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

Optimal. Leaf size=154 \[ -\frac{a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a (2 A b-3 a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-3 a B) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.105934, antiderivative size = 154, normalized size of antiderivative = 1., 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} \[ -\frac{a^2 (A b-a B)}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a (2 A b-3 a B)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (A b-3 a B) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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]

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]))))

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*}

Mathematica [A]  time = 0.0418458, size = 89, normalized size = 0.58 \[ \frac{a^2 b (3 A-4 B x)-5 a^3 B+4 a b^2 x (A+B x)+2 (a+b x)^2 (A b-3 a B) \log (a+b x)+2 b^3 B x^3}{2 b^4 (a+b x) \sqrt{(a+b x)^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 153, normalized size = 1. \begin{align*}{\frac{ \left ( 2\,A\ln \left ( bx+a \right ){x}^{2}{b}^{3}-6\,B\ln \left ( bx+a \right ){x}^{2}a{b}^{2}+2\,{b}^{3}B{x}^{3}+4\,A\ln \left ( bx+a \right ) xa{b}^{2}-12\,B\ln \left ( bx+a \right ) x{a}^{2}b+4\,B{x}^{2}a{b}^{2}+2\,A\ln \left ( bx+a \right ){a}^{2}b+4\,Aa{b}^{2}x-6\,B\ln \left ( bx+a \right ){a}^{3}-4\,B{a}^{2}bx+3\,Ab{a}^{2}-5\,B{a}^{3} \right ) \left ( bx+a \right ) }{2\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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

________________________________________________________________________________________

Maxima [A]  time = 0.992126, size = 266, normalized size = 1.73 \begin{align*} \frac{B x^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac{A \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} - \frac{3 \, B a \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} - \frac{9 \, B a^{3} b}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{3 \, A a^{2} b^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{6 \, B a^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, A a b x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, B a^{2}}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac{B a^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

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) + A*log(x + a/b)/(b^2)^(3/2) - 3*B*a*log(x + a/b)/((b^2)^(3/2)*b) -
9/2*B*a^3*b/((b^2)^(7/2)*(x + a/b)^2) + 3/2*A*a^2*b^2/((b^2)^(7/2)*(x + a/b)^2) - 6*B*a^2*x/((b^2)^(5/2)*(x +
a/b)^2) + 2*A*a*b*x/((b^2)^(5/2)*(x + a/b)^2) + 2*B*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - B*a^3/((b^2)^(3/
2)*b^3*(x + a/b)^2)

________________________________________________________________________________________

Fricas [A]  time = 1.55436, size = 278, normalized size = 1.81 \begin{align*} \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{align*}

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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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