3.2465 \(\int \frac{(A+B x) (d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=407 \[ \frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (B \left (-4 c e (9 a e+16 b d)+35 b^2 e^2+24 c^2 d^2\right )+40 A c e (2 c d-b e)\right )+8 A c e \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+64 c^2 d^2\right )+B \left (-8 c^2 d e (48 a e+47 b d)+20 b c e^2 (11 a e+18 b d)-105 b^3 e^3+96 c^3 d^3\right )\right )}{192 c^4}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (24 b^2 c e \left (-5 a B e^2+6 A c d e+6 B c d^2\right )-32 b c^2 \left (-3 a A e^3-9 a B d e^2+6 A c d^2 e+2 B c d^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )-3 a B e \left (4 c d^2-a e^2\right )\right )-40 b^3 c e^2 (A e+3 B d)+35 b^4 B e^3\right )}{128 c^{9/2}}+\frac{(d+e x)^2 \sqrt{a+b x+c x^2} (8 A c e-7 b B e+6 B c d)}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c} \]

[Out]

((6*B*c*d - 7*b*B*e + 8*A*c*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(24*c^2) + (B*(d + e*x)^3*Sqrt[a + b*x + c*x
^2])/(4*c) + ((8*A*c*e*(64*c^2*d^2 + 15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e)) + B*(96*c^3*d^3 - 105*b^3*e^3 + 20*b
*c*e^2*(18*b*d + 11*a*e) - 8*c^2*d*e*(47*b*d + 48*a*e)) + 2*c*e*(40*A*c*e*(2*c*d - b*e) + B*(24*c^2*d^2 + 35*b
^2*e^2 - 4*c*e*(16*b*d + 9*a*e)))*x)*Sqrt[a + b*x + c*x^2])/(192*c^4) + ((35*b^4*B*e^3 - 40*b^3*c*e^2*(3*B*d +
A*e) + 24*b^2*c*e*(6*B*c*d^2 + 6*A*c*d*e - 5*a*B*e^2) - 32*b*c^2*(2*B*c*d^3 + 6*A*c*d^2*e - 9*a*B*d*e^2 - 3*a
*A*e^3) + 16*c^2*(4*A*c*d*(2*c*d^2 - 3*a*e^2) - 3*a*B*e*(4*c*d^2 - a*e^2)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(128*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.683948, antiderivative size = 407, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {832, 779, 621, 206} \[ \frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (B \left (-4 c e (9 a e+16 b d)+35 b^2 e^2+24 c^2 d^2\right )+40 A c e (2 c d-b e)\right )+8 A c e \left (-2 c e (8 a e+27 b d)+15 b^2 e^2+64 c^2 d^2\right )+B \left (-8 c^2 d e (48 a e+47 b d)+20 b c e^2 (11 a e+18 b d)-105 b^3 e^3+96 c^3 d^3\right )\right )}{192 c^4}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (24 b^2 c e \left (-5 a B e^2+6 A c d e+6 B c d^2\right )-32 b c^2 \left (-3 a A e^3-9 a B d e^2+6 A c d^2 e+2 B c d^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )-3 a B e \left (4 c d^2-a e^2\right )\right )-40 b^3 c e^2 (A e+3 B d)+35 b^4 B e^3\right )}{128 c^{9/2}}+\frac{(d+e x)^2 \sqrt{a+b x+c x^2} (8 A c e-7 b B e+6 B c d)}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^3)/Sqrt[a + b*x + c*x^2],x]

[Out]

