3.227 \(\int \frac{(g+h x)^2 (d+e x+f x^2)}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=420 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (48 c^2 \left (a^2 f h^2+2 a b h (e h+2 f g)+b^2 \left (d h^2+2 e g h+f g^2\right )\right )-40 b^2 c h (3 a f h+b e h+2 b f g)-64 c^3 \left (a \left (d h^2+2 e g h+f g^2\right )+b g (2 d h+e g)\right )+35 b^4 f h^2+128 c^4 d g^2\right )}{128 c^{9/2}}-\frac{\sqrt{a+b x+c x^2} \left (-2 c h x \left (-4 c h (9 a f h+10 b e h+6 b f g)+35 b^2 f h^2-8 c^2 \left (f g^2-2 h (3 d h+2 e g)\right )\right )+8 c^2 h \left (16 a h (e h+2 f g)+b \left (18 h (d h+2 e g)+11 f g^2\right )\right )-20 b c h^2 (11 a f h+6 b (e h+2 f g))+105 b^3 f h^3+32 c^3 g \left (f g^2-4 h (3 d h+e g)\right )\right )}{192 c^4 h}-\frac{(g+h x)^2 \sqrt{a+b x+c x^2} (7 b f h-8 c e h+2 c f g)}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h} \]

[Out]

-((2*c*f*g - 8*c*e*h + 7*b*f*h)*(g + h*x)^2*Sqrt[a + b*x + c*x^2])/(24*c^2*h) + (f*(g + h*x)^3*Sqrt[a + b*x +
c*x^2])/(4*c*h) - ((105*b^3*f*h^3 + 32*c^3*g*(f*g^2 - 4*h*(e*g + 3*d*h)) - 20*b*c*h^2*(11*a*f*h + 6*b*(2*f*g +
e*h)) + 8*c^2*h*(16*a*h*(2*f*g + e*h) + b*(11*f*g^2 + 18*h*(2*e*g + d*h))) - 2*c*h*(35*b^2*f*h^2 - 4*c*h*(6*b
*f*g + 10*b*e*h + 9*a*f*h) - 8*c^2*(f*g^2 - 2*h*(2*e*g + 3*d*h)))*x)*Sqrt[a + b*x + c*x^2])/(192*c^4*h) + ((12
8*c^4*d*g^2 + 35*b^4*f*h^2 - 40*b^2*c*h*(2*b*f*g + b*e*h + 3*a*f*h) - 64*c^3*(b*g*(e*g + 2*d*h) + a*(f*g^2 + 2
*e*g*h + d*h^2)) + 48*c^2*(a^2*f*h^2 + 2*a*b*h*(2*f*g + e*h) + b^2*(f*g^2 + 2*e*g*h + d*h^2)))*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 1.01109, antiderivative size = 418, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1653, 832, 779, 621, 206} \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (48 c^2 \left (a^2 f h^2+2 a b h (e h+2 f g)+b^2 \left (h (d h+2 e g)+f g^2\right )\right )-40 b^2 c h (3 a f h+b e h+2 b f g)-64 c^3 \left (a h (d h+2 e g)+a f g^2+b g (2 d h+e g)\right )+35 b^4 f h^2+128 c^4 d g^2\right )}{128 c^{9/2}}-\frac{\sqrt{a+b x+c x^2} \left (-2 c h x \left (-4 c h (9 a f h+10 b e h+6 b f g)+35 b^2 f h^2-8 c^2 \left (f g^2-2 h (3 d h+2 e g)\right )\right )+8 c^2 h \left (16 a h (e h+2 f g)+18 b h (d h+2 e g)+11 b f g^2\right )-20 b c h^2 (11 a f h+6 b (e h+2 f g))+105 b^3 f h^3+32 c^3 \left (f g^3-4 g h (3 d h+e g)\right )\right )}{192 c^4 h}-\frac{(g+h x)^2 \sqrt{a+b x+c x^2} (7 b f h-8 c e h+2 c f g)}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*c*f*g - 8*c*e*h + 7*b*f*h)*(g + h*x)^2*Sqrt[a + b*x + c*x^2])/(24*c^2*h) + (f*(g + h*x)^3*Sqrt[a + b*x +
c*x^2])/(4*c*h) - ((105*b^3*f*h^3 + 32*c^3*(f*g^3 - 4*g*h*(e*g + 3*d*h)) - 20*b*c*h^2*(11*a*f*h + 6*b*(2*f*g +
e*h)) + 8*c^2*h*(11*b*f*g^2 + 18*b*h*(2*e*g + d*h) + 16*a*h*(2*f*g + e*h)) - 2*c*h*(35*b^2*f*h^2 - 4*c*h*(6*b
*f*g + 10*b*e*h + 9*a*f*h) - 8*c^2*(f*g^2 - 2*h*(2*e*g + 3*d*h)))*x)*Sqrt[a + b*x + c*x^2])/(192*c^4*h) + ((12
8*c^4*d*g^2 + 35*b^4*f*h^2 - 40*b^2*c*h*(2*b*f*g + b*e*h + 3*a*f*h) - 64*c^3*(a*f*g^2 + a*h*(2*e*g + d*h) + b*
g*(e*g + 2*d*h)) + 48*c^2*(a^2*f*h^2 + 2*a*b*h*(2*f*g + e*h) + b^2*(f*g^2 + h*(2*e*g + d*h))))*ArcTanh[(b + 2*
c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(9/2))

