\(\int \frac {(d+e x+f x^2)^2}{(a+b x+c x^2)^{5/2}} \, dx\) [159]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 27, antiderivative size = 440 \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)-b c \left (c^2 d^2-3 a^2 f^2+a c \left (e^2+2 d f\right )\right )-\left (2 c^4 d^2+b^4 f^2-2 b^2 c f (b e+2 a f)-2 c^3 \left (b d e+a \left (e^2+2 d f\right )\right )+c^2 \left (6 a b e f+2 a^2 f^2+b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \left (2 b^4 e f+48 a^2 c^2 e f-\frac {b^5 f^2}{c}+4 b^2 c e (2 c d-3 a f)+b^3 \left (10 a f^2-c \left (e^2+2 d f\right )\right )-4 b c \left (2 c^2 d^2+8 a^2 f^2+a c \left (e^2+2 d f\right )\right )-2 \left (8 c^4 d^2-2 b^4 f^2+b^2 c f (b e+14 a f)-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (12 a b e f+16 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )\right ) x\right )}{3 c^2 \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}+\frac {f^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \] Output:

2/3*(2*a*b^2*c*e*f-a*b^3*f^2+4*a*c^2*e*(-a*f+c*d)-b*c*(c^2*d^2-3*a^2*f^2+a 
*c*(2*d*f+e^2))-(2*c^4*d^2+b^4*f^2-2*b^2*c*f*(2*a*f+b*e)-2*c^3*(b*d*e+a*(2 
*d*f+e^2))+c^2*(6*a*b*e*f+2*a^2*f^2+b^2*(2*d*f+e^2)))*x)/c^3/(-4*a*c+b^2)/ 
(c*x^2+b*x+a)^(3/2)-2/3*(2*b^4*e*f+48*a^2*c^2*e*f-b^5*f^2/c+4*b^2*c*e*(-3* 
a*f+2*c*d)+b^3*(10*a*f^2-c*(2*d*f+e^2))-4*b*c*(2*c^2*d^2+8*a^2*f^2+a*c*(2* 
d*f+e^2))-2*(8*c^4*d^2-2*b^4*f^2+b^2*c*f*(14*a*f+b*e)-c^3*(8*b*d*e-4*a*(2* 
d*f+e^2))-c^2*(12*a*b*e*f+16*a^2*f^2-b^2*(2*d*f+e^2)))*x)/c^2/(-4*a*c+b^2) 
^2/(c*x^2+b*x+a)^(1/2)+f^2*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/ 
2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 4.03 (sec) , antiderivative size = 392, normalized size of antiderivative = 0.89 \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (-3 b^5 f^2 x^2-2 b^4 f^2 x \left (3 a+2 c x^2\right )+4 b c \left (5 a^3 f^2+2 c^3 d x^2 (3 d-2 e x)+2 a^2 c \left (e^2+2 d f-6 e f x\right )+3 a c^2 (d-e x) (d+x (-e+2 f x))\right )+b^3 \left (-3 a^2 f^2+18 a c f^2 x^2+c^2 \left (-d^2+6 d x (-e+f x)+e x^2 (3 e+2 f x)\right )\right )+8 c^2 \left (2 c^3 d^2 x^3-a^3 f (4 e+3 f x)+a c^2 x \left (3 d^2+e^2 x^2+2 d f x^2\right )-2 a^2 c \left (d e+f x^2 (3 e+2 f x)\right )\right )+2 b^2 c \left (21 a^2 f^2 x+c^2 x \left (3 d^2+e^2 x^2+2 d x (-6 e+f x)\right )-2 a c \left (d (e-6 f x)+x \left (-3 e^2+3 e f x-7 f^2 x^2\right )\right )\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}+\frac {2 f^2 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{5/2}} \] Input:

