### 3.193 $$\int \frac{\sqrt{a+b x+c x^2} (d+e x+f x^2)}{(g+h x)^4} \, dx$$

Optimal. Leaf size=603 $-\frac{\sqrt{a+b x+c x^2} \left (h x \left (h^2 \left (8 a^2 f h^2-2 a b h (10 f g-e h)+b^2 \left (11 f g^2-h (d h+e g)\right )\right )+2 c g h \left (2 a h (6 f g-e h)-b \left (12 f g^2-h (2 d h+e g)\right )\right )+4 c^2 \left (3 f g^4-d g^2 h^2\right )\right )+h^2 \left (4 a^2 e h^3-2 a b h \left (d h^2+2 e g h+3 f g^2\right )+b^2 g \left (d h^2+e g h+5 f g^2\right )\right )-2 c g h \left (-2 a d h^3-6 a f g^2 h+b d g h^2+7 b f g^3\right )+8 c^2 f g^5\right )}{8 h^3 (g+h x)^2 \left (a h^2-b g h+c g^2\right )^2}-\frac{\tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right ) \left (2 c h^2 \left (4 a^2 h^2 (4 f g-e h)-2 a b h \left (-d h^2-e g h+15 f g^2\right )+b^2 \left (d g h^2+15 f g^3\right )\right )-b h^3 \left (8 a^2 f h^2-2 a b h (e h+6 f g)+b^2 \left (d h^2+e g h+5 f g^2\right )\right )-8 c^2 g h \left (a d h^3-5 a f g^2 h+5 b f g^3\right )+16 c^3 f g^5\right )}{16 h^4 \left (a h^2-b g h+c g^2\right )^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (f g^2-h (e g-d h)\right )}{3 h (g+h x)^3 \left (a h^2-b g h+c g^2\right )}+\frac{\sqrt{c} f \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{h^4}$

[Out]

-((8*c^2*f*g^5 - 2*c*g*h*(7*b*f*g^3 - 6*a*f*g^2*h + b*d*g*h^2 - 2*a*d*h^3) + h^2*(4*a^2*e*h^3 + b^2*g*(5*f*g^2
+ e*g*h + d*h^2) - 2*a*b*h*(3*f*g^2 + 2*e*g*h + d*h^2)) + h*(4*c^2*(3*f*g^4 - d*g^2*h^2) + h^2*(8*a^2*f*h^2 -
2*a*b*h*(10*f*g - e*h) + b^2*(11*f*g^2 - h*(e*g + d*h))) + 2*c*g*h*(2*a*h*(6*f*g - e*h) - b*(12*f*g^2 - h*(e*
g + 2*d*h))))*x)*Sqrt[a + b*x + c*x^2])/(8*h^3*(c*g^2 - b*g*h + a*h^2)^2*(g + h*x)^2) - ((f*g^2 - h*(e*g - d*h
))*(a + b*x + c*x^2)^(3/2))/(3*h*(c*g^2 - b*g*h + a*h^2)*(g + h*x)^3) + (Sqrt[c]*f*ArcTanh[(b + 2*c*x)/(2*Sqrt
[c]*Sqrt[a + b*x + c*x^2])])/h^4 - ((16*c^3*f*g^5 - 8*c^2*g*h*(5*b*f*g^3 - 5*a*f*g^2*h + a*d*h^3) - b*h^3*(8*a
^2*f*h^2 - 2*a*b*h*(6*f*g + e*h) + b^2*(5*f*g^2 + e*g*h + d*h^2)) + 2*c*h^2*(4*a^2*h^2*(4*f*g - e*h) - 2*a*b*h
*(15*f*g^2 - e*g*h - d*h^2) + b^2*(15*f*g^3 + d*g*h^2)))*ArcTanh[(b*g - 2*a*h + (2*c*g - b*h)*x)/(2*Sqrt[c*g^2
- b*g*h + a*h^2]*Sqrt[a + b*x + c*x^2])])/(16*h^4*(c*g^2 - b*g*h + a*h^2)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 1.44877, antiderivative size = 601, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.188, Rules used = {1650, 810, 843, 621, 206, 724} $-\frac{\sqrt{a+b x+c x^2} \left (h x \left (8 a^2 f h^3-2 b \left (a h^2 (10 f g-e h)-c g h (2 d h+e g)+12 c f g^3\right )+4 a c g h (6 f g-e h)+b^2 h \left (11 f g^2-h (d h+e g)\right )+c^2 \left (\frac{12 f g^4}{h}-4 d g^2 h\right )\right )+4 a^2 e h^4-2 b \left (a h^2 \left (d h^2+2 e g h+3 f g^2\right )+c \left (d g^2 h^2+7 f g^4\right )\right )+4 a c g h \left (d h^2+3 f g^2\right )+b^2 g h \left (h (d h+e g)+5 f g^2\right )+\frac{8 c^2 f g^5}{h}\right )}{8 h^2 (g+h x)^2 \left (a h^2-b g h+c g^2\right )^2}-\frac{\tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right ) \left (2 c h^2 \left (4 a^2 h^2 (4 f g-e h)-2 a b h \left (-d h^2-e g h+15 f g^2\right )+b^2 \left (d g h^2+15 f g^3\right )\right )-b h^3 \left (8 a^2 f h^2-2 a b h (e h+6 f g)+b^2 \left (d h^2+e g h+5 f g^2\right )\right )-8 c^2 g h \left (a d h^3-5 a f g^2 h+5 b f g^3\right )+16 c^3 f g^5\right )}{16 h^4 \left (a h^2-b g h+c g^2\right )^{5/2}}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (f g^2-h (e g-d h)\right )}{3 h (g+h x)^3 \left (a h^2-b g h+c g^2\right )}+\frac{\sqrt{c} f \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{h^4}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2))/(g + h*x)^4,x]

