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

Optimal. Leaf size=186 \[ -\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (a+b x) \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 77} \begin {gather*} -\frac {(b d-a e) (-3 a B e+2 A b e+b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (a+b x) \log (a+b x) (-3 a B e+A b e+2 b B d)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*d - a*e)*(b*B*d + 2*A*b*e - 3*a*B*e))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - ((A*b - a*B)*(b*d - a*e)^2)
/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*e^2*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (
e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)*Log[a + b*x])/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(A+B x) (d+e x)^2}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {B e^2}{b^6}+\frac {(A b-a B) (b d-a e)^2}{b^6 (a+b x)^3}+\frac {(b d-a e) (b B d+2 A b e-3 a B e)}{b^6 (a+b x)^2}+\frac {e (2 b B d+A b e-3 a B e)}{b^6 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(b d-a e) (b B d+2 A b e-3 a B e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (2 b B d+A b e-3 a B e) (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 151, normalized size = 0.81 \begin {gather*} \frac {B \left (-5 a^3 e^2+2 a^2 b e (3 d-2 e x)+a b^2 \left (-d^2+8 d e x+4 e^2 x^2\right )+2 b^3 x \left (e^2 x^2-d^2\right )\right )+2 e (a+b x)^2 \log (a+b x) (-3 a B e+A b e+2 b B d)-A b (b d-a e) (3 a e+b (d+4 e x))}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(b*d - a*e)*(3*a*e + b*(d + 4*e*x))) + B*(-5*a^3*e^2 + 2*a^2*b*e*(3*d - 2*e*x) + 2*b^3*x*(-d^2 + e^2*x^
2) + a*b^2*(-d^2 + 8*d*e*x + 4*e^2*x^2)) + 2*e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^4*(a
+ b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 5.25, size = 4158, normalized size = 22.35 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[b^2]*d*(-(a*A*d) + A*b*d*x - 2*a*B*d*x - 4*a*A*e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - d*(-(a^2*A*b*d) +
a^3*B*d + 2*a^3*A*e + 2*a^2*b*B*d*x + 4*a^2*A*b*e*x - A*b^3*d*x^2 + 2*a*b^2*B*d*x^2 + 4*a*A*b^2*e*x^2))/(b*x^
2*(-2*a*b^3 - 2*b^4*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + b*Sqrt[b^2]*x^2*(2*a^2*b^2 + 4*a*b^3*x + 2*b^4*x^2)) +
((8*a^2*(b^2)^(3/2)*B*d^2*x)/b^4 + (16*a^2*A*(b^2)^(3/2)*d*e*x)/b^4 + (16*a^4*Sqrt[b^2]*B*e^2*x)/b^4 + (12*a*S
qrt[b^2]*B*d^2*x^2)/b + (24*a*A*Sqrt[b^2]*d*e*x^2)/b + (32*a^3*Sqrt[b^2]*B*e^2*x^2)/b^3 + 8*Sqrt[b^2]*B*d^2*x^
3 + 16*A*Sqrt[b^2]*d*e*x^3 + (8*a^2*(b^2)^(3/2)*B*e^2*x^3)/b^4 - (20*a*Sqrt[b^2]*B*e^2*x^4)/b - 8*Sqrt[b^2]*B*
e^2*x^5 - (4*a^2*B*d^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (8*a^2*A*d*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 -
(4*a^4*B*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (4*a*B*d^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (8*a*A*d*e*x
*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (12*a^3*B*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - 8*B*d^2*x^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2] - 16*A*d*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (20*a^2*B*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/b^2 + (12*a*B*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + 8*B*e^2*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (2
4*a^3*B*e^2*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (48*a^2*B*e^2*x^3*ArcTanh[(
-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 24*a*B*e^2*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/a] + (24*a^2*Sqrt[b^2]*B*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/a])/b^3 + (24*a*(b^2)^(3/2)*B*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqr
t[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4)/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((8*a^4*B*d*e)/(b^2)^(3/2) + (4*a^4*A*e^2)/(b^2)^(3/2) + (2
4*a^3*b*B*d*e*x)/(b^2)^(3/2) + (12*a^3*A*b*e^2*x)/(b^2)^(3/2) + (24*a^2*B*d*e*x^2)/Sqrt[b^2] + (12*a^2*A*e^2*x
^2)/Sqrt[b^2] - (24*a^2*B*d*e*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (12*a^2*A*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])/b^2 - (8*a^2*B*d*e*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (4*a^2*A*e^2*x
^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (16*a*b^3*B*d*e*x^3*Log[-a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (8*a*A*b^3*e^2*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/(b^2)^(3/2) - (8*b^4*B*d*e*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2)
- (4*A*b^4*e^2*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (8*a*B*d*e*x^2*Sqrt[a
^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + (4*a*A*e^2*x^2*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 8*B*d*e*x^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + 4*A*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log
[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (8*a^2*B*d*e*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/Sqrt[b^2] - (4*a^2*A*e^2*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (
16*a*b^3*B*d*e*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (8*a*A*b^3*e^2*x^3*Log[
a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (8*b^4*B*d*e*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^
2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (4*A*b^4*e^2*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/
(b^2)^(3/2) + (8*a*B*d*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]
])/b + (4*a*A*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b +
8*B*d*e*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + 4*A*e^2*x^3*S
qrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a
^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((-4*a^5*B*e^2)/(b^3*Sqrt[b^
2]) - (24*a^3*B*d*e*x)/(b*Sqrt[b^2]) - (12*a^3*A*e^2*x)/(b*Sqrt[b^2]) - (16*a^4*B*e^2*x)/(b^2)^(3/2) - (48*a^2
*B*d*e*x^2)/Sqrt[b^2] - (24*a^2*A*e^2*x^2)/Sqrt[b^2] - (16*a^3*B*e^2*x^2)/(b*Sqrt[b^2]) - (32*a*b*B*d*e*x^3)/S
qrt[b^2] - (16*a*A*b*e^2*x^3)/Sqrt[b^2] + (8*a^3*B*d*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (4*a^3*A*e^2*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/b^3 + (16*a^2*B*d*e*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (8*a^2*A*e^2*x*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/b^2 + (16*a^3*B*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (32*a*B*d*e*x^2*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/b + (16*a*A*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (16*a^2*B*d*e*x^2*ArcTanh[(-(Sqrt[b^2]
*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + (8*a^2*A*e^2*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/a])/b + 32*a*B*d*e*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 16*a*A*e^2*x^3*A
rcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 16*b*B*d*e*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/a] + 8*A*b*e^2*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (16*a*
B*d*e*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^2]
- (8*a*A*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/S
qrt[b^2] - (16*b*B*d*e*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/a])/Sqrt[b^2] - (8*A*b*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/a])/Sqrt[b^2] + (12*a^3*B*e^2*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b
^2]) + (24*a^2*B*e^2*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (12*a*b*B*e^2*x^4*
Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*B*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^2 - (12*a*B*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + (12*a^3*B*e^2*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*B*e^2*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])
/Sqrt[b^2] + (12*a*b*B*e^2*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*B*e^2
*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^2 - (12*a*B*e^2*x^3
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b)/((-a - Sqrt[b^2]*x + S
qrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)

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fricas [A]  time = 0.44, size = 258, normalized size = 1.39 \begin {gather*} \frac {2 \, B b^{3} e^{2} x^{3} + 4 \, B a b^{2} e^{2} x^{2} - {\left (B a b^{2} + A b^{3}\right )} d^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{2}\right )} d e - {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \, {\left (B b^{3} d^{2} - 2 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d e + 2 \, {\left (B a^{2} b - A a b^{2}\right )} e^{2}\right )} x + 2 \, {\left (2 \, B a^{2} b d e - {\left (3 \, B a^{3} - A a^{2} b\right )} e^{2} + {\left (2 \, B b^{3} d e - {\left (3 \, B a b^{2} - A b^{3}\right )} e^{2}\right )} x^{2} + 2 \, {\left (2 \, B a b^{2} d e - {\left (3 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*B*b^3*e^2*x^3 + 4*B*a*b^2*e^2*x^2 - (B*a*b^2 + A*b^3)*d^2 + 2*(3*B*a^2*b - A*a*b^2)*d*e - (5*B*a^3 - 3*
A*a^2*b)*e^2 - 2*(B*b^3*d^2 - 2*(2*B*a*b^2 - A*b^3)*d*e + 2*(B*a^2*b - A*a*b^2)*e^2)*x + 2*(2*B*a^2*b*d*e - (3
*B*a^3 - A*a^2*b)*e^2 + (2*B*b^3*d*e - (3*B*a*b^2 - A*b^3)*e^2)*x^2 + 2*(2*B*a*b^2*d*e - (3*B*a^2*b - A*a*b^2)
*e^2)*x)*log(b*x + a))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.07, size = 303, normalized size = 1.63 \begin {gather*} \frac {\left (2 A \,b^{3} e^{2} x^{2} \ln \left (b x +a \right )-6 B a \,b^{2} e^{2} x^{2} \ln \left (b x +a \right )+4 B \,b^{3} d e \,x^{2} \ln \left (b x +a \right )+2 B \,b^{3} e^{2} x^{3}+4 A a \,b^{2} e^{2} x \ln \left (b x +a \right )-12 B \,a^{2} b \,e^{2} x \ln \left (b x +a \right )+8 B a \,b^{2} d e x \ln \left (b x +a \right )+4 B a \,b^{2} e^{2} x^{2}+2 A \,a^{2} b \,e^{2} \ln \left (b x +a \right )+4 A a \,b^{2} e^{2} x -4 A \,b^{3} d e x -6 B \,a^{3} e^{2} \ln \left (b x +a \right )+4 B \,a^{2} b d e \ln \left (b x +a \right )-4 B \,a^{2} b \,e^{2} x +8 B a \,b^{2} d e x -2 B \,b^{3} d^{2} x +3 A \,a^{2} b \,e^{2}-2 A a \,b^{2} d e -A \,b^{3} d^{2}-5 B \,a^{3} e^{2}+6 B \,a^{2} b d e -B a \,b^{2} d^{2}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*ln(b*x+a)*x^2*b^3*e^2-6*B*ln(b*x+a)*x^2*a*b^2*e^2+4*B*ln(b*x+a)*x^2*b^3*d*e+2*B*b^3*e^2*x^3+4*A*ln(b*
x+a)*x*a*b^2*e^2-12*B*ln(b*x+a)*x*a^2*b*e^2+8*B*ln(b*x+a)*x*a*b^2*d*e+4*B*a*b^2*e^2*x^2+2*A*a^2*b*e^2*ln(b*x+a
)+4*A*a*b^2*e^2*x-4*A*b^3*d*e*x-6*B*a^3*e^2*ln(b*x+a)+4*B*a^2*b*d*e*ln(b*x+a)-4*B*a^2*b*e^2*x+8*B*a*b^2*d*e*x-
2*B*b^3*d^2*x+3*A*a^2*b*e^2-2*A*a*b^2*d*e-A*b^3*d^2-5*B*a^3*e^2+6*B*a^2*b*d*e-B*a*b^2*d^2)*(b*x+a)/b^4/((b*x+a
)^2)^(3/2)

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maxima [A]  time = 0.60, size = 277, normalized size = 1.49 \begin {gather*} \frac {B e^{2} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, B a e^{2} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, B a^{2} e^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {6 \, B a^{2} e^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (2 \, B d e + A e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {B d^{2} + 2 \, A d e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {A d^{2}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, B a^{3} e^{2}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, {\left (2 \, B d e + A e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, {\left (2 \, B d e + A e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e^2*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*B*a*e^2*log(x + a/b)/b^4 + 2*B*a^2*e^2/(sqrt(b^2*x^2 + 2*a*b
*x + a^2)*b^4) - 6*B*a^2*e^2*x/(b^5*(x + a/b)^2) + (2*B*d*e + A*e^2)*log(x + a/b)/b^3 - (B*d^2 + 2*A*d*e)/(sqr
t(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 1/2*A*d^2/(b^3*(x + a/b)^2) - 11/2*B*a^3*e^2/(b^6*(x + a/b)^2) + 2*(2*B*d*e
+ A*e^2)*a*x/(b^4*(x + a/b)^2) + 3/2*(2*B*d*e + A*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(B*d^2 + 2*A*d*e)*a/(b^4*(x
+ a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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