Integrate[(d + e*x + f*x^2)^2/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*(-3*b^5*f^2*x^2 - 2*b^4*f^2*x*(3*a + 2*c*x^2) + 4*b*c*(5*a^3*f^2 + 2*c^ 
3*d*x^2*(3*d - 2*e*x) + 2*a^2*c*(e^2 + 2*d*f - 6*e*f*x) + 3*a*c^2*(d - e*x 
)*(d + x*(-e + 2*f*x))) + b^3*(-3*a^2*f^2 + 18*a*c*f^2*x^2 + c^2*(-d^2 + 6 
*d*x*(-e + f*x) + e*x^2*(3*e + 2*f*x))) + 8*c^2*(2*c^3*d^2*x^3 - a^3*f*(4* 
e + 3*f*x) + a*c^2*x*(3*d^2 + e^2*x^2 + 2*d*f*x^2) - 2*a^2*c*(d*e + f*x^2* 
(3*e + 2*f*x))) + 2*b^2*c*(21*a^2*f^2*x + c^2*x*(3*d^2 + e^2*x^2 + 2*d*x*( 
-6*e + f*x)) - 2*a*c*(d*(e - 6*f*x) + x*(-3*e^2 + 3*e*f*x - 7*f^2*x^2))))) 
/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) + (2*f^2*ArcTanh[(Sqrt[c] 
*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/c^(5/2)
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2191, 27, 2191, 27, 1092, 219}

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.

\(\displaystyle \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2191

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {2 \int \frac {3 \left (4 a-\frac {b^2}{c}\right ) f^2 x^2-\frac {3 \left (b^2-4 a c\right ) f (2 c e-b f) x}{c^2}+\frac {f^2 b^4-c f (2 b e+a f) b^2+8 c^4 d^2-c^3 \left (8 b d e-4 a \left (e^2+2 d f\right )\right )-c^2 \left (4 a^2 f^2-b^2 \left (e^2+2 d f\right )\right )}{c^3}}{2 \left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\int \frac {\frac {f^2 b^4}{c^3}-\frac {f (2 b e+a f) b^2}{c^2}-8 d e b+8 c d^2+3 \left (4 a-\frac {b^2}{c}\right ) f^2 x^2+4 a \left (e^2+2 d f\right )-\frac {4 a^2 f^2-b^2 \left (e^2+2 d f\right )}{c}-\frac {3 \left (b^2-4 a c\right ) f (2 c e-b f) x}{c^2}}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 2191

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f-\left (b^2 \left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+4 b^2 c^2 e (2 c d-3 a f)+b^5 \left (-f^2\right )+2 b^4 c e f\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {3 \left (b^2-4 a c\right )^2 f^2}{2 c^2 \sqrt {c x^2+b x+a}}dx}{b^2-4 a c}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f-\left (b^2 \left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+4 b^2 c^2 e (2 c d-3 a f)+b^5 \left (-f^2\right )+2 b^4 c e f\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {3 f^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{c^2}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f-\left (b^2 \left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+4 b^2 c^2 e (2 c d-3 a f)+b^5 \left (-f^2\right )+2 b^4 c e f\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {6 f^2 \left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c^2}}{3 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )-a b^3 f^2+2 a b^2 c e f+4 a c^2 e (c d-a f)\right )}{3 c^3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac {\frac {2 \left (-2 c x \left (-c^2 \left (16 a^2 f^2+12 a b e f-\left (b^2 \left (2 d f+e^2\right )\right )\right )+b^2 c f (14 a f+b e)-c^3 \left (8 b d e-4 a \left (2 d f+e^2\right )\right )-2 b^4 f^2+8 c^4 d^2\right )-4 b c^2 \left (8 a^2 f^2+a c \left (2 d f+e^2\right )+2 c^2 d^2\right )+48 a^2 c^3 e f+b^3 c \left (10 a f^2-c \left (2 d f+e^2\right )\right )+4 b^2 c^2 e (2 c d-3 a f)+b^5 \left (-f^2\right )+2 b^4 c e f\right )}{c^3 \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {3 f^2 \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}}}{3 \left (b^2-4 a c\right )}\)

Input:

Int[(d + e*x + f*x^2)^2/(a + b*x + c*x^2)^(5/2),x]
 

Output:

