3.115 \(\int \frac{(d+e x+f x^2)^2}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=309 \[ \frac{2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )+2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)\right )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-12 c f (a f+2 b e)+15 b^2 f^2+8 c^2 \left (2 d f+e^2\right )\right )}{8 c^{7/2}}+\frac{f \sqrt{a+b x+c x^2} (8 c e-7 b f)}{4 c^3}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2} \]

[Out]

(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))/(c^3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (f*(8*c*e - 7*b*f)*Sqrt[a + b*x + c*x^2])/(4*c
^3) + (f^2*x*Sqrt[a + b*x + c*x^2])/(2*c^2) + ((15*b^2*f^2 - 12*c*f*(2*b*e + a*f) + 8*c^2*(e^2 + 2*d*f))*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.446654, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {1660, 1661, 640, 621, 206} \[ \frac{2 \left (-x \left (c^2 \left (2 a^2 f^2+6 a b e f+b^2 \left (2 d f+e^2\right )\right )-2 b^2 c f (2 a f+b e)-2 c^3 \left (a \left (2 d f+e^2\right )+b d e\right )+b^4 f^2+2 c^4 d^2\right )-b c \left (-3 a^2 f^2+a c \left (2 d f+e^2\right )+c^2 d^2\right )+2 a b^2 c e f-a b^3 f^2+4 a c^2 e (c d-a f)\right )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-12 c f (a f+2 b e)+15 b^2 f^2+8 c^2 \left (2 d f+e^2\right )\right )}{8 c^{7/2}}+\frac{f \sqrt{a+b x+c x^2} (8 c e-7 b f)}{4 c^3}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))/(c^3*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (f*(8*c*e - 7*b*f)*Sqrt[a + b*x + c*x^2])/(4*c
^3) + (f^2*x*Sqrt[a + b*x + c*x^2])/(2*c^2) + ((15*b^2*f^2 - 12*c*f*(2*b*e + a*f) + 8*c^2*(e^2 + 2*d*f))*ArcTa
nh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))

Rule 1660

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[Pq, 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] + Dist[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
]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[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]

Rule 640

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x+f x^2\right )^2}{\left (a+b x+c x^2\right )^{3/2}} \, 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 )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 \int \frac{-\frac{\left (b^2-4 a c\right ) \left (b^2 f^2-c f (2 b e+a f)+c^2 \left (e^2+2 d f\right )\right )}{2 c^3}-\frac{\left (b^2-4 a c\right ) f (2 c e-b f) x}{2 c^2}-\frac{\left (b^2-4 a c\right ) f^2 x^2}{2 c}}{\sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=\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 )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2}-\frac{\int \frac{-\frac{\left (b^2-4 a c\right ) \left (2 b^2 f^2-c f (4 b e+3 a f)+2 c^2 \left (e^2+2 d f\right )\right )}{2 c^2}-\frac{\left (b^2-4 a c\right ) f (8 c e-7 b f) x}{4 c}}{\sqrt{a+b x+c x^2}} \, dx}{c \left (b^2-4 a c\right )}\\ &=\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 )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f (8 c e-7 b f) \sqrt{a+b x+c x^2}}{4 c^3}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2}+\frac{\left (15 b^2 f^2-12 c f (2 b e+a f)+8 c^2 \left (e^2+2 d f\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 c^3}\\ &=\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 )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f (8 c e-7 b f) \sqrt{a+b x+c x^2}}{4 c^3}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2}+\frac{\left (15 b^2 f^2-12 c f (2 b e+a f)+8 c^2 \left (e^2+2 d f\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x}{\sqrt{a+b x+c x^2}}\right )}{4 c^3}\\ &=\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 )}{c^3 \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{f (8 c e-7 b f) \sqrt{a+b x+c x^2}}{4 c^3}+\frac{f^2 x \sqrt{a+b x+c x^2}}{2 c^2}+\frac{\left (15 b^2 f^2-12 c f (2 b e+a f)+8 c^2 \left (e^2+2 d f\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.756993, size = 288, normalized size = 0.93 \[ \frac{4 b c \left (-13 a^2 f^2+a c \left (4 d f+2 e^2+20 e f x-5 f^2 x^2\right )+2 c^2 d (d-2 e x)\right )+8 c^2 \left (a^2 f (8 e+3 f x)+a c \left (x \left (-2 e^2+4 e f x+f^2 x^2\right )-4 d (e+f x)\right )+2 c^2 d^2 x\right )-2 b^2 c \left (a f (12 e+31 f x)+c x \left (-8 d f-4 e^2+4 e f x+f^2 x^2\right )\right )+b^3 f (15 a f+c x (5 f x-24 e))+15 b^4 f^2 x}{4 c^3 \left (4 a c-b^2\right ) \sqrt{a+x (b+c x)}}+\frac{\log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right ) \left (-12 c f (a f+2 b e)+15 b^2 f^2+8 c^2 \left (2 d f+e^2\right )\right )}{8 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*b^4*f^2*x + b^3*f*(15*a*f + c*x*(-24*e + 5*f*x)) + 4*b*c*(-13*a^2*f^2 + 2*c^2*d*(d - 2*e*x) + a*c*(2*e^2 +
4*d*f + 20*e*f*x - 5*f^2*x^2)) - 2*b^2*c*(a*f*(12*e + 31*f*x) + c*x*(-4*e^2 - 8*d*f + 4*e*f*x + f^2*x^2)) + 8
*c^2*(2*c^2*d^2*x + a^2*f*(8*e + 3*f*x) + a*c*(-4*d*(e + f*x) + x*(-2*e^2 + 4*e*f*x + f^2*x^2))))/(4*c^3*(-b^2
+ 4*a*c)*Sqrt[a + x*(b + c*x)]) + ((15*b^2*f^2 - 12*c*f*(2*b*e + a*f) + 8*c^2*(e^2 + 2*d*f))*Log[b + 2*c*x +
2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(8*c^(7/2))

