3.194 \(\int \frac{\sqrt{a+b x+c x^2} (d+e x+f x^2)}{(g+h x)^5} \, dx\)

Optimal. Leaf size=497 \[ \frac{\sqrt{a+b x+c x^2} (-2 a h+x (2 c g-b h)+b g) \left (16 a^2 f h^2-4 c \left (a \left (d h^2-5 e g h+f g^2\right )+2 b g (2 d h+e g)\right )-8 a b h (e h+2 f g)+b^2 \left (5 d h^2+3 e g h+5 f g^2\right )+16 c^2 d g^2\right )}{64 (g+h x)^2 \left (a h^2-b g h+c g^2\right )^3}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right ) \left (16 a^2 f h^2-4 c \left (a \left (d h^2-5 e g h+f g^2\right )+2 b g (2 d h+e g)\right )-8 a b h (e h+2 f g)+b^2 \left (5 d h^2+3 e g h+5 f g^2\right )+16 c^2 d g^2\right )}{128 \left (a h^2-b g h+c g^2\right )^{7/2}}+\frac{\left (a+b x+c x^2\right )^{3/2} \left (h \left (8 a h (2 f g-e h)-b \left (-5 d h^2-3 e g h+11 f g^2\right )\right )+2 c g \left (h (e g-5 d h)+3 f g^2\right )\right )}{24 h (g+h x)^3 \left (a h^2-b g h+c g^2\right )^2}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (f g^2-h (e g-d h)\right )}{4 h (g+h x)^4 \left (a h^2-b g h+c g^2\right )} \]

[Out]

((16*c^2*d*g^2 + 16*a^2*f*h^2 - 8*a*b*h*(2*f*g + e*h) + b^2*(5*f*g^2 + 3*e*g*h + 5*d*h^2) - 4*c*(2*b*g*(e*g +
2*d*h) + a*(f*g^2 - 5*e*g*h + d*h^2)))*(b*g - 2*a*h + (2*c*g - b*h)*x)*Sqrt[a + b*x + c*x^2])/(64*(c*g^2 - b*g
*h + a*h^2)^3*(g + h*x)^2) - ((f*g^2 - h*(e*g - d*h))*(a + b*x + c*x^2)^(3/2))/(4*h*(c*g^2 - b*g*h + a*h^2)*(g
+ h*x)^4) + ((2*c*g*(3*f*g^2 + h*(e*g - 5*d*h)) + h*(8*a*h*(2*f*g - e*h) - b*(11*f*g^2 - 3*e*g*h - 5*d*h^2)))
*(a + b*x + c*x^2)^(3/2))/(24*h*(c*g^2 - b*g*h + a*h^2)^2*(g + h*x)^3) - ((b^2 - 4*a*c)*(16*c^2*d*g^2 + 16*a^2
*f*h^2 - 8*a*b*h*(2*f*g + e*h) + b^2*(5*f*g^2 + 3*e*g*h + 5*d*h^2) - 4*c*(2*b*g*(e*g + 2*d*h) + a*(f*g^2 - 5*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])]
)/(128*(c*g^2 - b*g*h + a*h^2)^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.85545, antiderivative size = 499, 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 = {1650, 806, 720, 724, 206} \[ \frac{\sqrt{a+b x+c x^2} (-2 a h+x (2 c g-b h)+b g) \left (16 a^2 f h^2-4 c \left (-a h (5 e g-d h)+a f g^2+2 b g (2 d h+e g)\right )-8 a b h (e h+2 f g)+b^2 \left (h (5 d h+3 e g)+5 f g^2\right )+16 c^2 d g^2\right )}{64 (g+h x)^2 \left (a h^2-b g h+c g^2\right )^3}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{-2 a h+x (2 c g-b h)+b g}{2 \sqrt{a+b x+c x^2} \sqrt{a h^2-b g h+c g^2}}\right ) \left (16 a^2 f h^2-4 c \left (-a h (5 e g-d h)+a f g^2+2 b g (2 d h+e g)\right )-8 a b h (e h+2 f g)+b^2 \left (h (5 d h+3 e g)+5 f g^2\right )+16 c^2 d g^2\right )}{128 \left (a h^2-b g h+c g^2\right )^{7/2}}+\frac{\left (a+b x+c x^2\right )^{3/2} \left (8 a h^2 (2 f g-e h)-b h \left (11 f g^2-h (5 d h+3 e g)\right )+2 c g h (e g-5 d h)+6 c f g^3\right )}{24 h (g+h x)^3 \left (a h^2-b g h+c g^2\right )^2}-\frac{\left (a+b x+c x^2\right )^{3/2} \left (f g^2-h (e g-d h)\right )}{4 h (g+h x)^4 \left (a h^2-b g h+c g^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2))/(g + h*x)^5,x]