(2*(2*a*b^2*c*e*f - a*b^3*f^2 + 4*a*c^2*e*(c*d - a*f) - b*c*(c^2*d^2 - 3*a 
^2*f^2 + a*c*(e^2 + 2*d*f)) - (2*c^4*d^2 + b^4*f^2 - 2*b^2*c*f*(b*e + 2*a* 
f) - 2*c^3*(b*d*e + a*(e^2 + 2*d*f)) + c^2*(6*a*b*e*f + 2*a^2*f^2 + b^2*(e 
^2 + 2*d*f)))*x))/(3*c^3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - ((2*(2*b 
^4*c*e*f + 48*a^2*c^3*e*f - b^5*f^2 + 4*b^2*c^2*e*(2*c*d - 3*a*f) + b^3*c* 
(10*a*f^2 - c*(e^2 + 2*d*f)) - 4*b*c^2*(2*c^2*d^2 + 8*a^2*f^2 + a*c*(e^2 + 
 2*d*f)) - 2*c*(8*c^4*d^2 - 2*b^4*f^2 + b^2*c*f*(b*e + 14*a*f) - c^3*(8*b* 
d*e - 4*a*(e^2 + 2*d*f)) - c^2*(12*a*b*e*f + 16*a^2*f^2 - b^2*(e^2 + 2*d*f 
)))*x))/(c^3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) - (3*(b^2 - 4*a*c)*f^2*A 
rcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(5/2))/(3*(b^2 - 
4*a*c))
 

Defintions of rubi rules used

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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 2191
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = 
PolynomialQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[P 
q, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x + 
c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^ 
(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c))   Int 
[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (2*p + 3)* 
(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^ 
2 - 4*a*c, 0] && LtQ[p, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1128\) vs. \(2(424)=848\).

Time = 2.54 (sec) , antiderivative size = 1129, normalized size of antiderivative = 2.57

method result size
default \(\text {Expression too large to display}\) \(1129\)

Input:

int((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

d^2*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2 
*c*x+b)/(c*x^2+b*x+a)^(1/2))+f^2*(-1/3*x^3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*( 
-x^2/c/(c*x^2+b*x+a)^(3/2)+1/2*b/c*(-1/2*x/c/(c*x^2+b*x+a)^(3/2)-1/4*b/c*( 
-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a 
)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+1/2*a/c*(2/3* 
(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/( 
c*x^2+b*x+a)^(1/2)))+2*a/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x 
+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+ 
b*x+a)^(1/2))))+1/c*(-x/c/(c*x^2+b*x+a)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^ 
(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2))+1/c^(3/2)*ln((1/2*b+c 
*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+(2*d*f+e^2)*(-1/2*x/c/(c*x^2+b*x+a)^(3/ 
2)-1/4*b/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/ 
(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+1 
/2*a/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2 
*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+2*d*e*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c 
*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c* 
x+b)/(c*x^2+b*x+a)^(1/2)))+2*e*f*(-x^2/c/(c*x^2+b*x+a)^(3/2)+1/2*b/c*(-1/2 
*x/c/(c*x^2+b*x+a)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3* 
(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/( 
c*x^2+b*x+a)^(1/2)))+1/2*a/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(...
 