((6*B*c*d - 7*b*B*e + 8*A*c*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(24*c^2) + (B*(d + e*x)^3*Sqrt[a + b*x + c*x
^2])/(4*c) + ((8*A*c*e*(64*c^2*d^2 + 15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e)) + B*(96*c^3*d^3 - 105*b^3*e^3 + 20*b
*c*e^2*(18*b*d + 11*a*e) - 8*c^2*d*e*(47*b*d + 48*a*e)) + 2*c*e*(40*A*c*e*(2*c*d - b*e) + B*(24*c^2*d^2 + 35*b
^2*e^2 - 4*c*e*(16*b*d + 9*a*e)))*x)*Sqrt[a + b*x + c*x^2])/(192*c^4) + ((35*b^4*B*e^3 - 40*b^3*c*e^2*(3*B*d +
A*e) + 24*b^2*c*e*(6*B*c*d^2 + 6*A*c*d*e - 5*a*B*e^2) - 32*b*c^2*(2*B*c*d^3 + 6*A*c*d^2*e - 9*a*B*d*e^2 - 3*a
*A*e^3) + 16*c^2*(4*A*c*d*(2*c*d^2 - 3*a*e^2) - 3*a*B*e*(4*c*d^2 - a*e^2)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(128*c^(9/2))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 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{(A+B x) (d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx &=\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c}+\frac{\int \frac{(d+e x)^2 \left (\frac{1}{2} (-b B d+8 A c d-6 a B e)+\frac{1}{2} (6 B c d-7 b B e+8 A c e) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{4 c}\\ &=\frac{(6 B c d-7 b B e+8 A c e) (d+e x)^2 \sqrt{a+b x+c x^2}}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c}+\frac{\int \frac{(d+e x) \left (\frac{1}{4} \left (7 b^2 B d e+4 c \left (12 A c d^2-15 a B d e-8 a A e^2\right )-4 b \left (3 B c d^2+2 A c d e-7 a B e^2\right )\right )+\frac{1}{4} \left (40 A c e (2 c d-b e)+B \left (24 c^2 d^2+35 b^2 e^2-4 c e (16 b d+9 a e)\right )\right ) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{12 c^2}\\ &=\frac{(6 B c d-7 b B e+8 A c e) (d+e x)^2 \sqrt{a+b x+c x^2}}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c}+\frac{\left (8 A c e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)\right )+B \left (96 c^3 d^3-105 b^3 e^3+20 b c e^2 (18 b d+11 a e)-8 c^2 d e (47 b d+48 a e)\right )+2 c e \left (40 A c e (2 c d-b e)+B \left (24 c^2 d^2+35 b^2 e^2-4 c e (16 b d+9 a e)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4}+\frac{\left (35 b^4 B e^3-40 b^3 c e^2 (3 B d+A e)+24 b^2 c e \left (6 B c d^2+6 A c d e-5 a B e^2\right )-32 b c^2 \left (2 B c d^3+6 A c d^2 e-9 a B d e^2-3 a A e^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )-3 a B e \left (4 c d^2-a e^2\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{128 c^4}\\ &=\frac{(6 B c d-7 b B e+8 A c e) (d+e x)^2 \sqrt{a+b x+c x^2}}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c}+\frac{\left (8 A c e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)\right )+B \left (96 c^3 d^3-105 b^3 e^3+20 b c e^2 (18 b d+11 a e)-8 c^2 d e (47 b d+48 a e)\right )+2 c e \left (40 A c e (2 c d-b e)+B \left (24 c^2 d^2+35 b^2 e^2-4 c e (16 b d+9 a e)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4}+\frac{\left (35 b^4 B e^3-40 b^3 c e^2 (3 B d+A e)+24 b^2 c e \left (6 B c d^2+6 A c d e-5 a B e^2\right )-32 b c^2 \left (2 B c d^3+6 A c d^2 e-9 a B d e^2-3 a A e^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )-3 a B e \left (4 c d^2-a e^2\right )\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 )}{64 c^4}\\ &=\frac{(6 B c d-7 b B e+8 A c e) (d+e x)^2 \sqrt{a+b x+c x^2}}{24 c^2}+\frac{B (d+e x)^3 \sqrt{a+b x+c x^2}}{4 c}+\frac{\left (8 A c e \left (64 c^2 d^2+15 b^2 e^2-2 c e (27 b d+8 a e)\right )+B \left (96 c^3 d^3-105 b^3 e^3+20 b c e^2 (18 b d+11 a e)-8 c^2 d e (47 b d+48 a e)\right )+2 c e \left (40 A c e (2 c d-b e)+B \left (24 c^2 d^2+35 b^2 e^2-4 c e (16 b d+9 a e)\right )\right ) x\right ) \sqrt{a+b x+c x^2}}{192 c^4}+\frac{\left (35 b^4 B e^3-40 b^3 c e^2 (3 B d+A e)+24 b^2 c e \left (6 B c d^2+6 A c d e-5 a B e^2\right )-32 b c^2 \left (2 B c d^3+6 A c d^2 e-9 a B d e^2-3 a A e^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )-3 a B e \left (4 c d^2-a e^2\right )\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.592163, size = 357, normalized size = 0.88 \[ \frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (8 A c e \left (-2 c e (8 a e+27 b d+5 b e x)+15 b^2 e^2+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+B \left (-8 c^2 e \left (3 a e (16 d+3 e x)+b \left (54 d^2+30 d e x+7 e^2 x^2\right )\right )+10 b c e^2 (22 a e+36 b d+7 b e x)-105 b^3 e^3+48 c^3 \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )\right )\right )+3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (24 b^2 c e \left (-5 a B e^2+6 A c d e+6 B c d^2\right )-32 b c^2 \left (-3 a A e^3-9 a B d e^2+6 A c d^2 e+2 B c d^3\right )+16 c^2 \left (4 A c d \left (2 c d^2-3 a e^2\right )+3 a B e \left (a e^2-4 c d^2\right )\right )-40 b^3 c e^2 (A e+3 B d)+35 b^4 B e^3\right )}{384 c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^3)/Sqrt[a + b*x + c*x^2],x]

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(8*A*c*e*(15*b^2*e^2 - 2*c*e*(27*b*d + 8*a*e + 5*b*e*x) + 4*c^2*(18*d^2 + 9*d
*e*x + 2*e^2*x^2)) + B*(-105*b^3*e^3 + 10*b*c*e^2*(36*b*d + 22*a*e + 7*b*e*x) + 48*c^3*(4*d^3 + 6*d^2*e*x + 4*
d*e^2*x^2 + e^3*x^3) - 8*c^2*e*(3*a*e*(16*d + 3*e*x) + b*(54*d^2 + 30*d*e*x + 7*e^2*x^2)))) + 3*(35*b^4*B*e^3
- 40*b^3*c*e^2*(3*B*d + A*e) + 24*b^2*c*e*(6*B*c*d^2 + 6*A*c*d*e - 5*a*B*e^2) - 32*b*c^2*(2*B*c*d^3 + 6*A*c*d^
2*e - 9*a*B*d*e^2 - 3*a*A*e^3) + 16*c^2*(4*A*c*d*(2*c*d^2 - 3*a*e^2) + 3*a*B*e*(-4*c*d^2 + a*e^2)))*ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(384*c^(9/2))

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Maple [B]  time = 0.015, size = 981, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x)

[Out]

3/8*B*e^3*a^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+
b*x+a)^(1/2))*B*d^3+1/4*B*e^3*x^3/c*(c*x^2+b*x+a)^(1/2)-35/64*B*e^3*b^3/c^4*(c*x^2+b*x+a)^(1/2)+35/128*B*e^3*b
^4/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-5/16*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(
1/2))*A*e^3-2/3*a/c^2*(c*x^2+b*x+a)^(1/2)*A*e^3+3/c*(c*x^2+b*x+a)^(1/2)*A*d^2*e-5/4*b/c^2*x*(c*x^2+b*x+a)^(1/2
)*B*d*e^2+9/4*b/c^(5/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e^2+1/3*x^2/c*(c*x^2+b*x+a)^(1/2)*A*
e^3+5/8*b^2/c^3*(c*x^2+b*x+a)^(1/2)*A*e^3-3/8*B*e^3*a/c^2*x*(c*x^2+b*x+a)^(1/2)+35/96*B*e^3*b^2/c^3*x*(c*x^2+b
*x+a)^(1/2)-15/16*B*e^3*b^2/c^(7/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+55/48*B*e^3*b/c^3*a*(c*x^2+b
*x+a)^(1/2)+1/c*(c*x^2+b*x+a)^(1/2)*B*d^3+A*d^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-7/24*B*e^3
*b/c^2*x^2*(c*x^2+b*x+a)^(1/2)+3/2*x/c*(c*x^2+b*x+a)^(1/2)*B*d^2*e-9/4*b/c^2*(c*x^2+b*x+a)^(1/2)*A*d*e^2-15/16
*b^3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e^2+3/4*b/c^(5/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2
+b*x+a)^(1/2))*A*e^3-2*a/c^2*(c*x^2+b*x+a)^(1/2)*B*d*e^2+x^2/c*(c*x^2+b*x+a)^(1/2)*B*d*e^2-5/12*b/c^2*x*(c*x^2
+b*x+a)^(1/2)*A*e^3+15/8*b^2/c^3*(c*x^2+b*x+a)^(1/2)*B*d*e^2+9/8*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))*B*d^2*e-3/2*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*d*e^2-3/2*a/c^(3/2)*ln((1/2*b+c
*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d^2*e+3/2*x/c*(c*x^2+b*x+a)^(1/2)*A*d*e^2-3/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1
/2)+(c*x^2+b*x+a)^(1/2))*A*d^2*e-9/4*b/c^2*(c*x^2+b*x+a)^(1/2)*B*d^2*e+9/8*b^2/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+
(c*x^2+b*x+a)^(1/2))*A*d*e^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((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.67075, size = 1835, normalized size = 4.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/768*(3*(64*(B*b*c^3 - 2*A*c^4)*d^3 - 48*(3*B*b^2*c^2 - 4*(B*a + A*b)*c^3)*d^2*e + 24*(5*B*b^3*c + 8*A*a*c^
3 - 6*(2*B*a*b + A*b^2)*c^2)*d*e^2 - (35*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2 + A*b^3)*c)*e^3)*sqr
t(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*(48*B*c^4*e^3*x
^3 + 192*B*c^4*d^3 - 144*(3*B*b*c^3 - 4*A*c^4)*d^2*e + 24*(15*B*b^2*c^2 - 2*(8*B*a + 9*A*b)*c^3)*d*e^2 - (105*
B*b^3*c + 128*A*a*c^3 - 20*(11*B*a*b + 6*A*b^2)*c^2)*e^3 + 8*(24*B*c^4*d*e^2 - (7*B*b*c^3 - 8*A*c^4)*e^3)*x^2
+ 2*(144*B*c^4*d^2*e - 24*(5*B*b*c^3 - 6*A*c^4)*d*e^2 + (35*B*b^2*c^2 - 4*(9*B*a + 10*A*b)*c^3)*e^3)*x)*sqrt(c
*x^2 + b*x + a))/c^5, 1/384*(3*(64*(B*b*c^3 - 2*A*c^4)*d^3 - 48*(3*B*b^2*c^2 - 4*(B*a + A*b)*c^3)*d^2*e + 24*(
5*B*b^3*c + 8*A*a*c^3 - 6*(2*B*a*b + A*b^2)*c^2)*d*e^2 - (35*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 40*(3*B*a*b^2
+ A*b^3)*c)*e^3)*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*(
48*B*c^4*e^3*x^3 + 192*B*c^4*d^3 - 144*(3*B*b*c^3 - 4*A*c^4)*d^2*e + 24*(15*B*b^2*c^2 - 2*(8*B*a + 9*A*b)*c^3)
*d*e^2 - (105*B*b^3*c + 128*A*a*c^3 - 20*(11*B*a*b + 6*A*b^2)*c^2)*e^3 + 8*(24*B*c^4*d*e^2 - (7*B*b*c^3 - 8*A*
c^4)*e^3)*x^2 + 2*(144*B*c^4*d^2*e - 24*(5*B*b*c^3 - 6*A*c^4)*d*e^2 + (35*B*b^2*c^2 - 4*(9*B*a + 10*A*b)*c^3)*
e^3)*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 \frac{\left (A + B x\right ) \left (d + e x\right )^{3}}{\sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**3/sqrt(a + b*x + c*x**2), x)

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Giac [A]  time = 1.17873, size = 556, normalized size = 1.37 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (\frac{6 \, B x e^{3}}{c} + \frac{24 \, B c^{3} d e^{2} - 7 \, B b c^{2} e^{3} + 8 \, A c^{3} e^{3}}{c^{4}}\right )} x + \frac{144 \, B c^{3} d^{2} e - 120 \, B b c^{2} d e^{2} + 144 \, A c^{3} d e^{2} + 35 \, B b^{2} c e^{3} - 36 \, B a c^{2} e^{3} - 40 \, A b c^{2} e^{3}}{c^{4}}\right )} x + \frac{192 \, B c^{3} d^{3} - 432 \, B b c^{2} d^{2} e + 576 \, A c^{3} d^{2} e + 360 \, B b^{2} c d e^{2} - 384 \, B a c^{2} d e^{2} - 432 \, A b c^{2} d e^{2} - 105 \, B b^{3} e^{3} + 220 \, B a b c e^{3} + 120 \, A b^{2} c e^{3} - 128 \, A a c^{2} e^{3}}{c^{4}}\right )} + \frac{{\left (64 \, B b c^{3} d^{3} - 128 \, A c^{4} d^{3} - 144 \, B b^{2} c^{2} d^{2} e + 192 \, B a c^{3} d^{2} e + 192 \, A b c^{3} d^{2} e + 120 \, B b^{3} c d e^{2} - 288 \, B a b c^{2} d e^{2} - 144 \, A b^{2} c^{2} d e^{2} + 192 \, A a c^{3} d e^{2} - 35 \, B b^{4} e^{3} + 120 \, B a b^{2} c e^{3} + 40 \, A b^{3} c e^{3} - 48 \, B a^{2} c^{2} e^{3} - 96 \, A a b c^{2} e^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*B*x*e^3/c + (24*B*c^3*d*e^2 - 7*B*b*c^2*e^3 + 8*A*c^3*e^3)/c^4)*x + (144*
B*c^3*d^2*e - 120*B*b*c^2*d*e^2 + 144*A*c^3*d*e^2 + 35*B*b^2*c*e^3 - 36*B*a*c^2*e^3 - 40*A*b*c^2*e^3)/c^4)*x +
(192*B*c^3*d^3 - 432*B*b*c^2*d^2*e + 576*A*c^3*d^2*e + 360*B*b^2*c*d*e^2 - 384*B*a*c^2*d*e^2 - 432*A*b*c^2*d*
e^2 - 105*B*b^3*e^3 + 220*B*a*b*c*e^3 + 120*A*b^2*c*e^3 - 128*A*a*c^2*e^3)/c^4) + 1/128*(64*B*b*c^3*d^3 - 128*
A*c^4*d^3 - 144*B*b^2*c^2*d^2*e + 192*B*a*c^3*d^2*e + 192*A*b*c^3*d^2*e + 120*B*b^3*c*d*e^2 - 288*B*a*b*c^2*d*
e^2 - 144*A*b^2*c^2*d*e^2 + 192*A*a*c^3*d*e^2 - 35*B*b^4*e^3 + 120*B*a*b^2*c*e^3 + 40*A*b^3*c*e^3 - 48*B*a^2*c
^2*e^3 - 96*A*a*b*c^2*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)