[Out]

-(((8*c^2*f*g^5)/h + 4*a^2*e*h^4 + 4*a*c*g*h*(3*f*g^2 + d*h^2) + b^2*g*h*(5*f*g^2 + h*(e*g + d*h)) - 2*b*(a*h^
2*(3*f*g^2 + 2*e*g*h + d*h^2) + c*(7*f*g^4 + d*g^2*h^2)) + h*(8*a^2*f*h^3 + 4*a*c*g*h*(6*f*g - e*h) + c^2*((12
*f*g^4)/h - 4*d*g^2*h) + b^2*h*(11*f*g^2 - h*(e*g + d*h)) - 2*b*(12*c*f*g^3 - c*g*h*(e*g + 2*d*h) + a*h^2*(10*
f*g - e*h)))*x)*Sqrt[a + b*x + c*x^2])/(8*h^2*(c*g^2 - b*g*h + a*h^2)^2*(g + h*x)^2) - ((f*g^2 - h*(e*g - d*h)
)*(a + b*x + c*x^2)^(3/2))/(3*h*(c*g^2 - b*g*h + a*h^2)*(g + h*x)^3) + (Sqrt[c]*f*ArcTanh[(b + 2*c*x)/(2*Sqrt[
c]*Sqrt[a + b*x + c*x^2])])/h^4 - ((16*c^3*f*g^5 - 8*c^2*g*h*(5*b*f*g^3 - 5*a*f*g^2*h + a*d*h^3) - b*h^3*(8*a^
2*f*h^2 - 2*a*b*h*(6*f*g + e*h) + b^2*(5*f*g^2 + e*g*h + d*h^2)) + 2*c*h^2*(4*a^2*h^2*(4*f*g - e*h) - 2*a*b*h*
(15*f*g^2 - e*g*h - d*h^2) + b^2*(15*f*g^3 + d*g*h^2)))*ArcTanh[(b*g - 2*a*h + (2*c*g - b*h)*x)/(2*Sqrt[c*g^2
- b*g*h + a*h^2]*Sqrt[a + b*x + c*x^2])])/(16*h^4*(c*g^2 - b*g*h + a*h^2)^(5/2))

Rule 1650

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomia
lQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*
x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^
(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R*(m + 1) - b*e*R*(m + p + 2)
- c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 843

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

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])

