3.2473 \(\int \frac{(A+B x) (d+e x)^3}{(a+b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=325 \[ -\frac{e \sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c (3 a B e+A b e+b B d)+8 A c^2 d+5 b^2 B e\right )+4 b c \left (-13 a B e^2+6 A c d e+4 B c d^2\right )-32 c^2 \left (-a A e^2-3 a B d e+A c d^2\right )-12 b^2 c e (A e+3 B d)+15 b^3 B e^2\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2}}+\frac{2 (d+e x)^2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.338476, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {818, 779, 621, 206} \[ -\frac{e \sqrt{a+b x+c x^2} \left (-2 c e x \left (-4 c (3 a B e+A b e+b B d)+8 A c^2 d+5 b^2 B e\right )+4 b c \left (-13 a B e^2+6 A c d e+4 B c d^2\right )-32 c^2 \left (-a A e^2-3 a B d e+A c d^2\right )-12 b^2 c e (A e+3 B d)+15 b^3 B e^2\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2}}+\frac{2 (d+e x)^2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac{2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}+\frac{2 \int \frac{(d+e x) \left (\frac{1}{2} e \left (b^2 B d+4 A b c d-12 a B c d+4 a b B e-8 a A c e\right )+\frac{1}{2} e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{c \left (b^2-4 a c\right )}\\ &=\frac{2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{8 c^3}\\ &=\frac{2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\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 )}{4 c^3}\\ &=\frac{2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2}}-\frac{e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt{a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac{3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\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 = 0.736187, size = 426, normalized size = 1.31 \[ \frac{-2 \sqrt{c} \left (4 A c \left (-4 c \left (2 a^2 e^3+a c e \left (-3 d^2-3 d e x+e^2 x^2\right )+c^2 d^3 x\right )+b^2 e^2 (3 a e+c x (e x-6 d))-2 b c \left (a e^2 (3 d+5 e x)+c d^2 (d-3 e x)\right )+3 b^3 e^3 x\right )+B \left (-4 a^2 c e^2 (6 c (4 d+e x)-13 b e)+a \left (2 b^2 c e^2 (18 d+31 e x)-15 b^3 e^3+4 b c^2 e \left (-6 d^2-30 d e x+5 e^2 x^2\right )+8 c^3 \left (6 d^2 e x+2 d^3-6 d e^2 x^2-e^3 x^3\right )\right )+b x \left (b^2 c e^2 (36 d-5 e x)-15 b^3 e^3+2 b c^2 e \left (-12 d^2+6 d e x+e^2 x^2\right )+8 c^3 d^3\right )\right )\right )-3 e \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c]*(4*A*c*(3*b^3*e^3*x + b^2*e^2*(3*a*e + c*x*(-6*d + e*x)) - 2*b*c*(c*d^2*(d - 3*e*x) + a*e^2*(3*d +
 5*e*x)) - 4*c*(2*a^2*e^3 + c^2*d^3*x + a*c*e*(-3*d^2 - 3*d*e*x + e^2*x^2))) + B*(-4*a^2*c*e^2*(-13*b*e + 6*c*
(4*d + e*x)) + b*x*(8*c^3*d^3 - 15*b^3*e^3 + b^2*c*e^2*(36*d - 5*e*x) + 2*b*c^2*e*(-12*d^2 + 6*d*e*x + e^2*x^2
)) + a*(-15*b^3*e^3 + 2*b^2*c*e^2*(18*d + 31*e*x) + 4*b*c^2*e*(-6*d^2 - 30*d*e*x + 5*e^2*x^2) + 8*c^3*(2*d^3 +
 6*d^2*e*x - 6*d*e^2*x^2 - e^3*x^3)))) - 3*(b^2 - 4*a*c)*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 + 5*b^2*e^2 -
 4*c*e*(3*b*d + a*e)))*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 = 0.016, size = 1451, normalized size = 4.5 \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)^(3/2),x)

[Out]

-9/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d*e^2+4*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*e^3+6*a/c^2
*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d*e^2+3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*d*e^2+2*a/c^2/(c*x^2+
b*x+a)^(1/2)*A*e^3+15/8*B*e^3*b^2/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/2*B*e^3*a/c^(5/2)*ln((
1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*B*e^3*x^3/c/(c*x^2+b*x+a)^(1/2)+15/16*B*e^3*b^3/c^4/(c*x^2+b*x+a)^
