### 3.188 $$\int (g+h x) \sqrt{a+b x+c x^2} (d+e x+f x^2) \, dx$$

Optimal. Leaf size=322 $\frac{\left (a+b x+c x^2\right )^{3/2} \left (-2 c h (16 a f h+25 b (e h+f g))+35 b^2 f h^2-6 c h x (7 b f h-10 c e h+6 c f g)-16 c^2 \left (3 f g^2-5 h (d h+e g)\right )\right )}{240 c^3 h}+\frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (-8 c^2 (a e h+a f g+2 b d h+2 b e g)+2 b c (6 a f h+5 b (e h+f g))-7 b^3 f h+32 c^3 d g\right )}{128 c^4}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-8 c^2 (a e h+a f g+2 b d h+2 b e g)+2 b c (6 a f h+5 b (e h+f g))-7 b^3 f h+32 c^3 d g\right )}{256 c^{9/2}}+\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}$

[Out]

((32*c^3*d*g - 7*b^3*f*h - 8*c^2*(2*b*e*g + a*f*g + 2*b*d*h + a*e*h) + 2*b*c*(6*a*f*h + 5*b*(f*g + e*h)))*(b +
2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) + (f*(g + h*x)^2*(a + b*x + c*x^2)^(3/2))/(5*c*h) + ((35*b^2*f*h^2 -
16*c^2*(3*f*g^2 - 5*h*(e*g + d*h)) - 2*c*h*(16*a*f*h + 25*b*(f*g + e*h)) - 6*c*h*(6*c*f*g - 10*c*e*h + 7*b*f*h
)*x)*(a + b*x + c*x^2)^(3/2))/(240*c^3*h) - ((b^2 - 4*a*c)*(32*c^3*d*g - 7*b^3*f*h - 8*c^2*(2*b*e*g + a*f*g +
2*b*d*h + a*e*h) + 2*b*c*(6*a*f*h + 5*b*(f*g + e*h)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/
(256*c^(9/2))

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Rubi [A]  time = 0.503646, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {1653, 779, 612, 621, 206} $\frac{\left (a+b x+c x^2\right )^{3/2} \left (-2 c h (16 a f h+25 b (e h+f g))+35 b^2 f h^2-6 c h x (7 b f h-10 c e h+6 c f g)+c^2 \left (-\left (48 f g^2-80 h (d h+e g)\right )\right )\right )}{240 c^3 h}+\frac{(b+2 c x) \sqrt{a+b x+c x^2} \left (-8 c^2 (a e h+a f g+2 b d h+2 b e g)+2 b c (6 a f h+5 b (e h+f g))-7 b^3 f h+32 c^3 d g\right )}{128 c^4}-\frac{\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (-8 c^2 (a e h+a f g+2 b d h+2 b e g)+2 b c (6 a f h+5 b (e h+f g))-7 b^3 f h+32 c^3 d g\right )}{256 c^{9/2}}+\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((32*c^3*d*g - 7*b^3*f*h - 8*c^2*(2*b*e*g + a*f*g + 2*b*d*h + a*e*h) + 2*b*c*(6*a*f*h + 5*b*(f*g + e*h)))*(b +
2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) + (f*(g + h*x)^2*(a + b*x + c*x^2)^(3/2))/(5*c*h) + ((35*b^2*f*h^2 -
c^2*(48*f*g^2 - 80*h*(e*g + d*h)) - 2*c*h*(16*a*f*h + 25*b*(f*g + e*h)) - 6*c*h*(6*c*f*g - 10*c*e*h + 7*b*f*h)
*x)*(a + b*x + c*x^2)^(3/2))/(240*c^3*h) - ((b^2 - 4*a*c)*(32*c^3*d*g - 7*b^3*f*h - 8*c^2*(2*b*e*g + a*f*g + 2
*b*d*h + a*e*h) + 2*b*c*(6*a*f*h + 5*b*(f*g + e*h)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(
256*c^(9/2))

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 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 612