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[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{(g+h x)^2 \left (d+e x+f x^2\right )}{\sqrt{a+b x+c x^2}} \, dx &=\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h}+\frac{\int \frac{(g+h x)^2 \left (-\frac{1}{2} h (b f g-8 c d h+6 a f h)-\frac{1}{2} h (2 c f g-8 c e h+7 b f h) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{4 c h^2}\\ &=-\frac{(2 c f g-8 c e h+7 b f h) (g+h x)^2 \sqrt{a+b x+c x^2}}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h}+\frac{\int \frac{(g+h x) \left (\frac{1}{4} h \left (7 b^2 f g h+28 a b f h^2-4 b c g (f g+2 e h)+4 c h (12 c d g-7 a f g-8 a e h)\right )+\frac{1}{4} h \left (35 b^2 f h^2-4 c h (6 b f g+10 b e h+9 a f h)-8 c^2 \left (f g^2-2 h (2 e g+3 d h)\right )\right ) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{12 c^2 h^2}\\ &=-\frac{(2 c f g-8 c e h+7 b f h) (g+h x)^2 \sqrt{a+b x+c x^2}}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h}-\frac{\left (105 b^3 f h^3+32 c^3 \left (f g^3-4 g h (e g+3 d h)\right )-20 b c h^2 (11 a f h+6 b (2 f g+e h))+8 c^2 h \left (11 b f g^2+18 b h (2 e g+d h)+16 a h (2 f g+e h)\right )-2 c h \left (35 b^2 f h^2-4 c h (6 b f g+10 b e h+9 a f h)-8 c^2 \left (f g^2-2 h (2 e g+3 d h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4 h}+\frac{\left (128 c^4 d g^2+35 b^4 f h^2-40 b^2 c h (2 b f g+b e h+3 a f h)-64 c^3 \left (a f g^2+a h (2 e g+d h)+b g (e g+2 d h)\right )+48 c^2 \left (a^2 f h^2+2 a b h (2 f g+e h)+b^2 \left (f g^2+h (2 e g+d h)\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{128 c^4}\\ &=-\frac{(2 c f g-8 c e h+7 b f h) (g+h x)^2 \sqrt{a+b x+c x^2}}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h}-\frac{\left (105 b^3 f h^3+32 c^3 \left (f g^3-4 g h (e g+3 d h)\right )-20 b c h^2 (11 a f h+6 b (2 f g+e h))+8 c^2 h \left (11 b f g^2+18 b h (2 e g+d h)+16 a h (2 f g+e h)\right )-2 c h \left (35 b^2 f h^2-4 c h (6 b f g+10 b e h+9 a f h)-8 c^2 \left (f g^2-2 h (2 e g+3 d h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4 h}+\frac{\left (128 c^4 d g^2+35 b^4 f h^2-40 b^2 c h (2 b f g+b e h+3 a f h)-64 c^3 \left (a f g^2+a h (2 e g+d h)+b g (e g+2 d h)\right )+48 c^2 \left (a^2 f h^2+2 a b h (2 f g+e h)+b^2 \left (f g^2+h (2 e g+d h)\right )\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 )}{64 c^4}\\ &=-\frac{(2 c f g-8 c e h+7 b f h) (g+h x)^2 \sqrt{a+b x+c x^2}}{24 c^2 h}+\frac{f (g+h x)^3 \sqrt{a+b x+c x^2}}{4 c h}-\frac{\left (105 b^3 f h^3+32 c^3 \left (f g^3-4 g h (e g+3 d h)\right )-20 b c h^2 (11 a f h+6 b (2 f g+e h))+8 c^2 h \left (11 b f g^2+18 b h (2 e g+d h)+16 a h (2 f g+e h)\right )-2 c h \left (35 b^2 f h^2-4 c h (6 b f g+10 b e h+9 a f h)-8 c^2 \left (f g^2-2 h (2 e g+3 d h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4 h}+\frac{\left (128 c^4 d g^2+35 b^4 f h^2-40 b^2 c h (2 b f g+b e h+3 a f h)-64 c^3 \left (a f g^2+a h (2 e g+d h)+b g (e g+2 d h)\right )+48 c^2 \left (a^2 f h^2+2 a b h (2 f g+e h)+b^2 \left (f g^2+h (2 e g+d h)\right )\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.669533, size = 343, normalized size = 0.82 \[ \frac{3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (48 c^2 \left (a^2 f h^2+2 a b h (e h+2 f g)+b^2 \left (h (d h+2 e g)+f g^2\right )\right )-40 b^2 c h (3 a f h+b e h+2 b f g)-64 c^3 \left (a h (d h+2 e g)+a f g^2+b g (2 d h+e g)\right )+35 b^4 f h^2+128 c^4 d g^2\right )+2 \sqrt{c} \sqrt{a+x (b+c x)} \left (-8 c^2 \left (a h (16 e h+32 f g+9 f h x)+2 b h (9 d h+18 e g+5 e h x)+b f \left (18 g^2+20 g h x+7 h^2 x^2\right )\right )+10 b c h (22 a f h+b (12 e h+24 f g+7 f h x))-105 b^3 f h^2+16 c^3 \left (6 d h (4 g+h x)+4 e \left (3 g^2+3 g h x+h^2 x^2\right )+f x \left (6 g^2+8 g h x+3 h^2 x^2\right )\right )\right )}{384 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-105*b^3*f*h^2 + 10*b*c*h*(22*a*f*h + b*(24*f*g + 12*e*h + 7*f*h*x)) + 16*c^
3*(6*d*h*(4*g + h*x) + 4*e*(3*g^2 + 3*g*h*x + h^2*x^2) + f*x*(6*g^2 + 8*g*h*x + 3*h^2*x^2)) - 8*c^2*(2*b*h*(18
*e*g + 9*d*h + 5*e*h*x) + a*h*(32*f*g + 16*e*h + 9*f*h*x) + b*f*(18*g^2 + 20*g*h*x + 7*h^2*x^2))) + 3*(128*c^4
*d*g^2 + 35*b^4*f*h^2 - 40*b^2*c*h*(2*b*f*g + b*e*h + 3*a*f*h) - 64*c^3*(a*f*g^2 + a*h*(2*e*g + d*h) + b*g*(e*
g + 2*d*h)) + 48*c^2*(a^2*f*h^2 + 2*a*b*h*(2*f*g + e*h) + b^2*(f*g^2 + h*(2*e*g + d*h))))*ArcTanh[(b + 2*c*x)/
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(384*c^(9/2))