Rule 724

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

Rubi steps

\begin{align*} \int \frac{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )}{(g+h x)^4} \, dx &=-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 h \left (c g^2-b g h+a h^2\right ) (g+h x)^3}-\frac{\int \frac{\left (-\frac{3}{2} \left (2 c d g-b e g-2 a f g+\frac{b f g^2}{h}-b d h+2 a e h\right )+3 f \left (b g-\frac{c g^2}{h}-a h\right ) x\right ) \sqrt{a+b x+c x^2}}{(g+h x)^3} \, dx}{3 \left (c g^2-b g h+a h^2\right )}\\ &=-\frac{\left (\frac{8 c^2 f g^5}{h}+4 a^2 e h^4+4 a c g h \left (3 f g^2+d h^2\right )+b^2 g h \left (5 f g^2+h (e g+d h)\right )-2 b \left (a h^2 \left (3 f g^2+2 e g h+d h^2\right )+c \left (7 f g^4+d g^2 h^2\right )\right )+h \left (8 a^2 f h^3+4 a c g h (6 f g-e h)+c^2 \left (\frac{12 f g^4}{h}-4 d g^2 h\right )+b^2 h \left (11 f g^2-h (e g+d h)\right )-2 b \left (12 c f g^3-c g h (e g+2 d h)+a h^2 (10 f g-e h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{8 h^2 \left (c g^2-b g h+a h^2\right )^2 (g+h x)^2}-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 h \left (c g^2-b g h+a h^2\right ) (g+h x)^3}+\frac{\int \frac{\frac{3 \left (b^3 h^2 \left (5 f g^2+h (e g+d h)\right )+4 b \left (2 c^2 f g^4+2 a^2 f h^4+a c h^2 \left (7 f g^2-e g h-d h^2\right )\right )-8 a c h \left (a h^2 (2 f g-e h)+c \left (f g^3-d g h^2\right )\right )-2 b^2 \left (a h^3 (6 f g+e h)+c \left (7 f g^3 h+d g h^3\right )\right )\right )}{4 h}+\frac{12 c f \left (c g^2-b g h+a h^2\right )^2 x}{h}}{(g+h x) \sqrt{a+b x+c x^2}} \, dx}{12 h^2 \left (c g^2-b g h+a h^2\right )^2}\\ &=-\frac{\left (\frac{8 c^2 f g^5}{h}+4 a^2 e h^4+4 a c g h \left (3 f g^2+d h^2\right )+b^2 g h \left (5 f g^2+h (e g+d h)\right )-2 b \left (a h^2 \left (3 f g^2+2 e g h+d h^2\right )+c \left (7 f g^4+d g^2 h^2\right )\right )+h \left (8 a^2 f h^3+4 a c g h (6 f g-e h)+c^2 \left (\frac{12 f g^4}{h}-4 d g^2 h\right )+b^2 h \left (11 f g^2-h (e g+d h)\right )-2 b \left (12 c f g^3-c g h (e g+2 d h)+a h^2 (10 f g-e h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{8 h^2 \left (c g^2-b g h+a h^2\right )^2 (g+h x)^2}-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 h \left (c g^2-b g h+a h^2\right ) (g+h x)^3}+\frac{(c f) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{h^4}-\frac{\left (16 c^3 f g^5-8 c^2 g h \left (5 b f g^3-5 a f g^2 h+a d h^3\right )-b h^3 \left (8 a^2 f h^2-2 a b h (6 f g+e h)+b^2 \left (5 f g^2+e g h+d h^2\right )\right )+2 c h^2 \left (4 a^2 h^2 (4 f g-e h)-2 a b h \left (15 f g^2-e g h-d h^2\right )+b^2 \left (15 f g^3+d g h^2\right )\right )\right ) \int \frac{1}{(g+h x) \sqrt{a+b x+c x^2}} \, dx}{16 h^4 \left (c g^2-b g h+a h^2\right )^2}\\ &=-\frac{\left (\frac{8 c^2 f g^5}{h}+4 a^2 e h^4+4 a c g h \left (3 f g^2+d h^2\right )+b^2 g h \left (5 f g^2+h (e g+d h)\right )-2 b \left (a h^2 \left (3 f g^2+2 e g h+d h^2\right )+c \left (7 f g^4+d g^2 h^2\right )\right )+h \left (8 a^2 f h^3+4 a c g h (6 f g-e h)+c^2 \left (\frac{12 f g^4}{h}-4 d g^2 h\right )+b^2 h \left (11 f g^2-h (e g+d h)\right )-2 b \left (12 c f g^3-c g h (e g+2 d h)+a h^2 (10 f g-e h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{8 h^2 \left (c g^2-b g h+a h^2\right )^2 (g+h x)^2}-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 h \left (c g^2-b g h+a h^2\right ) (g+h x)^3}+\frac{(2 c f) \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 )}{h^4}+\frac{\left (16 c^3 f g^5-8 c^2 g h \left (5 b f g^3-5 a f g^2 h+a d h^3\right )-b h^3 \left (8 a^2 f h^2-2 a b h (6 f g+e h)+b^2 \left (5 f g^2+e g h+d h^2\right )\right )+2 c h^2 \left (4 a^2 h^2 (4 f g-e h)-2 a b h \left (15 f g^2-e g h-d h^2\right )+b^2 \left (15 f g^3+d g h^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c g^2-4 b g h+4 a h^2-x^2} \, dx,x,\frac{-b g+2 a h-(2 c g-b h) x}{\sqrt{a+b x+c x^2}}\right )}{8 h^4 \left (c g^2-b g h+a h^2\right )^2}\\ &=-\frac{\left (\frac{8 c^2 f g^5}{h}+4 a^2 e h^4+4 a c g h \left (3 f g^2+d h^2\right )+b^2 g h \left (5 f g^2+h (e g+d h)\right )-2 b \left (a h^2 \left (3 f g^2+2 e g h+d h^2\right )+c \left (7 f g^4+d g^2 h^2\right )\right )+h \left (8 a^2 f h^3+4 a c g h (6 f g-e h)+c^2 \left (\frac{12 f g^4}{h}-4 d g^2 h\right )+b^2 h \left (11 f g^2-h (e g+d h)\right )-2 b \left (12 c f g^3-c g h (e g+2 d h)+a h^2 (10 f g-e h)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{8 h^2 \left (c g^2-b g h+a h^2\right )^2 (g+h x)^2}-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{3 h \left (c g^2-b g h+a h^2\right ) (g+h x)^3}+\frac{\sqrt{c} f \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{h^4}-\frac{\left (16 c^3 f g^5-8 c^2 g h \left (5 b f g^3-5 a f g^2 h+a d h^3\right )-b h^3 \left (8 a^2 f h^2-2 a b h (6 f g+e h)+b^2 \left (5 f g^2+e g h+d h^2\right )\right )+2 c h^2 \left (4 a^2 h^2 (4 f g-e h)-2 a b h \left (15 f g^2-e g h-d h^2\right )+b^2 \left (15 f g^3+d g h^2\right )\right )\right ) \tanh ^{-1}\left (\frac{b g-2 a h+(2 c g-b h) x}{2 \sqrt{c g^2-b g h+a h^2} \sqrt{a+b x+c x^2}}\right )}{16 h^4 \left (c g^2-b g h+a h^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 2.28668, size = 439, normalized size = 0.73 $\frac{\frac{\left (\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a h-b g+b h x-2 c g x}{2 \sqrt{a+x (b+c x)} \sqrt{h (a h-b g)+c g^2}}\right )}{8 \left (h (a h-b g)+c g^2\right )^{3/2}}+\frac{\sqrt{a+x (b+c x)} (-2 a h+b (g-h x)+2 c g x)}{4 (g+h x)^2 \left (h (a h-b g)+c g^2\right )}\right ) \left (2 a h^2 (e h-2 f g)-b h \left (h (d h+e g)-3 f g^2\right )+c \left (2 d g h^2-2 f g^3\right )\right )}{2 \left (h (a h-b g)+c g^2\right )}-\frac{h (a+x (b+c x))^{3/2} \left (h (d h-e g)+f g^2\right )}{3 (g+h x)^3 \left (h (a h-b g)+c g^2\right )}+\frac{f \left (\frac{(2 c g-b h) \tanh ^{-1}\left (\frac{2 a h-b g+b h x-2 c g x}{2 \sqrt{a+x (b+c x)} \sqrt{h (a h-b g)+c g^2}}\right )}{2 \sqrt{h (a h-b g)+c g^2}}-\frac{h \sqrt{a+x (b+c x)}}{g+h x}+\sqrt{c} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )}{h^2}}{h^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2))/(g + h*x)^4,x]