Fricas [A] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 1581, normalized size of antiderivative = 3.59 \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f^2*x^4 + 2*(b^5*c - 8*a*b^3 
*c^2 + 16*a^2*b*c^3)*f^2*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*f^2*x^2 + 2* 
(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*f^2*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a 
^4*c^2)*f^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + 
 a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*a^2*b*c^3*e^2 + 2*(8*c^6*d^2 - 8*b 
*c^5*d*e + (b^2*c^4 + 4*a*c^5)*e^2 - 2*(b^4*c^2 - 7*a*b^2*c^3 + 8*a^2*c^4) 
*f^2 + (2*(b^2*c^4 + 4*a*c^5)*d + (b^3*c^3 - 12*a*b*c^4)*e)*f)*x^3 - (b^3* 
c^3 - 12*a*b*c^4)*d^2 - 4*(a*b^2*c^3 + 4*a^2*c^4)*d*e - (3*a^2*b^3*c - 20* 
a^3*b*c^2)*f^2 + 3*(8*b*c^5*d^2 - 8*b^2*c^4*d*e + (b^3*c^3 + 4*a*b*c^4)*e^ 
2 - (b^5*c - 6*a*b^3*c^2)*f^2 + 2*((b^3*c^3 + 4*a*b*c^4)*d - 2*(a*b^2*c^3 
+ 4*a^2*c^4)*e)*f)*x^2 + 16*(a^2*b*c^3*d - 2*a^3*c^3*e)*f + 6*(2*a*b^2*c^3 
*e^2 + (b^2*c^4 + 4*a*c^5)*d^2 - (b^3*c^3 + 4*a*b*c^4)*d*e - (a*b^4*c - 7* 
a^2*b^2*c^2 + 4*a^3*c^3)*f^2 + 4*(a*b^2*c^3*d - 2*a^2*b*c^3*e)*f)*x)*sqrt( 
c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^4*c^5 + (b^4*c^5 - 8 
*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^5 + 16*a^2*b*c^6)*x^ 
3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^5*c^3 - 8*a^2*b^3*c^ 
4 + 16*a^3*b*c^5)*x), -1/3*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*f^2*x^ 
4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*f^2*x^3 + (b^6 - 6*a*b^4*c + 32 
*a^3*c^3)*f^2*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*f^2*x + (a^2*b^ 
4 - 8*a^3*b^2*c + 16*a^4*c^2)*f^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((f*x**2+e*x+d)**2/(c*x**2+b*x+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.32 \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {f^{2} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (8 \, c^{5} d^{2} - 8 \, b c^{4} d e + b^{2} c^{3} e^{2} + 4 \, a c^{4} e^{2} + 2 \, b^{2} c^{3} d f + 8 \, a c^{4} d f + b^{3} c^{2} e f - 12 \, a b c^{3} e f - 2 \, b^{4} c f^{2} + 14 \, a b^{2} c^{2} f^{2} - 16 \, a^{2} c^{3} f^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (8 \, b c^{4} d^{2} - 8 \, b^{2} c^{3} d e + b^{3} c^{2} e^{2} + 4 \, a b c^{3} e^{2} + 2 \, b^{3} c^{2} d f + 8 \, a b c^{3} d f - 4 \, a b^{2} c^{2} e f - 16 \, a^{2} c^{3} e f - b^{5} f^{2} + 6 \, a b^{3} c f^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {6 \, {\left (b^{2} c^{3} d^{2} + 4 \, a c^{4} d^{2} - b^{3} c^{2} d e - 4 \, a b c^{3} d e + 2 \, a b^{2} c^{2} e^{2} + 4 \, a b^{2} c^{2} d f - 8 \, a^{2} b c^{2} e f - a b^{4} f^{2} + 7 \, a^{2} b^{2} c f^{2} - 4 \, a^{3} c^{2} f^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {b^{3} c^{2} d^{2} - 12 \, a b c^{3} d^{2} + 4 \, a b^{2} c^{2} d e + 16 \, a^{2} c^{3} d e - 8 \, a^{2} b c^{2} e^{2} - 16 \, a^{2} b c^{2} d f + 32 \, a^{3} c^{2} e f + 3 \, a^{2} b^{3} f^{2} - 20 \, a^{3} b c f^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \] Input:

integrate((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")
 

Output:

-f^2*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2) + 
 2/3*(((2*(8*c^5*d^2 - 8*b*c^4*d*e + b^2*c^3*e^2 + 4*a*c^4*e^2 + 2*b^2*c^3 
*d*f + 8*a*c^4*d*f + b^3*c^2*e*f - 12*a*b*c^3*e*f - 2*b^4*c*f^2 + 14*a*b^2 
*c^2*f^2 - 16*a^2*c^3*f^2)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(8*b 
*c^4*d^2 - 8*b^2*c^3*d*e + b^3*c^2*e^2 + 4*a*b*c^3*e^2 + 2*b^3*c^2*d*f + 8 
*a*b*c^3*d*f - 4*a*b^2*c^2*e*f - 16*a^2*c^3*e*f - b^5*f^2 + 6*a*b^3*c*f^2) 
/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 6*(b^2*c^3*d^2 + 4*a*c^4*d^2 - 
b^3*c^2*d*e - 4*a*b*c^3*d*e + 2*a*b^2*c^2*e^2 + 4*a*b^2*c^2*d*f - 8*a^2*b* 
c^2*e*f - a*b^4*f^2 + 7*a^2*b^2*c*f^2 - 4*a^3*c^2*f^2)/(b^4*c^2 - 8*a*b^2* 
c^3 + 16*a^2*c^4))*x - (b^3*c^2*d^2 - 12*a*b*c^3*d^2 + 4*a*b^2*c^2*d*e + 1 
6*a^2*c^3*d*e - 8*a^2*b*c^2*e^2 - 16*a^2*b*c^2*d*f + 32*a^3*c^2*e*f + 3*a^ 
2*b^3*f^2 - 20*a^3*b*c*f^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + 
 b*x + a)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (f\,x^2+e\,x+d\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \] Input:

