3.7.43 \(\int \frac {x^2 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=154 \[ -\frac {a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (2 A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-3 a B) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.11, antiderivative size = 154, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used =
{770, 77} \begin {gather*} -\frac {a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (2 A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-3 a B) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B 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[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(a*(2*A*b - 3*a*B))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a^2*(A*b - a*B))/(2*b^4*(a + b*x)*Sqrt[a^2 + 2*a*b*
x + b^2*x^2]) + (B*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*(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 {x^2 (A+B x)}{\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 {x^2 (A+B x)}{\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}{b^6}-\frac {a^2 (-A b+a B)}{b^6 (a+b x)^3}+\frac {a (-2 A b+3 a B)}{b^6 (a+b x)^2}+\frac {A b-3 a B}{b^6 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {a (2 A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) (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 = 0.06, size = 89, normalized size = 0.58 \begin {gather*} \frac {-5 a^3 B+a^2 b (3 A-4 B x)+4 a b^2 x (A+B x)+2 (a+b x)^2 (A b-3 a B) \log (a+b x)+2 b^3 B x^3}{2 b^4 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(-5*a^3*B + 2*b^3*B*x^3 + a^2*b*(3*A - 4*B*x) + 4*a*b^2*x*(A + B*x) + 2*(A*b - 3*a*B)*(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 = 2.00, size = 2422, normalized size = 15.73 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
((-12*a^3*A*x)/(b*Sqrt[b^2]) - (24*a^2*A*x^2)/Sqrt[b^2] - (16*a*A*b*x^3)/Sqrt[b^2] + (4*a^3*A*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/b^3 + (8*a^2*A*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (16*a*A*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/b + (8*a^2*A*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 16*a*A*x^3*ArcTanh[(-(Sqrt
[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 8*A*b*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/a] - (8*a*A*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*b*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])/((-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^4*A*Sqrt[b^2])/b^4 + (12*a^3*A*Sqrt[b^2]*x)/b^3 + (16*a^4*Sqrt[b^2]*B*x)/b^4 + (12*a^2*A*
(b^2)^(3/2)*x^2)/b^4 + (32*a^3*Sqrt[b^2]*B*x^2)/b^3 + (8*a^2*(b^2)^(3/2)*B*x^3)/b^4 - (20*a*Sqrt[b^2]*B*x^4)/b
- 8*Sqrt[b^2]*B*x^5 - (4*a^4*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (12*a^2*A*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/b^2 - (12*a^3*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (20*a^2*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (1
2*a*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + 8*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (24*a^3*B*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*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/a])/b - 24*a*B*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*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])/b^3 + (
24*a*(b^2)^(3/2)*B*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])/b^4 - (4*a^2*A*(b^2)^(3/2)*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*A*Sqrt[b^
2]*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b - 4*A*Sqrt[b^2]*x^4*Log[-a - Sqrt[b^2]*x + Sqr
t[a^2 + 2*a*b*x + b^2*x^2]] + (4*a*A*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*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^2*A*(b^2)^(3/2)*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*A*Sqrt[b^2]*x^3*
Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b - 4*A*Sqrt[b^2]*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2
*a*b*x + b^2*x^2]] + (4*a*A*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*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]])/((-a - S
qrt[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)/(b^3*Sqrt[b^2]) - (16*a^4*B*x)/(b^2)^(3/2) - (16*a^3*B*x^2)/(b*Sqrt[b^2]) + (16*a^3*B*x*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/b^3 + (12*a^3*B*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*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*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*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*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*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*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*x^4*Log[a - Sqr
t[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (12*a^2*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqr
t[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^2 - (12*a*B*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 + 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)
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fricas [A] time = 0.41, size = 134, normalized size = 0.87 \begin {gather*} \frac {2 \, B b^{3} x^{3} + 4 \, B a b^{2} x^{2} - 5 \, B a^{3} + 3 \, A a^{2} b - 4 \, {\left (B a^{2} b - A a b^{2}\right )} x - 2 \, {\left (3 \, B a^{3} - A a^{2} b + {\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (3 \, B a^{2} b - A a b^{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(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
[Out]
1/2*(2*B*b^3*x^3 + 4*B*a*b^2*x^2 - 5*B*a^3 + 3*A*a^2*b - 4*(B*a^2*b - A*a*b^2)*x - 2*(3*B*a^3 - A*a^2*b + (3*B
*a*b^2 - A*b^3)*x^2 + 2*(3*B*a^2*b - A*a*b^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(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")
[Out]
sage0*x
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maple [A] time = 0.09, size = 153, normalized size = 0.99 \begin {gather*} \frac {\left (2 A \,b^{3} x^{2} \ln \left (b x +a \right )-6 B a \,b^{2} x^{2} \ln \left (b x +a \right )+2 B \,b^{3} x^{3}+4 A a \,b^{2} x \ln \left (b x +a \right )-12 B \,a^{2} b x \ln \left (b x +a \right )+4 B a \,b^{2} x^{2}+2 A \,a^{2} b \ln \left (b x +a \right )+4 A a \,b^{2} x -6 B \,a^{3} \ln \left (b x +a \right )-4 B \,a^{2} b x +3 A \,a^{2} b -5 B \,a^{3}\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(x^2*(B*x+A)/(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^3-6*B*ln(b*x+a)*x^2*a*b^2+2*B*b^3*x^3+4*A*ln(b*x+a)*x*a*b^2-12*B*ln(b*x+a)*x*a^2*b+4*
B*a*b^2*x^2+2*A*a^2*b*ln(b*x+a)+4*A*a*b^2*x-6*B*a^3*ln(b*x+a)-4*B*a^2*b*x+3*A*a^2*b-5*B*a^3)*(b*x+a)/b^4/((b*x
+a)^2)^(3/2)
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maxima [A] time = 0.53, size = 154, normalized size = 1.00 \begin {gather*} \frac {B x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {3 \, B a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {A \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, B a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {6 \, B a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, A a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, B a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, A a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
[Out]
B*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 3*B*a*log(x + a/b)/b^4 + A*log(x + a/b)/b^3 + 2*B*a^2/(sqrt(b^2*x^
2 + 2*a*b*x + a^2)*b^4) - 6*B*a^2*x/(b^5*(x + a/b)^2) + 2*A*a*x/(b^4*(x + a/b)^2) - 11/2*B*a^3/(b^6*(x + a/b)^
2) + 3/2*A*a^2/(b^5*(x + a/b)^2)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (A+B\,x\right )}{{\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((x^2*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
[Out]
int((x^2*(A + B*x))/(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 {x^{2} \left (A + B x\right )}{\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(x**2*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
[Out]
Integral(x**2*(A + B*x)/((a + b*x)**2)**(3/2), x)
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