________________________________________________________________________________________

Maple [B]  time = 0.063, size = 1069, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)^2*(f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x)

[Out]

-a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e*g*h+3/8*h^2*f*a^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x
^2+b*x+a)^(1/2))+1/2*x/c*(c*x^2+b*x+a)^(1/2)*d*h^2+1/2*x/c*(c*x^2+b*x+a)^(1/2)*f*g^2-3/4*b/c^2*(c*x^2+b*x+a)^(
1/2)*d*h^2+1/c*(c*x^2+b*x+a)^(1/2)*e*g^2+g^2*d*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)+35/96*h^2*f
*b^2/c^3*x*(c*x^2+b*x+a)^(1/2)-15/16*h^2*f*b^2/c^(7/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-7/24*h^2*
f*b/c^2*x^2*(c*x^2+b*x+a)^(1/2)-35/64*h^2*f*b^3/c^4*(c*x^2+b*x+a)^(1/2)+1/4*h^2*f*x^3/c*(c*x^2+b*x+a)^(1/2)+2/
3*x^2/c*(c*x^2+b*x+a)^(1/2)*g*h*f+1/3*x^2/c*(c*x^2+b*x+a)^(1/2)*h^2*e+5/8*b^2/c^3*(c*x^2+b*x+a)^(1/2)*h^2*e-5/
16*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*h^2*e-2/3*a/c^2*(c*x^2+b*x+a)^(1/2)*h^2*e+35/128*h^
2*f*b^4/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/8*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))*d*h^2+3/8*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f*g^2-1/2*a/c^(3/2)*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*h^2-1/2*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f*g^2+2/c*(c*x^2
+b*x+a)^(1/2)*g*h*d-1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e*g^2-3/4*b/c^2*(c*x^2+b*x+a)^(1
/2)*f*g^2-5/6*b/c^2*x*(c*x^2+b*x+a)^(1/2)*g*h*f-3/8*h^2*f*a/c^2*x*(c*x^2+b*x+a)^(1/2)+3/2*b/c^(5/2)*a*ln((1/2*
b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*g*h*f-3/2*b/c^2*(c*x^2+b*x+a)^(1/2)*e*g*h+3/4*b^2/c^(5/2)*ln((1/2*b+c*x)/c
^(1/2)+(c*x^2+b*x+a)^(1/2))*e*g*h+55/48*h^2*f*b/c^3*a*(c*x^2+b*x+a)^(1/2)+x/c*(c*x^2+b*x+a)^(1/2)*e*g*h-b/c^(3
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*g*h*d-5/12*b/c^2*x*(c*x^2+b*x+a)^(1/2)*h^2*e+5/4*b^2/c^3*(c*x^
2+b*x+a)^(1/2)*g*h*f-5/8*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*g*h*f+3/4*b/c^(5/2)*a*ln((1/2
*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*h^2*e-4/3*a/c^2*(c*x^2+b*x+a)^(1/2)*g*h*f

________________________________________________________________________________________

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((h*x+g)^2*(f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.17739, size = 1953, normalized size = 4.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^2*(f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/768*(3*(16*(8*c^4*d - 4*b*c^3*e + (3*b^2*c^2 - 4*a*c^3)*f)*g^2 - 16*(8*b*c^3*d - 2*(3*b^2*c^2 - 4*a*c^3)*e
+ (5*b^3*c - 12*a*b*c^2)*f)*g*h + (16*(3*b^2*c^2 - 4*a*c^3)*d - 8*(5*b^3*c - 12*a*b*c^2)*e + (35*b^4 - 120*a*b
^2*c + 48*a^2*c^2)*f)*h^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*(48*c^4*f*h^2*x^3 + 48*(4*c^4*e - 3*b*c^3*f)*g^2 + 16*(24*c^4*d - 18*b*c^3*e + (15*b^2*c^2 - 16
*a*c^3)*f)*g*h - (144*b*c^3*d - 8*(15*b^2*c^2 - 16*a*c^3)*e + 5*(21*b^3*c - 44*a*b*c^2)*f)*h^2 + 8*(16*c^4*f*g
*h + (8*c^4*e - 7*b*c^3*f)*h^2)*x^2 + 2*(48*c^4*f*g^2 + 16*(6*c^4*e - 5*b*c^3*f)*g*h + (48*c^4*d - 40*b*c^3*e
+ (35*b^2*c^2 - 36*a*c^3)*f)*h^2)*x)*sqrt(c*x^2 + b*x + a))/c^5, -1/384*(3*(16*(8*c^4*d - 4*b*c^3*e + (3*b^2*c
^2 - 4*a*c^3)*f)*g^2 - 16*(8*b*c^3*d - 2*(3*b^2*c^2 - 4*a*c^3)*e + (5*b^3*c - 12*a*b*c^2)*f)*g*h + (16*(3*b^2*
c^2 - 4*a*c^3)*d - 8*(5*b^3*c - 12*a*b*c^2)*e + (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*f)*h^2)*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*(48*c^4*f*h^2*x^3 + 48*(4*c^4*e - 3*
b*c^3*f)*g^2 + 16*(24*c^4*d - 18*b*c^3*e + (15*b^2*c^2 - 16*a*c^3)*f)*g*h - (144*b*c^3*d - 8*(15*b^2*c^2 - 16*
a*c^3)*e + 5*(21*b^3*c - 44*a*b*c^2)*f)*h^2 + 8*(16*c^4*f*g*h + (8*c^4*e - 7*b*c^3*f)*h^2)*x^2 + 2*(48*c^4*f*g
^2 + 16*(6*c^4*e - 5*b*c^3*f)*g*h + (48*c^4*d - 40*b*c^3*e + (35*b^2*c^2 - 36*a*c^3)*f)*h^2)*x)*sqrt(c*x^2 + b
*x + a))/c^5]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)**2*(f*x**2+e*x+d)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((g + h*x)**2*(d + e*x + f*x**2)/sqrt(a + b*x + c*x**2), x)