[Out]

(-(h*(f*g^2 + h*(-(e*g) + d*h))*(a + x*(b + c*x))^(3/2))/(3*(c*g^2 + h*(-(b*g) + a*h))*(g + h*x)^3) + ((2*a*h^
2*(-2*f*g + e*h) + c*(-2*f*g^3 + 2*d*g*h^2) - b*h*(-3*f*g^2 + h*(e*g + d*h)))*((Sqrt[a + x*(b + c*x)]*(-2*a*h
+ 2*c*g*x + b*(g - h*x)))/(4*(c*g^2 + h*(-(b*g) + a*h))*(g + h*x)^2) + ((b^2 - 4*a*c)*ArcTanh[(-(b*g) + 2*a*h
- 2*c*g*x + b*h*x)/(2*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*Sqrt[a + x*(b + c*x)])])/(8*(c*g^2 + h*(-(b*g) + a*h))^(3
/2))))/(2*(c*g^2 + h*(-(b*g) + a*h))) + (f*(-((h*Sqrt[a + x*(b + c*x)])/(g + h*x)) + Sqrt[c]*ArcTanh[(b + 2*c*
x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + ((2*c*g - b*h)*ArcTanh[(-(b*g) + 2*a*h - 2*c*g*x + b*h*x)/(2*Sqrt[c*g^
2 + h*(-(b*g) + a*h)]*Sqrt[a + x*(b + c*x)])])/(2*Sqrt[c*g^2 + h*(-(b*g) + a*h)])))/h^2)/h^2

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Maple [B]  time = 0.3, size = 19321, normalized size = 32. \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2)/(h*x+g)^4,x)

[Out]

result too large to display

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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)*(c*x^2+b*x+a)^(1/2)/(h*x+g)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2)/(h*x+g)^4,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}{\left (g + h x\right )^{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e*x+d)*(c*x**2+b*x+a)**(1/2)/(h*x+g)**4,x)

[Out]

Integral(sqrt(a + b*x + c*x**2)*(d + e*x + f*x**2)/(g + h*x)**4, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2)/(h*x+g)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError