*x^2 
+d),x, algorithm="giac")
 

Output:

-(8*c^5*d^2*x^3*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b* 
x + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) + 6*b^2*c^3*d*f*x^ 
3*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f 
*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) - 8*a*c^4*d*f*x^3*e^(-2*(3*c^2*d 
*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f*log(c*x^2 + b*x 
 + a))/(4*c^2*d + b^2*f)) + b^4*c*f^2*x^3*e^(-2*(3*c^2*d*log(c*x^2 + b*x + 
 a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + 
b^2*f)) - 2*a*b^2*c^2*f^2*x^3*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f* 
log(c*x^2 + b*x + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) + 12 
*b*c^4*d^2*x^2*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x 
 + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) + 7*b^3*c^2*d*f*x^2 
*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f* 
log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) - 4*a*b*c^3*d*f*x^2*e^(-2*(3*c^2* 
d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f*log(c*x^2 + b* 
x + a))/(4*c^2*d + b^2*f)) + b^5*f^2*x^2*e^(-2*(3*c^2*d*log(c*x^2 + b*x + 
a) + b^2*f*log(c*x^2 + b*x + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b 
^2*f)) - a*b^3*c*f^2*x^2*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c 
*x^2 + b*x + a) - a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) + 4*b^2*c 
^3*d^2*x*e^(-2*(3*c^2*d*log(c*x^2 + b*x + a) + b^2*f*log(c*x^2 + b*x + a) 
- a*c*f*log(c*x^2 + b*x + a))/(4*c^2*d + b^2*f)) + 8*a*c^4*d^2*x*e^(-2*...
 

Mupad [B] (verification not implemented)

Time = 16.37 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.32 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=\frac {\frac {x^3\,\left (f\,b^2+4\,d\,c^2\right )}{4\,a\,c-b^2}+\frac {x\,\left (f\,b^2+4\,d\,c^2\right )\,\left (-2\,f\,a^2\,c+2\,f\,a\,b^2+2\,d\,a\,c^2+d\,b^2\,c\right )}{c\,\left (4\,a\,c-b^2\right )\,\left (f\,b^2+2\,d\,c^2-2\,a\,f\,c\right )}+\frac {b\,x^2\,\left (f\,b^2+4\,d\,c^2\right )\,\left (f\,b^2+3\,d\,c^2-a\,f\,c\right )}{c\,\left (4\,a\,c-b^2\right )\,\left (f\,b^2+2\,d\,c^2-2\,a\,f\,c\right )}+\frac {a\,b\,\left (f\,b^2+4\,d\,c^2\right )\,\left (a\,f+c\,d\right )}{c\,\left (4\,a\,c-b^2\right )\,\left (f\,b^2+2\,d\,c^2-2\,a\,f\,c\right )}}{{\left (c\,x^2+b\,x+a\right )}^{\frac {2\,f\,b^2+6\,d\,c^2-2\,a\,f\,c}{f\,b^2+4\,d\,c^2}}} \] Input:

int((d + f*x^2)/(a + b*x + c*x^2)^((6*c^2*d + 2*b^2*f - 2*a*c*f)/(4*c^2*d 
+ b^2*f)),x)
 

Output:

((x^3*(4*c^2*d + b^2*f))/(4*a*c - b^2) + (x*(4*c^2*d + b^2*f)*(2*a*c^2*d + 
 2*a*b^2*f + b^2*c*d - 2*a^2*c*f))/(c*(4*a*c - b^2)*(2*c^2*d + b^2*f - 2*a 
*c*f)) + (b*x^2*(4*c^2*d + b^2*f)*(3*c^2*d + b^2*f - a*c*f))/(c*(4*a*c - b 
^2)*(2*c^2*d + b^2*f - 2*a*c*f)) + (a*b*(4*c^2*d + b^2*f)*(a*f + c*d))/(c* 
(4*a*c - b^2)*(2*c^2*d + b^2*f - 2*a*c*f)))/(a + b*x + c*x^2)^((6*c^2*d + 
2*b^2*f - 2*a*c*f)/(4*c^2*d + b^2*f))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.31 \[ \int \left (a+b x+c x^2\right )^{\frac {-6 c^2 d-2 b^2 f+2 a c f}{4 c^2 d+b^2 f}} \left (d+f x^2\right ) \, dx=\frac {\left (c \,x^{2}+b x +a \right )^{\frac {2 a c f +2 c^{2} d}{b^{2} f +4 c^{2} d}} \left (2 a \,b^{2} c \,f^{2} x +8 a \,c^{3} d f x -b^{4} f^{2} x -6 b^{2} c^{2} d f x -8 c^{4} d^{2} x -a \,b^{3} f^{2}-4 a b \,c^{2} d f -b^{3} c d f -4 b \,c^{3} d^{2}\right )}{c \left (8 a^{2} c^{3} f \,x^{2}-6 a \,b^{2} c^{2} f \,x^{2}-8 a \,c^{4} d \,x^{2}+b^{4} c f \,x^{2}+2 b^{2} c^{3} d \,x^{2}+8 a^{2} b \,c^{2} f x -6 a \,b^{3} c f x -8 a b \,c^{3} d x +b^{5} f x +2 b^{3} c^{2} d x +8 a^{3} c^{2} f -6 a^{2} b^{2} c f -8 a^{2} c^{3} d +a \,b^{4} f +2 a \,b^{2} c^{2} d \right )} \] Input:

int((c*x^2+b*x+a)^((2*a*c*f-2*b^2*f-6*c^2*d)/(b^2*f+4*c^2*d))*(f*x^2+d),x)
 

Output:

((a + b*x + c*x**2)**((2*a*c*f + 2*c**2*d)/(b**2*f + 4*c**2*d))*( - a*b**3 
*f**2 + 2*a*b**2*c*f**2*x - 4*a*b*c**2*d*f + 8*a*c**3*d*f*x - b**4*f**2*x 
- b**3*c*d*f - 6*b**2*c**2*d*f*x - 4*b*c**3*d**2 - 8*c**4*d**2*x))/(c*(8*a 
**3*c**2*f - 6*a**2*b**2*c*f + 8*a**2*b*c**2*f*x - 8*a**2*c**3*d + 8*a**2* 
c**3*f*x**2 + a*b**4*f - 6*a*b**3*c*f*x + 2*a*b**2*c**2*d - 6*a*b**2*c**2* 
f*x**2 - 8*a*b*c**3*d*x - 8*a*c**4*d*x**2 + b**5*f*x + b**4*c*f*x**2 + 2*b 
**3*c**2*d*x + 2*b**2*c**3*d*x**2))