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

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 (g+h x) \sqrt{a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx &=\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}+\frac{\int (g+h x) \left (-\frac{1}{2} h (3 b f g-10 c d h+4 a f h)-\frac{1}{2} h (6 c f g-10 c e h+7 b f h) x\right ) \sqrt{a+b x+c x^2} \, dx}{5 c h^2}\\ &=\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}+\frac{\left (35 b^2 f h^2-c^2 \left (48 f g^2-80 h (e g+d h)\right )-2 c h (16 a f h+25 b (f g+e h))-6 c h (6 c f g-10 c e h+7 b f h) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3 h}+\frac{\left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right ) \int \sqrt{a+b x+c x^2} \, dx}{32 c^3}\\ &=\frac{\left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}+\frac{\left (35 b^2 f h^2-c^2 \left (48 f g^2-80 h (e g+d h)\right )-2 c h (16 a f h+25 b (f g+e h))-6 c h (6 c f g-10 c e h+7 b f h) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3 h}-\frac{\left (\left (b^2-4 a c\right ) \left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{256 c^4}\\ &=\frac{\left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}+\frac{\left (35 b^2 f h^2-c^2 \left (48 f g^2-80 h (e g+d h)\right )-2 c h (16 a f h+25 b (f g+e h))-6 c h (6 c f g-10 c e h+7 b f h) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3 h}-\frac{\left (\left (b^2-4 a c\right ) \left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e 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 )}{128 c^4}\\ &=\frac{\left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{128 c^4}+\frac{f (g+h x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c h}+\frac{\left (35 b^2 f h^2-c^2 \left (48 f g^2-80 h (e g+d h)\right )-2 c h (16 a f h+25 b (f g+e h))-6 c h (6 c f g-10 c e h+7 b f h) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3 h}-\frac{\left (b^2-4 a c\right ) \left (32 c^3 d g-7 b^3 f h-8 c^2 (2 b e g+a f g+2 b d h+a e h)+2 b c (6 a f h+5 b (f g+e h))\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.493249, size = 258, normalized size = 0.8 $\frac{\frac{(a+x (b+c x))^{3/2} \left (-2 c h (16 a f h+b (25 e h+25 f g+21 f h x))+35 b^2 f h^2+c^2 (20 h (4 d h+4 e g+3 e h x)-12 f g (4 g+3 h x))\right )}{48 c^2}-\frac{5 h \left (2 \sqrt{c} (b+2 c x) \sqrt{a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right ) \left (8 c^2 (a e h+a f g+2 b d h+2 b e g)-2 b c (6 a f h+5 b (e h+f g))+7 b^3 f h-32 c^3 d g\right )}{256 c^{7/2}}+f (g+h x)^2 (a+x (b+c x))^{3/2}}{5 c h}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(f*(g + h*x)^2*(a + x*(b + c*x))^(3/2) + ((a + x*(b + c*x))^(3/2)*(35*b^2*f*h^2 + c^2*(-12*f*g*(4*g + 3*h*x) +
20*h*(4*e*g + 4*d*h + 3*e*h*x)) - 2*c*h*(16*a*f*h + b*(25*f*g + 25*e*h + 21*f*h*x))))/(48*c^2) - (5*h*(-32*c^
3*d*g + 7*b^3*f*h + 8*c^2*(2*b*e*g + a*f*g + 2*b*d*h + a*e*h) - 2*b*c*(6*a*f*h + 5*b*(f*g + e*h)))*(2*Sqrt[c]*
(b + 2*c*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(25
6*c^(7/2)))/(5*c*h)

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Maple [B]  time = 0.053, size = 1117, normalized size = 3.5 \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)*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x)

[Out]

-1/4*b/c*x*(c*x^2+b*x+a)^(1/2)*e*g+3/16*h*f*b/c^2*a*x*(c*x^2+b*x+a)^(1/2)+1/3*(c*x^2+b*x+a)^(3/2)/c*d*h+1/3*(c
*x^2+b*x+a)^(3/2)/c*e*g+1/2*d*g*x*(c*x^2+b*x+a)^(1/2)-1/8*b^2/c^2*(c*x^2+b*x+a)^(1/2)*d*h-1/8*b^2/c^2*(c*x^2+b
*x+a)^(1/2)*e*g+1/4*x*(c*x^2+b*x+a)^(3/2)/c*f*g-5/24*b/c^2*(c*x^2+b*x+a)^(3/2)*e*h-5/24*b/c^2*(c*x^2+b*x+a)^(3
/2)*f*g+1/4*d*g/c*(c*x^2+b*x+a)^(1/2)*b+1/2*d*g/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/8*d*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*x*(c*x^2+b*x+a)^(3/2)/c*e*h-5/128*b^4/c^(7/2)*ln((
1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e*h+1/16*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*h-1
/4*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*d*h-1/4*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))*a*e*g-1/16*a/c^2*(c*x^2+b*x+a)^(1/2)*b*f*g-1/8*a^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/
2))*f*g-5/128*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*f*g-1/8*a^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/
2)+(c*x^2+b*x+a)^(1/2))*e*h+1/16*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*e*g+1/5*h*f*x^2*(c*x^
2+b*x+a)^(3/2)/c+7/48*h*f*b^2/c^3*(c*x^2+b*x+a)^(3/2)-7/128*h*f*b^4/c^4*(c*x^2+b*x+a)^(1/2)+7/256*h*f*b^5/c^(9
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/15*h*f*a/c^2*(c*x^2+b*x+a)^(3/2)+5/64*b^3/c^3*(c*x^2+b*x+a)^
(1/2)*e*h+5/64*b^3/c^3*(c*x^2+b*x+a)^(1/2)*f*g+3/16*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*
f*g-1/8*a/c*x*(c*x^2+b*x+a)^(1/2)*e*h-1/8*a/c*x*(c*x^2+b*x+a)^(1/2)*f*g-1/16*a/c^2*(c*x^2+b*x+a)^(1/2)*b*e*h-1
/4*b/c*x*(c*x^2+b*x+a)^(1/2)*d*h-7/40*h*f*b/c^2*x*(c*x^2+b*x+a)^(3/2)-7/64*h*f*b^3/c^3*x*(c*x^2+b*x+a)^(1/2)-5
/32*h*f*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+3/32*h*f*b^2/c^3*a*(c*x^2+b*x+a)^(1/2)+3/16*
h*f*b/c^(5/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+5/32*b^2/c^2*x*(c*x^2+b*x+a)^(1/2)*f*g+3/16*b^2/
c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a*e*h+5/32*b^2/c^2*x*(c*x^2+b*x+a)^(1/2)*e*h

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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)*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.69966, size = 2275, normalized size = 7.07 \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)*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/7680*(15*(2*(16*(b^2*c^3 - 4*a*c^4)*d - 8*(b^3*c^2 - 4*a*b*c^3)*e + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*
f)*g - (16*(b^3*c^2 - 4*a*b*c^3)*d - 2*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*e + (7*b^5 - 40*a*b^3*c + 48*a^2*
b*c^2)*f)*h)*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
*(384*c^5*f*h*x^4 + 48*(10*c^5*f*g + (10*c^5*e + b*c^4*f)*h)*x^3 + 8*(10*(8*c^5*e + b*c^4*f)*g + (80*c^5*d + 1
0*b*c^4*e - (7*b^2*c^3 - 16*a*c^4)*f)*h)*x^2 + 10*(48*b*c^4*d - 8*(3*b^2*c^3 - 8*a*c^4)*e + (15*b^3*c^2 - 52*a
*b*c^3)*f)*g - (80*(3*b^2*c^3 - 8*a*c^4)*d - 10*(15*b^3*c^2 - 52*a*b*c^3)*e + (105*b^4*c - 460*a*b^2*c^2 + 256
*a^2*c^3)*f)*h + 2*(10*(48*c^5*d + 8*b*c^4*e - (5*b^2*c^3 - 12*a*c^4)*f)*g + (80*b*c^4*d - 10*(5*b^2*c^3 - 12*
a*c^4)*e + (35*b^3*c^2 - 116*a*b*c^3)*f)*h)*x)*sqrt(c*x^2 + b*x + a))/c^5, 1/3840*(15*(2*(16*(b^2*c^3 - 4*a*c^
4)*d - 8*(b^3*c^2 - 4*a*b*c^3)*e + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*f)*g - (16*(b^3*c^2 - 4*a*b*c^3)*d -
2*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*e + (7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*f)*h)*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*(384*c^5*f*h*x^4 + 48*(10*c^5*f*g + (10*c^
5*e + b*c^4*f)*h)*x^3 + 8*(10*(8*c^5*e + b*c^4*f)*g + (80*c^5*d + 10*b*c^4*e - (7*b^2*c^3 - 16*a*c^4)*f)*h)*x^
2 + 10*(48*b*c^4*d - 8*(3*b^2*c^3 - 8*a*c^4)*e + (15*b^3*c^2 - 52*a*b*c^3)*f)*g - (80*(3*b^2*c^3 - 8*a*c^4)*d
- 10*(15*b^3*c^2 - 52*a*b*c^3)*e + (105*b^4*c - 460*a*b^2*c^2 + 256*a^2*c^3)*f)*h + 2*(10*(48*c^5*d + 8*b*c^4*
e - (5*b^2*c^3 - 12*a*c^4)*f)*g + (80*b*c^4*d - 10*(5*b^2*c^3 - 12*a*c^4)*e + (35*b^3*c^2 - 116*a*b*c^3)*f)*h)
*x)*sqrt(c*x^2 + b*x + a))/c^5]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.34907, size = 668, normalized size = 2.07 \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, f h x + \frac{10 \, c^{4} f g + b c^{3} f h + 10 \, c^{4} h e}{c^{4}}\right )} x + \frac{10 \, b c^{3} f g + 80 \, c^{4} d h - 7 \, b^{2} c^{2} f h + 16 \, a c^{3} f h + 80 \, c^{4} g e + 10 \, b c^{3} h e}{c^{4}}\right )} x + \frac{480 \, c^{4} d g - 50 \, b^{2} c^{2} f g + 120 \, a c^{3} f g + 80 \, b c^{3} d h + 35 \, b^{3} c f h - 116 \, a b c^{2} f h + 80 \, b c^{3} g e - 50 \, b^{2} c^{2} h e + 120 \, a c^{3} h e}{c^{4}}\right )} x + \frac{480 \, b c^{3} d g + 150 \, b^{3} c f g - 520 \, a b c^{2} f g - 240 \, b^{2} c^{2} d h + 640 \, a c^{3} d h - 105 \, b^{4} f h + 460 \, a b^{2} c f h - 256 \, a^{2} c^{2} f h - 240 \, b^{2} c^{2} g e + 640 \, a c^{3} g e + 150 \, b^{3} c h e - 520 \, a b c^{2} h e}{c^{4}}\right )} + \frac{{\left (32 \, b^{2} c^{3} d g - 128 \, a c^{4} d g + 10 \, b^{4} c f g - 48 \, a b^{2} c^{2} f g + 32 \, a^{2} c^{3} f g - 16 \, b^{3} c^{2} d h + 64 \, a b c^{3} d h - 7 \, b^{5} f h + 40 \, a b^{3} c f h - 48 \, a^{2} b c^{2} f h - 16 \, b^{3} c^{2} g e + 64 \, a b c^{3} g e + 10 \, b^{4} c h e - 48 \, a b^{2} c^{2} h e + 32 \, a^{2} c^{3} h e\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{9}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(8*f*h*x + (10*c^4*f*g + b*c^3*f*h + 10*c^4*h*e)/c^4)*x + (10*b*c^3*f*g
+ 80*c^4*d*h - 7*b^2*c^2*f*h + 16*a*c^3*f*h + 80*c^4*g*e + 10*b*c^3*h*e)/c^4)*x + (480*c^4*d*g - 50*b^2*c^2*f*
g + 120*a*c^3*f*g + 80*b*c^3*d*h + 35*b^3*c*f*h - 116*a*b*c^2*f*h + 80*b*c^3*g*e - 50*b^2*c^2*h*e + 120*a*c^3*
h*e)/c^4)*x + (480*b*c^3*d*g + 150*b^3*c*f*g - 520*a*b*c^2*f*g - 240*b^2*c^2*d*h + 640*a*c^3*d*h - 105*b^4*f*h
+ 460*a*b^2*c*f*h - 256*a^2*c^2*f*h - 240*b^2*c^2*g*e + 640*a*c^3*g*e + 150*b^3*c*h*e - 520*a*b*c^2*h*e)/c^4)
+ 1/256*(32*b^2*c^3*d*g - 128*a*c^4*d*g + 10*b^4*c*f*g - 48*a*b^2*c^2*f*g + 32*a^2*c^3*f*g - 16*b^3*c^2*d*h +
64*a*b*c^3*d*h - 7*b^5*f*h + 40*a*b^3*c*f*h - 48*a^2*b*c^2*f*h - 16*b^3*c^2*g*e + 64*a*b*c^3*g*e + 10*b^4*c*h
*e - 48*a*b^2*c^2*h*e + 32*a^2*c^3*h*e)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)