3.234 \(\int \frac{(g+h x)^2 (d+e x+f x^2)}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=289 \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-12 c h (a f h+b e h+2 b f g)+15 b^2 f h^2+8 c^2 \left (h (d h+2 e g)+f g^2\right )\right )}{8 c^{7/2}}+\frac{2 (g+h x)^2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{h \sqrt{a+b x+c x^2} \left (2 c h x \left (-12 a c f+5 b^2 f-4 b c e+8 c^2 d\right )-8 c^2 (4 a e h+8 a f g+b d h+2 b e g)+4 b c (13 a f h+3 b e h+6 b f g)-15 b^3 f h+32 c^3 d g\right )}{4 c^3 \left (b^2-4 a c\right )} \]

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x)*(g + h*x)^2)/(c*(b^2 - 4*a*c)*Sqrt[a
+ b*x + c*x^2]) + (h*(32*c^3*d*g - 15*b^3*f*h - 8*c^2*(2*b*e*g + 8*a*f*g + b*d*h + 4*a*e*h) + 4*b*c*(6*b*f*g +
3*b*e*h + 13*a*f*h) + 2*c*(8*c^2*d - 4*b*c*e + 5*b^2*f - 12*a*c*f)*h*x)*Sqrt[a + b*x + c*x^2])/(4*c^3*(b^2 -
4*a*c)) + ((15*b^2*f*h^2 - 12*c*h*(2*b*f*g + b*e*h + a*f*h) + 8*c^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])])/(8*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.391658, antiderivative size = 288, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1644, 779, 621, 206} \[ \frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-12 c h (a f h+b e h+2 b f g)+15 b^2 f h^2+8 c^2 \left (h (d h+2 e g)+f g^2\right )\right )}{8 c^{7/2}}+\frac{2 (g+h x)^2 \left (c \left (2 a e-b \left (\frac{a f}{c}+d\right )\right )-x \left (-2 a c f+b^2 f-b c e+2 c^2 d\right )\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{h \sqrt{a+b x+c x^2} \left (2 h x \left (-12 a c f+5 b^2 f-4 b c e+8 c^2 d\right )-8 c (4 a e h+8 a f g+b d h+2 b e g)+4 b (13 a f h+3 b e h+6 b f g)-\frac{15 b^3 f h}{c}+32 c^2 d g\right )}{4 c^2 \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

Int[((g + h*x)^2*(d + e*x + f*x^2))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*(c*(2*a*e - b*(d + (a*f)/c)) - (2*c^2*d - b*c*e + b^2*f - 2*a*c*f)*x)*(g + h*x)^2)/(c*(b^2 - 4*a*c)*Sqrt[a
+ b*x + c*x^2]) + (h*(32*c^2*d*g - (15*b^3*f*h)/c - 8*c*(2*b*e*g + 8*a*f*g + b*d*h + 4*a*e*h) + 4*b*(6*b*f*g +
3*b*e*h + 13*a*f*h) + 2*(8*c^2*d - 4*b*c*e + 5*b^2*f - 12*a*c*f)*h*x)*Sqrt[a + b*x + c*x^2])/(4*c^2*(b^2 - 4*
a*c)) + ((15*b^2*f*h^2 - 12*c*h*(2*b*f*g + b*e*h + a*f*h) + 8*c^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])])/(8*c^(7/2))

