Integrand size = 32, antiderivative size = 336 \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=-\frac {\left (f g^2-h (e g-d h)\right ) \sqrt {a+b x+c x^2}}{2 h \left (c g^2-b g h+a h^2\right ) (g+h x)^2}+\frac {\left (2 c g \left (f g^2+h (e g-3 d h)\right )+h \left (4 a h (2 f g-e h)-b \left (5 f g^2-e g h-3 d h^2\right )\right )\right ) \sqrt {a+b x+c x^2}}{4 h \left (c g^2-b g h+a h^2\right )^2 (g+h x)}+\frac {\left (8 c^2 d g^2+8 a^2 f h^2-4 a b h (2 f g+e h)+b^2 \left (3 f g^2+e g h+3 d h^2\right )-4 c \left (b g (e g+2 d h)+a \left (f g^2-3 e g h+d h^2\right )\right )\right ) \text {arctanh}\left (\frac {b g-2 a h+(2 c g-b h) x}{2 \sqrt {c g^2-b g h+a h^2} \sqrt {a+b x+c x^2}}\right )}{8 \left (c g^2-b g h+a h^2\right )^{5/2}} \]
1/8*(8*c^2*d*g^2+8*a^2*f*h^2-4*a*b*h*(e*h+2*f*g)+b^2*(3*d*h^2+e*g*h+3*f*g^ 2)-4*c*(b*g*(2*d*h+e*g)+a*(d*h^2-3*e*g*h+f*g^2)))*arctanh(1/2*(b*g-2*a*h+( -b*h+2*c*g)*x)/(a*h^2-b*g*h+c*g^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/(a*h^2-b*g*h +c*g^2)^(5/2)-1/2*(f*g^2-h*(-d*h+e*g))*(c*x^2+b*x+a)^(1/2)/h/(a*h^2-b*g*h+ c*g^2)/(h*x+g)^2+1/4*(2*c*g*(f*g^2+h*(-3*d*h+e*g))+h*(4*a*h*(-e*h+2*f*g)-b *(-3*d*h^2-e*g*h+5*f*g^2)))*(c*x^2+b*x+a)^(1/2)/h/(a*h^2-b*g*h+c*g^2)^2/(h *x+g)
Time = 11.21 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.40 \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=-\frac {\frac {4 h \left (f g^2+h (-e g+d h)\right ) \sqrt {a+x (b+c x)}}{\left (c g^2+h (-b g+a h)\right ) (g+h x)^2}+\frac {8 h (-2 f g+e h) \sqrt {a+x (b+c x)}}{\left (c g^2+h (-b g+a h)\right ) (g+h x)}+\frac {4 (2 c g-b h) (2 f g-e h) \text {arctanh}\left (\frac {-2 a h+2 c g x+b (g-h x)}{2 \sqrt {c g^2+h (-b g+a h)} \sqrt {a+x (b+c x)}}\right )}{\left (c g^2+h (-b g+a h)\right )^{3/2}}-\frac {8 f \text {arctanh}\left (\frac {-2 a h+2 c g x+b (g-h x)}{2 \sqrt {c g^2+h (-b g+a h)} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c g^2+h (-b g+a h)}}+\frac {\left (f g^2+h (-e g+d h)\right ) \left (6 h (2 c g-b h) \sqrt {c g^2+h (-b g+a h)} \sqrt {a+x (b+c x)}-\left (8 c^2 g^2+3 b^2 h^2-4 c h (2 b g+a h)\right ) (g+h x) \text {arctanh}\left (\frac {-2 a h+2 c g x+b (g-h x)}{2 \sqrt {c g^2+h (-b g+a h)} \sqrt {a+x (b+c x)}}\right )\right )}{\left (c g^2+h (-b g+a h)\right )^{5/2} (g+h x)}}{8 h^2} \]