________________________________________________________________________________________

Maple [B]  time = 0.067, size = 1011, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e*x+d)^2/(c*x^2+b*x+a)^(3/2),x)

[Out]

8*e*f*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+2*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*f-3*e*f*b^3/c^2/(4*a
*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+4*e*f*a/c^2*b^2/(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))*e^2-13/2*f^2*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+3*e*f*b/c^2*x/(c*x^2+b*x+a)^(1/
2)-3/2*e*f*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-4*d*e*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-2*d*e*b^2/c/(4*a*
c-b^2)/(c*x^2+b*x+a)^(1/2)+b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*e^2+b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
*d*f+15/8*f^2*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-2*x/c/(c*x^2+b*x+a)^(1/2)*d*f+2*e*f*x^2/c/(c*x^2+b*x+a
)^(1/2)-5/4*f^2*b/c^2*x^2/(c*x^2+b*x+a)^(1/2)-15/8*f^2*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)+15/16*f^2*b^5/c^4/(4*a*c-
b^2)/(c*x^2+b*x+a)^(1/2)-3/2*e*f*b^2/c^3/(c*x^2+b*x+a)^(1/2)-3*e*f*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))+4*e*f*a/c^2/(c*x^2+b*x+a)^(1/2)+b/c^2/(c*x^2+b*x+a)^(1/2)*d*f+1/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)
^(1/2)*e^2+3/2*f^2*a/c^2*x/(c*x^2+b*x+a)^(1/2)-13/4*f^2*b/c^3*a/(c*x^2+b*x+a)^(1/2)+2/c^(3/2)*ln((1/2*b+c*x)/c
^(1/2)+(c*x^2+b*x+a)^(1/2))*d*f-x/c/(c*x^2+b*x+a)^(1/2)*e^2+1/2*b/c^2/(c*x^2+b*x+a)^(1/2)*e^2+15/8*f^2*b^2/c^(
7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/2*f^2*a/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
+1/2*f^2*x^3/c/(c*x^2+b*x+a)^(1/2)+15/16*f^2*b^3/c^4/(c*x^2+b*x+a)^(1/2)-2*d*e/c/(c*x^2+b*x+a)^(1/2)+2*d^2*(2*
c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-13/4*f^2*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 9.59512, size = 2750, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((8*(a*b^2*c^2 - 4*a^2*c^3)*e^2 + 3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*f^2 + (8*(b^2*c^3 - 4*a*c^4)*
e^2 + 3*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*f^2 + 8*(2*(b^2*c^3 - 4*a*c^4)*d - 3*(b^3*c^2 - 4*a*b*c^3)*e)*f)
*x^2 + 8*(2*(a*b^2*c^2 - 4*a^2*c^3)*d - 3*(a*b^3*c - 4*a^2*b*c^2)*e)*f + (8*(b^3*c^2 - 4*a*b*c^3)*e^2 + 3*(5*b
^5 - 24*a*b^3*c + 16*a^2*b*c^2)*f^2 + 8*(2*(b^3*c^2 - 4*a*b*c^3)*d - 3*(b^4*c - 4*a*b^2*c^2)*e)*f)*x)*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*b*c^4*d^2 - 32*a*
c^4*d*e + 8*a*b*c^3*e^2 - 2*(b^2*c^3 - 4*a*c^4)*f^2*x^3 + (15*a*b^3*c - 52*a^2*b*c^2)*f^2 - (8*(b^2*c^3 - 4*a*
c^4)*e*f - 5*(b^3*c^2 - 4*a*b*c^3)*f^2)*x^2 + 8*(2*a*b*c^3*d - (3*a*b^2*c^2 - 8*a^2*c^3)*e)*f + (16*c^5*d^2 -
16*b*c^4*d*e + 8*(b^2*c^3 - 2*a*c^4)*e^2 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*f^2 + 8*(2*(b^2*c^3 - 2*a*c^
4)*d - (3*b^3*c^2 - 10*a*b*c^3)*e)*f)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x
^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*((8*(a*b^2*c^2 - 4*a^2*c^3)*e^2 + 3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*
f^2 + (8*(b^2*c^3 - 4*a*c^4)*e^2 + 3*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*f^2 + 8*(2*(b^2*c^3 - 4*a*c^4)*d -
3*(b^3*c^2 - 4*a*b*c^3)*e)*f)*x^2 + 8*(2*(a*b^2*c^2 - 4*a^2*c^3)*d - 3*(a*b^3*c - 4*a^2*b*c^2)*e)*f + (8*(b^3*
c^2 - 4*a*b*c^3)*e^2 + 3*(5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*f^2 + 8*(2*(b^3*c^2 - 4*a*b*c^3)*d - 3*(b^4*c - 4
*a*b^2*c^2)*e)*f)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) +
2*(8*b*c^4*d^2 - 32*a*c^4*d*e + 8*a*b*c^3*e^2 - 2*(b^2*c^3 - 4*a*c^4)*f^2*x^3 + (15*a*b^3*c - 52*a^2*b*c^2)*f
^2 - (8*(b^2*c^3 - 4*a*c^4)*e*f - 5*(b^3*c^2 - 4*a*b*c^3)*f^2)*x^2 + 8*(2*a*b*c^3*d - (3*a*b^2*c^2 - 8*a^2*c^3
)*e)*f + (16*c^5*d^2 - 16*b*c^4*d*e + 8*(b^2*c^3 - 2*a*c^4)*e^2 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*f^2 +
8*(2*(b^2*c^3 - 2*a*c^4)*d - (3*b^3*c^2 - 10*a*b*c^3)*e)*f)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5
+ (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x + f x^{2}\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e*x+d)**2/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x + f*x**2)**2/(a + b*x + c*x**2)**(3/2), x)