________________________________________________________________________________________

Giac [A]  time = 1.24578, size = 617, normalized size = 1.47 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (\frac{6 \, f h^{2} x}{c} + \frac{16 \, c^{3} f g h - 7 \, b c^{2} f h^{2} + 8 \, c^{3} h^{2} e}{c^{4}}\right )} x + \frac{48 \, c^{3} f g^{2} - 80 \, b c^{2} f g h + 48 \, c^{3} d h^{2} + 35 \, b^{2} c f h^{2} - 36 \, a c^{2} f h^{2} + 96 \, c^{3} g h e - 40 \, b c^{2} h^{2} e}{c^{4}}\right )} x - \frac{144 \, b c^{2} f g^{2} - 384 \, c^{3} d g h - 240 \, b^{2} c f g h + 256 \, a c^{2} f g h + 144 \, b c^{2} d h^{2} + 105 \, b^{3} f h^{2} - 220 \, a b c f h^{2} - 192 \, c^{3} g^{2} e + 288 \, b c^{2} g h e - 120 \, b^{2} c h^{2} e + 128 \, a c^{2} h^{2} e}{c^{4}}\right )} - \frac{{\left (128 \, c^{4} d g^{2} + 48 \, b^{2} c^{2} f g^{2} - 64 \, a c^{3} f g^{2} - 128 \, b c^{3} d g h - 80 \, b^{3} c f g h + 192 \, a b c^{2} f g h + 48 \, b^{2} c^{2} d h^{2} - 64 \, a c^{3} d h^{2} + 35 \, b^{4} f h^{2} - 120 \, a b^{2} c f h^{2} + 48 \, a^{2} c^{2} f h^{2} - 64 \, b c^{3} g^{2} e + 96 \, b^{2} c^{2} g h e - 128 \, a c^{3} g h e - 40 \, b^{3} c h^{2} e + 96 \, a b c^{2} h^{2} e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)^2*(f*x^2+e*x+d)/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*f*h^2*x/c + (16*c^3*f*g*h - 7*b*c^2*f*h^2 + 8*c^3*h^2*e)/c^4)*x + (48*c^3
*f*g^2 - 80*b*c^2*f*g*h + 48*c^3*d*h^2 + 35*b^2*c*f*h^2 - 36*a*c^2*f*h^2 + 96*c^3*g*h*e - 40*b*c^2*h^2*e)/c^4)
*x - (144*b*c^2*f*g^2 - 384*c^3*d*g*h - 240*b^2*c*f*g*h + 256*a*c^2*f*g*h + 144*b*c^2*d*h^2 + 105*b^3*f*h^2 -
220*a*b*c*f*h^2 - 192*c^3*g^2*e + 288*b*c^2*g*h*e - 120*b^2*c*h^2*e + 128*a*c^2*h^2*e)/c^4) - 1/128*(128*c^4*d
*g^2 + 48*b^2*c^2*f*g^2 - 64*a*c^3*f*g^2 - 128*b*c^3*d*g*h - 80*b^3*c*f*g*h + 192*a*b*c^2*f*g*h + 48*b^2*c^2*d
*h^2 - 64*a*c^3*d*h^2 + 35*b^4*f*h^2 - 120*a*b^2*c*f*h^2 + 48*a^2*c^2*f*h^2 - 64*b*c^3*g^2*e + 96*b^2*c^2*g*h*
e - 128*a*c^3*g*h*e - 40*b^3*c*h^2*e + 96*a*b*c^2*h^2*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c
) - b))/c^(9/2)