-1/8*((4*h*(f*g^2 + h*(-(e*g) + d*h))*Sqrt[a + x*(b + c*x)])/((c*g^2 + h*( -(b*g) + a*h))*(g + h*x)^2) + (8*h*(-2*f*g + e*h)*Sqrt[a + x*(b + c*x)])/( (c*g^2 + h*(-(b*g) + a*h))*(g + h*x)) + (4*(2*c*g - b*h)*(2*f*g - e*h)*Arc Tanh[(-2*a*h + 2*c*g*x + b*(g - h*x))/(2*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*Sq rt[a + x*(b + c*x)])])/(c*g^2 + h*(-(b*g) + a*h))^(3/2) - (8*f*ArcTanh[(-2 *a*h + 2*c*g*x + b*(g - h*x))/(2*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*Sqrt[a + x *(b + c*x)])])/Sqrt[c*g^2 + h*(-(b*g) + a*h)] + ((f*g^2 + h*(-(e*g) + d*h) )*(6*h*(2*c*g - b*h)*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*Sqrt[a + x*(b + c*x)] - (8*c^2*g^2 + 3*b^2*h^2 - 4*c*h*(2*b*g + a*h))*(g + h*x)*ArcTanh[(-2*a*h + 2*c*g*x + b*(g - h*x))/(2*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*Sqrt[a + x*(b + c*x)])]))/((c*g^2 + h*(-(b*g) + a*h))^(5/2)*(g + h*x)))/h^2
Time = 0.69 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2181, 27, 1228, 1154, 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.
\(\displaystyle \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx\) |
\(\Big \downarrow \) 2181 |
\(\displaystyle -\frac {\int -\frac {\frac {b f g^2}{h}+4 c d g-b e g-4 a f g-3 b d h+4 a e h+2 \left (\frac {c f g^2}{h}+c e g-2 b f g-c d h+2 a f h\right ) x}{2 (g+h x)^2 \sqrt {c x^2+b x+a}}dx}{2 \left (a h^2-b g h+c g^2\right )}-\frac {\sqrt {a+b x+c x^2} \left (f g^2-h (e g-d h)\right )}{2 h (g+h x)^2 \left (a h^2-b g h+c g^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 c d g-4 a f g+4 a e h-b \left (-\frac {f g^2}{h}+e g+3 d h\right )-2 \left (2 b f g-2 a f h-c \left (\frac {f g^2}{h}+e g-d h\right )\right ) x}{(g+h x)^2 \sqrt {c x^2+b x+a}}dx}{4 \left (a h^2-b g h+c g^2\right )}-\frac {\sqrt {a+b x+c x^2} \left (f g^2-h (e g-d h)\right )}{2 h (g+h x)^2 \left (a h^2-b g h+c g^2\right )}\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {\frac {\left (8 a^2 f h^2-4 c \left (-a h (3 e g-d h)+a f g^2+b g (2 d h+e g)\right )-4 a b h (e h+2 f g)+b^2 \left (h (3 d h+e g)+3 f g^2\right )+8 c^2 d g^2\right ) \int \frac {1}{(g+h x) \sqrt {c x^2+b x+a}}dx}{2 \left (a h^2-b g h+c g^2\right )}+\frac {\sqrt {a+b x+c x^2} \left (2 c \left (g h (e g-3 d h)+f g^3\right )-h \left (-4 a h (2 f g-e h)-b h (3 d h+e g)+5 b f g^2\right )\right )}{h (g+h x) \left (a h^2-b g h+c g^2\right )}}{4 \left (a h^2-b g h+c g^2\right )}-\frac {\sqrt {a+b x+c x^2} \left (f g^2-h (e g-d h)\right )}{2 h (g+h x)^2 \left (a h^2-b g h+c g^2\right )}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\frac {\sqrt {a+b x+c x^2} \left (2 c \left (g h (e g-3 d h)+f g^3\right )-h \left (-4 a h (2 f g-e h)-b h (3 d h+e g)+5 b f g^2\right )\right )}{h (g+h x) \left (a h^2-b g h+c g^2\right )}-\frac {\left (8 a^2 f h^2-4 c \left (-a h (3 e g-d h)+a f g^2+b g (2 d h+e g)\right )-4 a b h (e h+2 f g)+b^2 \left (h (3 d h+e g)+3 f g^2\right )+8 c^2 d g^2\right ) \int \frac {1}{4 \left (c g^2-b h g+a h^2\right )-\frac {(b g-2 a h+(2 c g-b h) x)^2}{c x^2+b x+a}}d\left (-\frac {b g-2 a h+(2 c g-b h) x}{\sqrt {c x^2+b x+a}}\right )}{a h^2-b g h+c g^2}}{4 \left (a h^2-b g h+c g^2\right )}-\frac {\sqrt {a+b x+c x^2} \left (f g^2-h (e g-d h)\right )}{2 h (g+h x)^2 \left (a h^2-b g h+c g^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {-2 a h+x (2 c g-b h)+b g}{2 \sqrt {a+b x+c x^2} \sqrt {a h^2-b g h+c g^2}}\right ) \left (8 a^2 f h^2-4 c \left (-a h (3 e g-d h)+a f g^2+b g (2 d h+e g)\right )-4 a b h (e h+2 f g)+b^2 \left (h (3 d h+e g)+3 f g^2\right )+8 c^2 d g^2\right )}{2 \left (a h^2-b g h+c g^2\right )^{3/2}}+\frac {\sqrt {a+b x+c x^2} \left (2 c \left (g h (e g-3 d h)+f g^3\right )-h \left (-4 a h (2 f g-e h)-b h (3 d h+e g)+5 b f g^2\right )\right )}{h (g+h x) \left (a h^2-b g h+c g^2\right )}}{4 \left (a h^2-b g h+c g^2\right )}-\frac {\sqrt {a+b x+c x^2} \left (f g^2-h (e g-d h)\right )}{2 h (g+h x)^2 \left (a h^2-b g h+c g^2\right )}\) |
-1/2*((f*g^2 - h*(e*g - d*h))*Sqrt[a + b*x + c*x^2])/(h*(c*g^2 - b*g*h + a *h^2)*(g + h*x)^2) + (((2*c*(f*g^3 + g*h*(e*g - 3*d*h)) - h*(5*b*f*g^2 - b *h*(e*g + 3*d*h) - 4*a*h*(2*f*g - e*h)))*Sqrt[a + b*x + c*x^2])/(h*(c*g^2 - b*g*h + a*h^2)*(g + h*x)) + ((8*c^2*d*g^2 + 8*a^2*f*h^2 - 4*a*b*h*(2*f*g + e*h) - 4*c*(a*f*g^2 - a*h*(3*e*g - d*h) + b*g*(e*g + 2*d*h)) + b^2*(3*f *g^2 + h*(e*g + 3*d*h)))*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])])/(2*(c*g^2 - b*g*h + a*h^2)^ (3/2)))/(4*(c*g^2 - b*g*h + a*h^2))
3.3.32.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-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]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_ ), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = Polynomi alRemainder[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] + Simp[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)*Qx + 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]
Leaf count of result is larger than twice the leaf count of optimal. \(1012\) vs. \(2(318)=636\).