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Giac [A]  time = 1.65767, size = 549, normalized size = 1.78 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (b^{2} c^{2} f^{2} - 4 \, a c^{3} f^{2}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} - \frac{5 \, b^{3} c f^{2} - 20 \, a b c^{2} f^{2} - 8 \, b^{2} c^{2} f e + 32 \, a c^{3} f e}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{16 \, c^{4} d^{2} + 16 \, b^{2} c^{2} d f - 32 \, a c^{3} d f + 15 \, b^{4} f^{2} - 62 \, a b^{2} c f^{2} + 24 \, a^{2} c^{2} f^{2} - 16 \, b c^{3} d e - 24 \, b^{3} c f e + 80 \, a b c^{2} f e + 8 \, b^{2} c^{2} e^{2} - 16 \, a c^{3} e^{2}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{8 \, b c^{3} d^{2} + 16 \, a b c^{2} d f + 15 \, a b^{3} f^{2} - 52 \, a^{2} b c f^{2} - 32 \, a c^{3} d e - 24 \, a b^{2} c f e + 64 \, a^{2} c^{2} f e + 8 \, a b c^{2} e^{2}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt{c x^{2} + b x + a}} - \frac{{\left (16 \, c^{2} d f + 15 \, b^{2} f^{2} - 12 \, a c f^{2} - 24 \, b c f e + 8 \, c^{2} e^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((2*(b^2*c^2*f^2 - 4*a*c^3*f^2)*x/(b^2*c^3 - 4*a*c^4) - (5*b^3*c*f^2 - 20*a*b*c^2*f^2 - 8*b^2*c^2*f*e + 3
2*a*c^3*f*e)/(b^2*c^3 - 4*a*c^4))*x - (16*c^4*d^2 + 16*b^2*c^2*d*f - 32*a*c^3*d*f + 15*b^4*f^2 - 62*a*b^2*c*f^
2 + 24*a^2*c^2*f^2 - 16*b*c^3*d*e - 24*b^3*c*f*e + 80*a*b*c^2*f*e + 8*b^2*c^2*e^2 - 16*a*c^3*e^2)/(b^2*c^3 - 4
*a*c^4))*x - (8*b*c^3*d^2 + 16*a*b*c^2*d*f + 15*a*b^3*f^2 - 52*a^2*b*c*f^2 - 32*a*c^3*d*e - 24*a*b^2*c*f*e + 6
4*a^2*c^2*f*e + 8*a*b*c^2*e^2)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 1/8*(16*c^2*d*f + 15*b^2*f^2 - 12*
a*c*f^2 - 24*b*c*f*e + 8*c^2*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)