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

Optimal. Leaf size=321 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (4 c h (2 c g-b h) (b f g-2 c d h)-\left (-4 c h (b g-a h)-b^2 h^2+8 c^2 g^2\right ) (b f h-2 c e h+2 c f g)\right )}{16 c^{5/2} h^4}-\frac{\sqrt{a+b x+c x^2} (4 c h (b f g-2 c d h)+2 c h x (b f h-2 c e h+2 c f g)-(4 c g-b h) (b f h-2 c e h+2 c f g))}{8 c^2 h^3}+\frac{\sqrt{a h^2-b g h+c g^2} \left (d h^2-e g h+f g^2\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right )}{h^4}+\frac{f \left (a+b x+c x^2\right )^{3/2}}{3 c h} \]

[Out]

-((4*c*h*(b*f*g - 2*c*d*h) - (4*c*g - b*h)*(2*c*f*g - 2*c*e*h + b*f*h) + 2*c*h*(2*c*f*g - 2*c*e*h + b*f*h)*x)*
Sqrt[a + b*x + c*x^2])/(8*c^2*h^3) + (f*(a + b*x + c*x^2)^(3/2))/(3*c*h) + ((4*c*h*(2*c*g - b*h)*(b*f*g - 2*c*
d*h) - (2*c*f*g - 2*c*e*h + b*f*h)*(8*c^2*g^2 - b^2*h^2 - 4*c*h*(b*g - a*h)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S
qrt[a + b*x + c*x^2])])/(16*c^(5/2)*h^4) + (Sqrt[c*g^2 - b*g*h + a*h^2]*(f*g^2 - e*g*h + d*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])])/h^4

________________________________________________________________________________________

Rubi [A]  time = 0.778152, antiderivative size = 321, 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 = {1653, 814, 843, 621, 206, 724} \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (4 c h (2 c g-b h) (b f g-2 c d h)-\left (-4 c h (b g-a h)-b^2 h^2+8 c^2 g^2\right ) (b f h-2 c e h+2 c f g)\right )}{16 c^{5/2} h^4}-\frac{\sqrt{a+b x+c x^2} (4 c h (b f g-2 c d h)+2 c h x (b f h-2 c e h+2 c f g)-(4 c g-b h) (b f h-2 c e h+2 c f g))}{8 c^2 h^3}+\frac{\sqrt{a h^2-b g h+c g^2} \left (d h^2-e g h+f g^2\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right )}{h^4}+\frac{f \left (a+b x+c x^2\right )^{3/2}}{3 c h} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*c*h*(b*f*g - 2*c*d*h) - (4*c*g - b*h)*(2*c*f*g - 2*c*e*h + b*f*h) + 2*c*h*(2*c*f*g - 2*c*e*h + b*f*h)*x)*
Sqrt[a + b*x + c*x^2])/(8*c^2*h^3) + (f*(a + b*x + c*x^2)^(3/2))/(3*c*h) + ((4*c*h*(2*c*g - b*h)*(b*f*g - 2*c*
d*h) - (2*c*f*g - 2*c*e*h + b*f*h)*(8*c^2*g^2 - b^2*h^2 - 4*c*h*(b*g - a*h)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S
qrt[a + b*x + c*x^2])])/(16*c^(5/2)*h^4) + (Sqrt[c*g^2 - b*g*h + a*h^2]*(f*g^2 - e*g*h + d*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])])/h^4

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 814

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

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

Mathematica [A]  time = 0.782375, size = 331, normalized size = 1.03 \[ \frac{2 \sqrt{c} \left (h \sqrt{a+x (b+c x)} \left (2 c h (4 a f h+b (3 e h-3 f g+f h x))-3 b^2 f h^2+4 c^2 \left (3 h (2 d h-2 e g+e h x)+f \left (6 g^2-3 g h x+2 h^2 x^2\right )\right )\right )-24 c^2 \sqrt{h (a h-b g)+c g^2} \left (h (d h-e g)+f g^2\right ) \tanh ^{-1}\left (\frac{2 a h-b g+b h x-2 c g x}{2 \sqrt{a+x (b+c x)} \sqrt{h (a h-b g)+c g^2}}\right )\right )-3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-8 c^2 h \left (a h (e h-f g)+b h (d h-e g)+b f g^2\right )+2 b c h^2 (2 a f h+b e h-b f g)-b^3 f h^3+16 c^3 g \left (h (d h-e g)+f g^2\right )\right )}{48 c^{5/2} h^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-(b^3*f*h^3) + 2*b*c*h^2*(-(b*f*g) + b*e*h + 2*a*f*h) + 16*c^3*g*(f*g^2 + h*(-(e*g) + d*h)) - 8*c^2*h*(b*
f*g^2 + b*h*(-(e*g) + d*h) + a*h*(-(f*g) + e*h)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + 2*S
qrt[c]*(h*Sqrt[a + x*(b + c*x)]*(-3*b^2*f*h^2 + 2*c*h*(4*a*f*h + b*(-3*f*g + 3*e*h + f*h*x)) + 4*c^2*(3*h*(-2*
e*g + 2*d*h + e*h*x) + f*(6*g^2 - 3*g*h*x + 2*h^2*x^2))) - 24*c^2*Sqrt[c*g^2 + h*(-(b*g) + a*h)]*(f*g^2 + h*(-
(e*g) + d*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
)])]))/(48*c^(5/2)*h^4)

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Maple [B]  time = 0.352, size = 2549, normalized size = 7.9 \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)*(c*x^2+b*x+a)^(1/2)/(h*x+g),x)

[Out]

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

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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),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),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 )}{g + h x}\, 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),x)

[Out]

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

[Out]

Exception raised: TypeError