Time = 0.99 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.01
-f/h^3/((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*ln((2*(a*h^2-b*g*h+c*g^2)/h^2+(b*h- 2*c*g)/h*(x+1/h*g)+2*((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*((x+1/h*g)^2*c+(b*h-2 *c*g)/h*(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2))/(x+1/h*g))+(e*h-2*f*g)/h ^4*(-1/(a*h^2-b*g*h+c*g^2)*h^2/(x+1/h*g)*((x+1/h*g)^2*c+(b*h-2*c*g)/h*(x+1 /h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2)+1/2*(b*h-2*c*g)*h/(a*h^2-b*g*h+c*g^2) /((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*ln((2*(a*h^2-b*g*h+c*g^2)/h^2+(b*h-2*c*g) /h*(x+1/h*g)+2*((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*((x+1/h*g)^2*c+(b*h-2*c*g)/ h*(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2))/(x+1/h*g)))+(d*h^2-e*g*h+f*g^2 )/h^5*(-1/2/(a*h^2-b*g*h+c*g^2)*h^2/(x+1/h*g)^2*((x+1/h*g)^2*c+(b*h-2*c*g) /h*(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2)-3/4*(b*h-2*c*g)*h/(a*h^2-b*g*h +c*g^2)*(-1/(a*h^2-b*g*h+c*g^2)*h^2/(x+1/h*g)*((x+1/h*g)^2*c+(b*h-2*c*g)/h *(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2)+1/2*(b*h-2*c*g)*h/(a*h^2-b*g*h+c *g^2)/((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*ln((2*(a*h^2-b*g*h+c*g^2)/h^2+(b*h-2 *c*g)/h*(x+1/h*g)+2*((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*((x+1/h*g)^2*c+(b*h-2* c*g)/h*(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2))/(x+1/h*g)))+1/2*c/(a*h^2- b*g*h+c*g^2)*h^2/((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*ln((2*(a*h^2-b*g*h+c*g^2) /h^2+(b*h-2*c*g)/h*(x+1/h*g)+2*((a*h^2-b*g*h+c*g^2)/h^2)^(1/2)*((x+1/h*g)^ 2*c+(b*h-2*c*g)/h*(x+1/h*g)+(a*h^2-b*g*h+c*g^2)/h^2)^(1/2))/(x+1/h*g)))
Leaf count of result is larger than twice the leaf count of optimal. 996 vs. \(2 (318) = 636\).
Time = 34.05 (sec) , antiderivative size = 2034, normalized size of antiderivative = 6.05 \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Too large to display} \]
[1/16*(((8*c^2*d - 4*b*c*e + (3*b^2 - 4*a*c)*f)*g^4 - (8*b*c*d + 8*a*b*f - (b^2 + 12*a*c)*e)*g^3*h - (4*a*b*e - 8*a^2*f - (3*b^2 - 4*a*c)*d)*g^2*h^2 + ((8*c^2*d - 4*b*c*e + (3*b^2 - 4*a*c)*f)*g^2*h^2 - (8*b*c*d + 8*a*b*f - (b^2 + 12*a*c)*e)*g*h^3 - (4*a*b*e - 8*a^2*f - (3*b^2 - 4*a*c)*d)*h^4)*x^ 2 + 2*((8*c^2*d - 4*b*c*e + (3*b^2 - 4*a*c)*f)*g^3*h - (8*b*c*d + 8*a*b*f - (b^2 + 12*a*c)*e)*g^2*h^2 - (4*a*b*e - 8*a^2*f - (3*b^2 - 4*a*c)*d)*g*h^ 3)*x)*sqrt(c*g^2 - b*g*h + a*h^2)*log((8*a*b*g*h - 8*a^2*h^2 - (b^2 + 4*a* c)*g^2 - (8*c^2*g^2 - 8*b*c*g*h + (b^2 + 4*a*c)*h^2)*x^2 - 4*sqrt(c*g^2 - b*g*h + a*h^2)*sqrt(c*x^2 + b*x + a)*(b*g - 2*a*h + (2*c*g - b*h)*x) - 2*( 4*b*c*g^2 + 4*a*b*h^2 - (3*b^2 + 4*a*c)*g*h)*x)/(h^2*x^2 + 2*g*h*x + g^2)) - 4*(2*a^2*d*h^5 - (4*c^2*e - 3*b*c*f)*g^5 + (8*c^2*d + 5*b*c*e - 3*(b^2 + 2*a*c)*f)*g^4*h - (13*b*c*d - 9*a*b*f + (b^2 + 2*a*c)*e)*g^3*h^2 - (a*b* e + 6*a^2*f - 5*(b^2 + 