(1/2)+2*A*d^3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/4*b^2/c^3/(c*x^2+b*x+a)^(1/2)*A*e^3-3/2*b/c^(5/2)*ln
((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*e^3+3/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*A*d*e^2+
x^2/c/(c*x^2+b*x+a)^(1/2)*A*e^3+3/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/c/(c*x^2+b*x+a
)^(1/2)*A*d^2*e-3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*d^2*e+15/8*B*e^3*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(
1/2)*x-13/4*B*e^3*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+9/2*b/c^2*x/(c*x^2+b*x+a)^(1/2)*B*d*e^2-3/2*b^3/c^
2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*e^3-9/4*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d*e^2+3/2*b^3/c^2/(4*a
*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*d*e^2+3/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d^2*e-6*b/(4*a*c-b^2)/(c*x^2
+b*x+a)^(1/2)*x*A*d^2*e+2*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*e^3-5/4*B*e^3*b/c^2*x^2/(c*x^2+b*x+a)^(1
/2)-13/4*B*e^3*b/c^3*a/(c*x^2+b*x+a)^(1/2)+3/2*B*e^3*a/c^2*x/(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)*A*e^3-9/2*b/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e^2+6*a/c^2/(c*x^2+b*x+a)^(
1/2)*B*d*e^2+3*x^2/c/(c*x^2+b*x+a)^(1/2)*B*d*e^2+3/2*b/c^2*x/(c*x^2+b*x+a)^(1/2)*A*e^3-9/4*b^2/c^3/(c*x^2+b*x+
a)^(1/2)*B*d*e^2-15/8*B*e^3*b^2/c^3*x/(c*x^2+b*x+a)^(1/2)+15/16*B*e^3*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-
3*x/c/(c*x^2+b*x+a)^(1/2)*A*d*e^2-3*x/c/(c*x^2+b*x+a)^(1/2)*B*d^2*e+3/2*b/c^2/(c*x^2+b*x+a)^(1/2)*A*d*e^2+3/2*
b/c^2/(c*x^2+b*x+a)^(1/2)*B*d^2*e-2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d^3-b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^
(1/2)*B*d^3+12*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d*e^2-1/c/(c*x^2+b*x+a)^(1/2)*B*d^3+3*b^2/c/(4*a*c-b^
2)/(c*x^2+b*x+a)^(1/2)*x*B*d^2*e-13/2*B*e^3*b^2/c^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 27.6188, size = 3428, normalized size = 10.55 \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(8*(B*a*b^2*c^2 - 4*B*a^2*c^3)*d^2*e - 4*(3*B*a*b^3*c + 8*A*a^2*c^3 - 2*(6*B*a^2*b + A*a*b^2)*c^2)*d
*e^2 + (5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 - 4*(6*B*a^2*b^2 + A*a*b^3)*c)*e^3 + (8*(B*b^2*c^3 - 4*B*a*c^4)*d
^2*e - 4*(3*B*b^3*c^2 + 8*A*a*c^4 - 2*(6*B*a*b + A*b^2)*c^3)*d*e^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(
6*B*a*b^2 + A*b^3)*c^2)*e^3)*x^2 + (8*(B*b^3*c^2 - 4*B*a*b*c^3)*d^2*e - 4*(3*B*b^4*c + 8*A*a*b*c^3 - 2*(6*B*a*
b^2 + A*b^3)*c^2)*d*e^2 + (5*B*b^5 + 16*(B*a^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*e^3)*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*(8*(2*B*a - A*b)*c^4*d^
3 + 2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 - 24*(B*a*b*c^3 - 2*A*a*c^4)*d^2*e + 12*(3*B*a*b^2*c^2 - 2*(4*B*a^2 + A*
a*b)*c^3)*d*e^2 - (15*B*a*b^3*c + 32*A*a^2*c^3 - 4*(13*B*a^2*b + 3*A*a*b^2)*c^2)*e^3 + (12*(B*b^2*c^3 - 4*B*a*
c^4)*d*e^2 - (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*b^2)*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(
B*b^2*c^3 - (2*B*a + A*b)*c^4)*d^2*e + 12*(3*B*b^3*c^2 + 4*A*a*c^4 - 2*(5*B*a*b + A*b^2)*c^3)*d*e^2 - (15*B*b^
4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^3)*c^2)*e^3)*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*(3*(8*(B*a*b^2*c^2 - 4*B*a^2*c^3)*d^2*e - 4
*(3*B*a*b^3*c + 8*A*a^2*c^3 - 2*(6*B*a^2*b + A*a*b^2)*c^2)*d*e^2 + (5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 - 4*(
6*B*a^2*b^2 + A*a*b^3)*c)*e^3 + (8*(B*b^2*c^3 - 4*B*a*c^4)*d^2*e - 4*(3*B*b^3*c^2 + 8*A*a*c^4 - 2*(6*B*a*b + A
*b^2)*c^3)*d*e^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(6*B*a*b^2 + A*b^3)*c^2)*e^3)*x^2 + (8*(B*b^3*c^2 -
 4*B*a*b*c^3)*d^2*e - 4*(3*B*b^4*c + 8*A*a*b*c^3 - 2*(6*B*a*b^2 + A*b^3)*c^2)*d*e^2 + (5*B*b^5 + 16*(B*a^2*b +
 A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*e^3)*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*(8*(2*B*a - A*b)*c^4*d^3 + 2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 - 24*(B*a*b*c^3 -
2*A*a*c^4)*d^2*e + 12*(3*B*a*b^2*c^2 - 2*(4*B*a^2 + A*a*b)*c^3)*d*e^2 - (15*B*a*b^3*c + 32*A*a^2*c^3 - 4*(13*B
*a^2*b + 3*A*a*b^2)*c^2)*e^3 + (12*(B*b^2*c^3 - 4*B*a*c^4)*d*e^2 - (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*
b^2)*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(B*b^2*c^3 - (2*B*a + A*b)*c^4)*d^2*e + 12*(3*B*b^3*c^2 +
 4*A*a*c^4 - 2*(5*B*a*b + A*b^2)*c^3)*d*e^2 - (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^
3)*c^2)*e^3)*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 (A + B x\right ) \left (d + e x\right )^{3}}{\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((B*x+A)*(e*x+d)**3/(c*x**2+b*x+a)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.26229, size = 716, normalized size = 2.2 \begin{align*} \frac{{\left ({\left (\frac{2 \,{\left (B b^{2} c^{2} e^{3} - 4 \, B a c^{3} e^{3}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac{12 \, B b^{2} c^{2} d e^{2} - 48 \, B a c^{3} d e^{2} - 5 \, B b^{3} c e^{3} + 20 \, B a b c^{2} e^{3} + 4 \, A b^{2} c^{2} e^{3} - 16 \, A a c^{3} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x + \frac{8 \, B b c^{3} d^{3} - 16 \, A c^{4} d^{3} - 24 \, B b^{2} c^{2} d^{2} e + 48 \, B a c^{3} d^{2} e + 24 \, A b c^{3} d^{2} e + 36 \, B b^{3} c d e^{2} - 120 \, B a b c^{2} d e^{2} - 24 \, A b^{2} c^{2} d e^{2} + 48 \, A a c^{3} d e^{2} - 15 \, B b^{4} e^{3} + 62 \, B a b^{2} c e^{3} + 12 \, A b^{3} c e^{3} - 24 \, B a^{2} c^{2} e^{3} - 40 \, A a b c^{2} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x + \frac{16 \, B a c^{3} d^{3} - 8 \, A b c^{3} d^{3} - 24 \, B a b c^{2} d^{2} e + 48 \, A a c^{3} d^{2} e + 36 \, B a b^{2} c d e^{2} - 96 \, B a^{2} c^{2} d e^{2} - 24 \, A a b c^{2} d e^{2} - 15 \, B a b^{3} e^{3} + 52 \, B a^{2} b c e^{3} + 12 \, A a b^{2} c e^{3} - 32 \, A a^{2} c^{2} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt{c x^{2} + b x + a}} - \frac{3 \,{\left (8 \, B c^{2} d^{2} e - 12 \, B b c d e^{2} + 8 \, A c^{2} d e^{2} + 5 \, B b^{2} e^{3} - 4 \, B a c e^{3} - 4 \, A b c e^{3}\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((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/4*(((2*(B*b^2*c^2*e^3 - 4*B*a*c^3*e^3)*x/(b^2*c^3 - 4*a*c^4) + (12*B*b^2*c^2*d*e^2 - 48*B*a*c^3*d*e^2 - 5*B*
b^3*c*e^3 + 20*B*a*b*c^2*e^3 + 4*A*b^2*c^2*e^3 - 16*A*a*c^3*e^3)/(b^2*c^3 - 4*a*c^4))*x + (8*B*b*c^3*d^3 - 16*
A*c^4*d^3 - 24*B*b^2*c^2*d^2*e + 48*B*a*c^3*d^2*e + 24*A*b*c^3*d^2*e + 36*B*b^3*c*d*e^2 - 120*B*a*b*c^2*d*e^2
- 24*A*b^2*c^2*d*e^2 + 48*A*a*c^3*d*e^2 - 15*B*b^4*e^3 + 62*B*a*b^2*c*e^3 + 12*A*b^3*c*e^3 - 24*B*a^2*c^2*e^3
- 40*A*a*b*c^2*e^3)/(b^2*c^3 - 4*a*c^4))*x + (16*B*a*c^3*d^3 - 8*A*b*c^3*d^3 - 24*B*a*b*c^2*d^2*e + 48*A*a*c^3
*d^2*e + 36*B*a*b^2*c*d*e^2 - 96*B*a^2*c^2*d*e^2 - 24*A*a*b*c^2*d*e^2 - 15*B*a*b^3*e^3 + 52*B*a^2*b*c*e^3 + 12
*A*a*b^2*c*e^3 - 32*A*a^2*c^2*e^3)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 3/8*(8*B*c^2*d^2*e - 12*B*b*c*
d*e^2 + 8*A*c^2*d*e^2 + 5*B*b^2*e^3 - 4*B*a*c*e^3 - 4*A*b*c*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)*sqrt(c) - b))/c^(7/2)