Rule 1644

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[Po
lynomialRemainder[Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g +
(2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x
+ c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*(d + e*x)*Q + g*(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c
*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[p] ||  !IntegerQ[m
] ||  !RationalQ[a, b, c, d, e]) &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2,
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 )}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)^2}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{2 \int \frac{(g+h x) \left (-\frac{b^2 f g+4 b (c d+a f) h-4 a c (f g+2 e h)}{2 c}-\frac{1}{2} \left (8 c d-4 b e-12 a f+\frac{5 b^2 f}{c}\right ) h x\right )}{\sqrt{a+b x+c x^2}} \, dx}{b^2-4 a c}\\ &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)^2}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{h \left (32 c^2 d g-\frac{15 b^3 f h}{c}-8 c (2 b e g+8 a f g+b d h+4 a e h)+4 b (6 b f g+3 b e h+13 a f h)+2 \left (8 c^2 d-4 b c e+5 b^2 f-12 a c f\right ) h x\right ) \sqrt{a+b x+c x^2}}{4 c^2 \left (b^2-4 a c\right )}+\frac{\left (15 b^2 f h^2-12 c h (2 b f g+b e h+a f h)+8 c^2 \left (f g^2+h (2 e g+d h)\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 c^3}\\ &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)^2}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{h \left (32 c^2 d g-\frac{15 b^3 f h}{c}-8 c (2 b e g+8 a f g+b d h+4 a e h)+4 b (6 b f g+3 b e h+13 a f h)+2 \left (8 c^2 d-4 b c e+5 b^2 f-12 a c f\right ) h x\right ) \sqrt{a+b x+c x^2}}{4 c^2 \left (b^2-4 a c\right )}+\frac{\left (15 b^2 f h^2-12 c h (2 b f g+b e h+a f h)+8 c^2 \left (f g^2+h (2 e g+d 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 )}{4 c^3}\\ &=\frac{2 \left (c \left (2 a e-b \left (d+\frac{a f}{c}\right )\right )-\left (2 c^2 d-b c e+b^2 f-2 a c f\right ) x\right ) (g+h x)^2}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{h \left (32 c^2 d g-\frac{15 b^3 f h}{c}-8 c (2 b e g+8 a f g+b d h+4 a e h)+4 b (6 b f g+3 b e h+13 a f h)+2 \left (8 c^2 d-4 b c e+5 b^2 f-12 a c f\right ) h x\right ) \sqrt{a+b x+c x^2}}{4 c^2 \left (b^2-4 a c\right )}+\frac{\left (15 b^2 f h^2-12 c h (2 b f g+b e h+a f h)+8 c^2 \left (f g^2+h (2 e g+d h)\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.832222, size = 412, normalized size = 1.43 \[ \frac{2 \sqrt{c} \left (4 b c \left (-13 a^2 f h^2+a c \left (2 h (d h+2 e g+5 e h x)+f \left (2 g^2+20 g h x-5 h^2 x^2\right )\right )+2 c^2 g (d (g-2 h x)-e g x)\right )+8 c^2 \left (a^2 h (4 e h+8 f g+3 f h x)+a c \left (-2 d h (2 g+h x)-2 e \left (g^2+2 g h x-h^2 x^2\right )+f x \left (-2 g^2+4 g h x+h^2 x^2\right )\right )+2 c^2 d g^2 x\right )-2 b^2 c \left (a h (6 e h+12 f g+31 f h x)+c x \left (2 h (-2 d h-4 e g+e h x)+f \left (-4 g^2+4 g h x+h^2 x^2\right )\right )\right )+b^3 h (15 a f h+c x (-12 e h-24 f g+5 f h x))+15 b^4 f h^2 x\right )-\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-12 c h (a f h+b e h+2 b f g)+15 b^2 f h^2+8 c^2 \left (h (d h+2 e g)+f g^2\right )\right )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((g + h*x)^2*(d + e*x + f*x^2))/(a + b*x + c*x^2)^(3/2),x]

[Out]

(2*Sqrt[c]*(15*b^4*f*h^2*x + b^3*h*(15*a*f*h + c*x*(-24*f*g - 12*e*h + 5*f*h*x)) + 4*b*c*(-13*a^2*f*h^2 + 2*c^
2*g*(-(e*g*x) + d*(g - 2*h*x)) + a*c*(2*h*(2*e*g + d*h + 5*e*h*x) + f*(2*g^2 + 20*g*h*x - 5*h^2*x^2))) - 2*b^2
*c*(a*h*(12*f*g + 6*e*h + 31*f*h*x) + c*x*(2*h*(-4*e*g - 2*d*h + e*h*x) + f*(-4*g^2 + 4*g*h*x + h^2*x^2))) + 8
*c^2*(2*c^2*d*g^2*x + a^2*h*(8*f*g + 4*e*h + 3*f*h*x) + a*c*(-2*d*h*(2*g + h*x) - 2*e*(g^2 + 2*g*h*x - h^2*x^2
) + f*x*(-2*g^2 + 4*g*h*x + h^2*x^2)))) - (b^2 - 4*a*c)*(15*b^2*f*h^2 - 12*c*h*(2*b*f*g + b*e*h + a*f*h) + 8*c
^2*(f*g^2 + h*(2*e*g + d*h)))*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8
*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.063, size = 1557, normalized size = 5.4 \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)^(3/2),x)

[Out]

-3/2*h^2*f*a/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+15/8*h^2*f*b^2/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)
+(c*x^2+b*x+a)^(1/2))-x/c/(c*x^2+b*x+a)^(1/2)*d*h^2-x/c/(c*x^2+b*x+a)^(1/2)*f*g^2+1/2*b/c^2/(c*x^2+b*x+a)^(1/2
)*d*h^2+1/2*b/c^2/(c*x^2+b*x+a)^(1/2)*f*g^2-4*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*g*h*d-2*b^2/c/(4*a*c-b^2)/(c
*x^2+b*x+a)^(1/2)*g*h*d+8*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*g*h*f+2*g^2*d*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b
*x+a)^(1/2)+x^2/c/(c*x^2+b*x+a)^(1/2)*h^2*e-3/4*b^2/c^3/(c*x^2+b*x+a)^(1/2)*h^2*e-3/2*b/c^(5/2)*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))*h^2*e+2*a/c^2/(c*x^2+b*x+a)^(1/2)*h^2*e+1/2*h^2*f*x^3/c/(c*x^2+b*x+a)^(1/2)+15/1
6*h^2*f*b^3/c^4/(c*x^2+b*x+a)^(1/2)+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*h^2+1/c^(3/2)*ln((
1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f*g^2-1/c/(c*x^2+b*x+a)^(1/2)*e*g^2+2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))*e*g*h+b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*d*h^2+b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*
x*f*g^2+b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*e*g*h+15/8*h^2*f*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-3/2
*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*h^2*e+4*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*g*h*f+2*b^2/c/(4*
a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*e*g*h-3*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*g*h*f+4*a/c*b/(4*a*c-b^2)/(c*
x^2+b*x+a)^(1/2)*x*h^2*e-13/2*h^2*f*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-13/4*h^2*f*b^3/c^3*a/(4*a*c-b^
2)/(c*x^2+b*x+a)^(1/2)-3/2*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*g*h*f+2*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)
^(1/2)*h^2*e+3*b/c^2*x/(c*x^2+b*x+a)^(1/2)*g*h*f-2/c/(c*x^2+b*x+a)^(1/2)*g*h*d-2*x/c/(c*x^2+b*x+a)^(1/2)*e*g*h
+b/c^2/(c*x^2+b*x+a)^(1/2)*e*g*h-3*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*g*h*f+4*a/c^2/(c*x^2+
b*x+a)^(1/2)*g*h*f-b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*e*g^2+15/16*h^2*f*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(
1/2)-3/4*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*h^2*e+3/2*b/c^2*x/(c*x^2+b*x+a)^(1/2)*h^2*e-3/2*b^2/c^3/(c*x^
2+b*x+a)^(1/2)*g*h*f-13/4*h^2*f*b/c^3*a/(c*x^2+b*x+a)^(1/2)+3/2*h^2*f*a/c^2*x/(c*x^2+b*x+a)^(1/2)-2*b/(4*a*c-b
^2)/(c*x^2+b*x+a)^(1/2)*x*e*g^2+2*x^2/c/(c*x^2+b*x+a)^(1/2)*g*h*f-15/8*h^2*f*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)+1/2
*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*d*h^2+1/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*f*g^2-5/4*h^2*f*b/c
^2*x^2/(c*x^2+b*x+a)^(1/2)

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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((h*x+g)^2*(f*x^2+e*x+d)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 90.9787, size = 3730, normalized size = 12.91 \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*((8*(a*b^2*c^2 - 4*a^2*c^3)*f*g^2 + 8*(2*(a*b^2*c^2 - 4*a^2*c^3)*e - 3*(a*b^3*c - 4*a^2*b*c^2)*f)*g*h +
(8*(a*b^2*c^2 - 4*a^2*c^3)*d - 12*(a*b^3*c - 4*a^2*b*c^2)*e + 3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*f)*h^2
+ (8*(b^2*c^3 - 4*a*c^4)*f*g^2 + 8*(2*(b^2*c^3 - 4*a*c^4)*e - 3*(b^3*c^2 - 4*a*b*c^3)*f)*g*h + (8*(b^2*c^3 - 4
*a*c^4)*d - 12*(b^3*c^2 - 4*a*b*c^3)*e + 3*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*f)*h^2)*x^2 + (8*(b^3*c^2 - 4
*a*b*c^3)*f*g^2 + 8*(2*(b^3*c^2 - 4*a*b*c^3)*e - 3*(b^4*c - 4*a*b^2*c^2)*f)*g*h + (8*(b^3*c^2 - 4*a*b*c^3)*d -
12*(b^4*c - 4*a*b^2*c^2)*e + 3*(5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*f)*h^2)*x)*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*(2*(b^2*c^3 - 4*a*c^4)*f*h^2*x^3 - 8*(b*c^4
*d - 2*a*c^4*e + a*b*c^3*f)*g^2 + 8*(4*a*c^4*d - 2*a*b*c^3*e + (3*a*b^2*c^2 - 8*a^2*c^3)*f)*g*h - (8*a*b*c^3*d
- 4*(3*a*b^2*c^2 - 8*a^2*c^3)*e + (15*a*b^3*c - 52*a^2*b*c^2)*f)*h^2 + (8*(b^2*c^3 - 4*a*c^4)*f*g*h + (4*(b^2
*c^3 - 4*a*c^4)*e - 5*(b^3*c^2 - 4*a*b*c^3)*f)*h^2)*x^2 - (8*(2*c^5*d - b*c^4*e + (b^2*c^3 - 2*a*c^4)*f)*g^2 -
8*(2*b*c^4*d - 2*(b^2*c^3 - 2*a*c^4)*e + (3*b^3*c^2 - 10*a*b*c^3)*f)*g*h + (8*(b^2*c^3 - 2*a*c^4)*d - 4*(3*b^
3*c^2 - 10*a*b*c^3)*e + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*f)*h^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 -
4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*((8*(a*b^2*c^2 - 4*a^2*c^3)*f*g^2 + 8*(2*
(a*b^2*c^2 - 4*a^2*c^3)*e - 3*(a*b^3*c - 4*a^2*b*c^2)*f)*g*h + (8*(a*b^2*c^2 - 4*a^2*c^3)*d - 12*(a*b^3*c - 4*
a^2*b*c^2)*e + 3*(5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*f)*h^2 + (8*(b^2*c^3 - 4*a*c^4)*f*g^2 + 8*(2*(b^2*c^3 -
4*a*c^4)*e - 3*(b^3*c^2 - 4*a*b*c^3)*f)*g*h + (8*(b^2*c^3 - 4*a*c^4)*d - 12*(b^3*c^2 - 4*a*b*c^3)*e + 3*(5*b^
4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*f)*h^2)*x^2 + (8*(b^3*c^2 - 4*a*b*c^3)*f*g^2 + 8*(2*(b^3*c^2 - 4*a*b*c^3)*e -
3*(b^4*c - 4*a*b^2*c^2)*f)*g*h + (8*(b^3*c^2 - 4*a*b*c^3)*d - 12*(b^4*c - 4*a*b^2*c^2)*e + 3*(5*b^5 - 24*a*b^
3*c + 16*a^2*b*c^2)*f)*h^2)*x)*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*(2*(b^2*c^3 - 4*a*c^4)*f*h^2*x^3 - 8*(b*c^4*d - 2*a*c^4*e + a*b*c^3*f)*g^2 + 8*(4*a*c^4*d - 2*a*b
*c^3*e + (3*a*b^2*c^2 - 8*a^2*c^3)*f)*g*h - (8*a*b*c^3*d - 4*(3*a*b^2*c^2 - 8*a^2*c^3)*e + (15*a*b^3*c - 52*a^
2*b*c^2)*f)*h^2 + (8*(b^2*c^3 - 4*a*c^4)*f*g*h + (4*(b^2*c^3 - 4*a*c^4)*e - 5*(b^3*c^2 - 4*a*b*c^3)*f)*h^2)*x^
2 - (8*(2*c^5*d - b*c^4*e + (b^2*c^3 - 2*a*c^4)*f)*g^2 - 8*(2*b*c^4*d - 2*(b^2*c^3 - 2*a*c^4)*e + (3*b^3*c^2 -
10*a*b*c^3)*f)*g*h + (8*(b^2*c^3 - 2*a*c^4)*d - 4*(3*b^3*c^2 - 10*a*b*c^3)*e + (15*b^4*c - 62*a*b^2*c^2 + 24*
a^2*c^3)*f)*h^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b
*c^5)*x)]

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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 )}{\left (a + b x + c x^{2}\right )^{\frac{3}{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)**(3/2),x)

[Out]

Integral((g + h*x)**2*(d + e*x + f*x**2)/(a + b*x + c*x**2)**(3/2), x)

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Giac [B]  time = 1.33493, size = 783, normalized size = 2.71 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (b^{2} c^{2} f h^{2} - 4 \, a c^{3} f h^{2}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac{8 \, b^{2} c^{2} f g h - 32 \, a c^{3} f g h - 5 \, b^{3} c f h^{2} + 20 \, a b c^{2} f h^{2} + 4 \, b^{2} c^{2} h^{2} e - 16 \, a c^{3} h^{2} e}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{16 \, c^{4} d g^{2} + 8 \, b^{2} c^{2} f g^{2} - 16 \, a c^{3} f g^{2} - 16 \, b c^{3} d g h - 24 \, b^{3} c f g h + 80 \, a b c^{2} f g h + 8 \, b^{2} c^{2} d h^{2} - 16 \, a c^{3} d h^{2} + 15 \, b^{4} f h^{2} - 62 \, a b^{2} c f h^{2} + 24 \, a^{2} c^{2} f h^{2} - 8 \, b c^{3} g^{2} e + 16 \, b^{2} c^{2} g h e - 32 \, a c^{3} g h e - 12 \, b^{3} c h^{2} e + 40 \, a b c^{2} h^{2} e}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac{8 \, b c^{3} d g^{2} + 8 \, a b c^{2} f g^{2} - 32 \, a c^{3} d g h - 24 \, a b^{2} c f g h + 64 \, a^{2} c^{2} f g h + 8 \, a b c^{2} d h^{2} + 15 \, a b^{3} f h^{2} - 52 \, a^{2} b c f h^{2} - 16 \, a c^{3} g^{2} e + 16 \, a b c^{2} g h e - 12 \, a b^{2} c h^{2} e + 32 \, a^{2} c^{2} h^{2} e}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt{c x^{2} + b x + a}} - \frac{{\left (8 \, c^{2} f g^{2} - 24 \, b c f g h + 8 \, c^{2} d h^{2} + 15 \, b^{2} f h^{2} - 12 \, a c f h^{2} + 16 \, c^{2} g h e - 12 \, b c 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 )}{8 \, c^{\frac{7}{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)^(3/2),x, algorithm="giac")

[Out]

1/4*(((2*(b^2*c^2*f*h^2 - 4*a*c^3*f*h^2)*x/(b^2*c^3 - 4*a*c^4) + (8*b^2*c^2*f*g*h - 32*a*c^3*f*g*h - 5*b^3*c*f
*h^2 + 20*a*b*c^2*f*h^2 + 4*b^2*c^2*h^2*e - 16*a*c^3*h^2*e)/(b^2*c^3 - 4*a*c^4))*x - (16*c^4*d*g^2 + 8*b^2*c^2
*f*g^2 - 16*a*c^3*f*g^2 - 16*b*c^3*d*g*h - 24*b^3*c*f*g*h + 80*a*b*c^2*f*g*h + 8*b^2*c^2*d*h^2 - 16*a*c^3*d*h^
2 + 15*b^4*f*h^2 - 62*a*b^2*c*f*h^2 + 24*a^2*c^2*f*h^2 - 8*b*c^3*g^2*e + 16*b^2*c^2*g*h*e - 32*a*c^3*g*h*e - 1
2*b^3*c*h^2*e + 40*a*b*c^2*h^2*e)/(b^2*c^3 - 4*a*c^4))*x - (8*b*c^3*d*g^2 + 8*a*b*c^2*f*g^2 - 32*a*c^3*d*g*h -
24*a*b^2*c*f*g*h + 64*a^2*c^2*f*g*h + 8*a*b*c^2*d*h^2 + 15*a*b^3*f*h^2 - 52*a^2*b*c*f*h^2 - 16*a*c^3*g^2*e +
16*a*b*c^2*g*h*e - 12*a*b^2*c*h^2*e + 32*a^2*c^2*h^2*e)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 1/8*(8*c^
2*f*g^2 - 24*b*c*f*g*h + 8*c^2*d*h^2 + 15*b^2*f*h^2 - 12*a*c*f*h^2 + 16*c^2*g*h*e - 12*b*c*h^2*e)*log(abs(-2*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)