[Out]

((16*c^2*d*g^2 + 16*a^2*f*h^2 - 8*a*b*h*(2*f*g + e*h) - 4*c*(a*f*g^2 - a*h*(5*e*g - d*h) + 2*b*g*(e*g + 2*d*h)
) + b^2*(5*f*g^2 + h*(3*e*g + 5*d*h)))*(b*g - 2*a*h + (2*c*g - b*h)*x)*Sqrt[a + b*x + c*x^2])/(64*(c*g^2 - b*g
*h + a*h^2)^3*(g + h*x)^2) - ((f*g^2 - h*(e*g - d*h))*(a + b*x + c*x^2)^(3/2))/(4*h*(c*g^2 - b*g*h + a*h^2)*(g
+ h*x)^4) + ((6*c*f*g^3 + 2*c*g*h*(e*g - 5*d*h) + 8*a*h^2*(2*f*g - e*h) - b*h*(11*f*g^2 - h*(3*e*g + 5*d*h)))
*(a + b*x + c*x^2)^(3/2))/(24*h*(c*g^2 - b*g*h + a*h^2)^2*(g + h*x)^3) - ((b^2 - 4*a*c)*(16*c^2*d*g^2 + 16*a^2
*f*h^2 - 8*a*b*h*(2*f*g + e*h) - 4*c*(a*f*g^2 - a*h*(5*e*g - d*h) + 2*b*g*(e*g + 2*d*h)) + b^2*(5*f*g^2 + h*(3
*e*g + 5*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])]
)/(128*(c*g^2 - b*g*h + a*h^2)^(7/2))

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 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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] - Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 720

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

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{\sqrt{a+b x+c x^2} \left (d+e x+f x^2\right )}{(g+h x)^5} \, dx &=-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{4 h \left (c g^2-b g h+a h^2\right ) (g+h x)^4}-\frac{\int \frac{\left (\frac{1}{2} \left (-8 c d g+3 b e g+8 a f g-\frac{3 b f g^2}{h}+5 b d h-8 a e h\right )-\left (c e g-4 b f g+\frac{3 c f g^2}{h}-c d h+4 a f h\right ) x\right ) \sqrt{a+b x+c x^2}}{(g+h x)^4} \, dx}{4 \left (c g^2-b g h+a h^2\right )}\\ &=-\frac{\left (f g^2-h (e g-d h)\right ) \left (a+b x+c x^2\right )^{3/2}}{4 h \left (c g^2-b g h+a h^2\right ) (g+h x)^4}+\frac{\left (6 c f g^3+2 c g h (e g-5 d h)+8 a h^2 (2 f g-e h)-b h \left (11 f g^2-h (3 e g+5 d h)\right )\right ) \left (a+b x+c x^2\right )^{3/2}}{24 h \left (c g^2-b g h+a h^2\right )^2 (g+h x)^3}+\frac{\left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\right )\right ) \int \frac{\sqrt{a+b x+c x^2}}{(g+h x)^3} \, dx}{16 \left (c g^2-b g h+a h^2\right )^2}\\ &=\frac{\left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\right )\right ) (b g-2 a h+(2 c g-b h) x) \sqrt{a+b x+c x^2}}{64 \left (c g^2-b g h+a h^2\right )^3 (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}}{4 h \left (c g^2-b g h+a h^2\right ) (g+h x)^4}+\frac{\left (6 c f g^3+2 c g h (e g-5 d h)+8 a h^2 (2 f g-e h)-b h \left (11 f g^2-h (3 e g+5 d h)\right )\right ) \left (a+b x+c x^2\right )^{3/2}}{24 h \left (c g^2-b g h+a h^2\right )^2 (g+h x)^3}-\frac{\left (\left (b^2-4 a c\right ) \left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\right )\right )\right ) \int \frac{1}{(g+h x) \sqrt{a+b x+c x^2}} \, dx}{128 \left (c g^2-b g h+a h^2\right )^3}\\ &=\frac{\left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\right )\right ) (b g-2 a h+(2 c g-b h) x) \sqrt{a+b x+c x^2}}{64 \left (c g^2-b g h+a h^2\right )^3 (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}}{4 h \left (c g^2-b g h+a h^2\right ) (g+h x)^4}+\frac{\left (6 c f g^3+2 c g h (e g-5 d h)+8 a h^2 (2 f g-e h)-b h \left (11 f g^2-h (3 e g+5 d h)\right )\right ) \left (a+b x+c x^2\right )^{3/2}}{24 h \left (c g^2-b g h+a h^2\right )^2 (g+h x)^3}+\frac{\left (\left (b^2-4 a c\right ) \left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\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 )}{64 \left (c g^2-b g h+a h^2\right )^3}\\ &=\frac{\left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\right )\right ) (b g-2 a h+(2 c g-b h) x) \sqrt{a+b x+c x^2}}{64 \left (c g^2-b g h+a h^2\right )^3 (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}}{4 h \left (c g^2-b g h+a h^2\right ) (g+h x)^4}+\frac{\left (6 c f g^3+2 c g h (e g-5 d h)+8 a h^2 (2 f g-e h)-b h \left (11 f g^2-h (3 e g+5 d h)\right )\right ) \left (a+b x+c x^2\right )^{3/2}}{24 h \left (c g^2-b g h+a h^2\right )^2 (g+h x)^3}-\frac{\left (b^2-4 a c\right ) \left (16 c^2 d g^2+16 a^2 f h^2-8 a b h (2 f g+e h)-4 c \left (a f g^2-a h (5 e g-d h)+2 b g (e g+2 d h)\right )+b^2 \left (5 f g^2+h (3 e g+5 d h)\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 )}{128 \left (c g^2-b g h+a h^2\right )^{7/2}}\\ \end{align*}