int((d + e*x + f*x^2)^2/(a + b*x + c*x^2)^(5/2),x)
 

Output:

int((d + e*x + f*x^2)^2/(a + b*x + c*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 2663, normalized size of antiderivative = 6.05 \[ \int \frac {\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(5/2),x)
 

Output:

(40*sqrt(a + b*x + c*x**2)*a**3*b*c**2*f**2 - 64*sqrt(a + b*x + c*x**2)*a* 
*3*c**3*e*f - 48*sqrt(a + b*x + c*x**2)*a**3*c**3*f**2*x - 6*sqrt(a + b*x 
+ c*x**2)*a**2*b**3*c*f**2 + 84*sqrt(a + b*x + c*x**2)*a**2*b**2*c**2*f**2 
*x + 32*sqrt(a + b*x + c*x**2)*a**2*b*c**3*d*f + 16*sqrt(a + b*x + c*x**2) 
*a**2*b*c**3*e**2 - 96*sqrt(a + b*x + c*x**2)*a**2*b*c**3*e*f*x - 32*sqrt( 
a + b*x + c*x**2)*a**2*c**4*d*e - 96*sqrt(a + b*x + c*x**2)*a**2*c**4*e*f* 
x**2 - 64*sqrt(a + b*x + c*x**2)*a**2*c**4*f**2*x**3 - 12*sqrt(a + b*x + c 
*x**2)*a*b**4*c*f**2*x + 36*sqrt(a + b*x + c*x**2)*a*b**3*c**2*f**2*x**2 - 
 8*sqrt(a + b*x + c*x**2)*a*b**2*c**3*d*e + 48*sqrt(a + b*x + c*x**2)*a*b* 
*2*c**3*d*f*x + 24*sqrt(a + b*x + c*x**2)*a*b**2*c**3*e**2*x - 24*sqrt(a + 
 b*x + c*x**2)*a*b**2*c**3*e*f*x**2 + 56*sqrt(a + b*x + c*x**2)*a*b**2*c** 
3*f**2*x**3 + 24*sqrt(a + b*x + c*x**2)*a*b*c**4*d**2 - 48*sqrt(a + b*x + 
c*x**2)*a*b*c**4*d*e*x + 48*sqrt(a + b*x + c*x**2)*a*b*c**4*d*f*x**2 + 24* 
sqrt(a + b*x + c*x**2)*a*b*c**4*e**2*x**2 - 48*sqrt(a + b*x + c*x**2)*a*b* 
c**4*e*f*x**3 + 48*sqrt(a + b*x + c*x**2)*a*c**5*d**2*x + 32*sqrt(a + b*x 
+ c*x**2)*a*c**5*d*f*x**3 + 16*sqrt(a + b*x + c*x**2)*a*c**5*e**2*x**3 - 6 
*sqrt(a + b*x + c*x**2)*b**5*c*f**2*x**2 - 8*sqrt(a + b*x + c*x**2)*b**4*c 
**2*f**2*x**3 - 2*sqrt(a + b*x + c*x**2)*b**3*c**3*d**2 - 12*sqrt(a + b*x 
+ c*x**2)*b**3*c**3*d*e*x + 12*sqrt(a + b*x + c*x**2)*b**3*c**3*d*f*x**2 + 
 6*sqrt(a + b*x + c*x**2)*b**3*c**3*e**2*x**2 + 4*sqrt(a + b*x + c*x**2...