2*a*c)*d)*g^2*h^3 - (7*a*b*d - 2*a^2*e)*g*h^4 - (2* c^2*f*g^5 + (2*c^2*e - 7*b*c*f)*g^4*h - (6*c^2*d + b*c*e - 5*(b^2 + 2*a*c) *f)*g^3*h^2 + (9*b*c*d - 13*a*b*f - (b^2 + 2*a*c)*e)*g^2*h^3 + (5*a*b*e + 8*a^2*f - 3*(b^2 + 2*a*c)*d)*g*h^4 + (3*a*b*d - 4*a^2*e)*h^5)*x)*sqrt(c*x^ 2 + b*x + a))/(c^3*g^8 - 3*b*c^2*g^7*h - 3*a^2*b*g^3*h^5 + a^3*g^2*h^6 + 3 *(b^2*c + a*c^2)*g^6*h^2 - (b^3 + 6*a*b*c)*g^5*h^3 + 3*(a*b^2 + a^2*c)*g^4 *h^4 + (c^3*g^6*h^2 - 3*b*c^2*g^5*h^3 - 3*a^2*b*g*h^7 + a^3*h^8 + 3*(b^2*c + a*c^2)*g^4*h^4 - (b^3 + 6*a*b*c)*g^3*h^5 + 3*(a*b^2 + a^2*c)*g^2*h^6...
\[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=\int \frac {d + e x + f x^{2}}{\left (g + h x\right )^{3} \sqrt {a + b x + c x^{2}}}\, dx \]
Exception generated. \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*h^2-b*g*h>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 2279 vs. \(2 (318) = 636\).
Time = 0.33 (sec) , antiderivative size = 2279, normalized size of antiderivative = 6.78 \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Too large to display} \]
1/4*(8*c^2*d*g^2 - 4*b*c*e*g^2 + 3*b^2*f*g^2 - 4*a*c*f*g^2 - 8*b*c*d*g*h + b^2*e*g*h + 12*a*c*e*g*h - 8*a*b*f*g*h + 3*b^2*d*h^2 - 4*a*c*d*h^2 - 4*a* b*e*h^2 + 8*a^2*f*h^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x + a))*h + sq rt(c)*g)/sqrt(-c*g^2 + b*g*h - a*h^2))/((c^2*g^4 - 2*b*c*g^3*h + b^2*g^2*h ^2 + 2*a*c*g^2*h^2 - 2*a*b*g*h^3 + a^2*h^4)*sqrt(-c*g^2 + b*g*h - a*h^2)) + 1/4*(8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^2*f*g^4*h - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*f*g^3*h^2 - 8*(sqrt(c)*x - sqrt(c*x^2 + b* x + a))^3*c^2*d*g^2*h^3 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*e*g^ 2*h^3 + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*f*g^2*h^3 + 20*(sqrt(c )*x - sqrt(c*x^2 + b*x + a))^3*a*c*f*g^2*h^3 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*d*g*h^4 - (sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*e*g*h ^4 - 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*e*g*h^4 - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*f*g*h^4 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*d*h^5 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c*d*h^5 + 4*(s qrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*e*h^5 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*f*g^5 + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5 /2)*e*g^4*h - 16*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2)*f*g^4*h - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*d*g^3*h^2 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2)*e*g^3*h^2 - (sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*sqrt(c)*f*g^3*h^2 + 28*(sqrt(c)*x - sqrt(c*x^2 + b*x ...
Timed out. \[ \int \frac {d+e x+f x^2}{(g+h x)^3 \sqrt {a+b x+c x^2}} \, dx=\int \frac {f\,x^2+e\,x+d}{{\left (g+h\,x\right )}^3\,\sqrt {c\,x^2+b\,x+a}} \,d x \]