Mathematica [A]  time = 4.79482, size = 447, normalized size = 0.9 \[ \frac{\frac{\frac{3}{2} c h \left (\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a h-b g+b h x-2 c g x}{2 \sqrt{a+x (b+c x)} \sqrt{h (a h-b g)+c g^2}}\right )}{8 \left (h (a h-b g)+c g^2\right )^{3/2}}+\frac{\sqrt{a+x (b+c x)} (-2 a h+b (g-h x)+2 c g x)}{4 (g+h x)^2 \left (h (a h-b g)+c g^2\right )}\right ) \left (16 a^2 f h^2-4 c \left (a h (d h-5 e g)+a f g^2+2 b g (2 d h+e g)\right )-8 a b h (e h+2 f g)+b^2 \left (h (5 d h+3 e g)+5 f g^2\right )+16 c^2 d g^2\right )+\frac{c (a+x (b+c x))^{3/2} \left (-8 a h^2 (e h-2 f g)+b h \left (h (5 d h+3 e g)-11 f g^2\right )+2 c g h (e g-5 d h)+6 c f g^3\right )}{(g+h x)^3}}{24 \left (h (a h-b g)+c g^2\right )^2}+\frac{(a+x (b+c x))^{3/2} \left (4 f h (a h-b g)+c h (e g-d h)+3 c f g^2\right )}{4 (g+h x)^4 \left (h (a h-b g)+c g^2\right )}-\frac{f (a+x (b+c x))^{3/2}}{(g+h x)^4}}{c h} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^2))/(g + h*x)^5,x]

[Out]

(-((f*(a + x*(b + c*x))^(3/2))/(g + h*x)^4) + ((3*c*f*g^2 + 4*f*h*(-(b*g) + a*h) + c*h*(e*g - d*h))*(a + x*(b
+ c*x))^(3/2))/(4*(c*g^2 + h*(-(b*g) + a*h))*(g + h*x)^4) + ((c*(6*c*f*g^3 + 2*c*g*h*(e*g - 5*d*h) - 8*a*h^2*(
-2*f*g + e*h) + b*h*(-11*f*g^2 + h*(3*e*g + 5*d*h)))*(a + x*(b + c*x))^(3/2))/(g + h*x)^3 + (3*c*h*(16*c^2*d*g
^2 + 16*a^2*f*h^2 - 8*a*b*h*(2*f*g + e*h) - 4*c*(a*f*g^2 + a*h*(-5*e*g + d*h) + 2*b*g*(e*g + 2*d*h)) + b^2*(5*
f*g^2 + h*(3*e*g + 5*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)/(24*(c*g^2 + h*(-(b*g) + a*h))^2))/(c*h)

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Maple [B]  time = 0.284, size = 29161, normalized size = 58.7 \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)^5,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)^5,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)^5,x, algorithm="fricas")

[Out]

Timed out

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Sympy [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)**5,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError