3.724 \(\int \frac{x^2 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{x^4 (A b-a B)}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0505231, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {769, 646, 37} \[ \frac{A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{x^4 (A b-a B)}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{\left (2 A b^2-2 a b B\right ) \int \frac{x^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx}{2 a b}\\ &=\frac{A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{\left (b^3 \left (2 A b^2-2 a b B\right ) \left (a b+b^2 x\right )\right ) \int \frac{x^3}{\left (a b+b^2 x\right )^5} \, dx}{2 a \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A x^3}{3 a^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac{(A b-a B) x^4}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0330062, size = 73, normalized size = 0.95 \[ \frac{-a^2 b (A+12 B x)-3 a^3 B-2 a b^2 x (2 A+9 B x)-6 b^3 x^2 (A+2 B x)}{12 b^4 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-3*a^3*B - 6*b^3*x^2*(A + 2*B*x) - 2*a*b^2*x*(2*A + 9*B*x) - a^2*b*(A + 12*B*x))/(12*b^4*(a + b*x)^3*Sqrt[(a
+ b*x)^2])

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Maple [A]  time = 0.007, size = 77, normalized size = 1. \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 12\,{b}^{3}B{x}^{3}+6\,A{b}^{3}{x}^{2}+18\,B{x}^{2}a{b}^{2}+4\,Aa{b}^{2}x+12\,B{a}^{2}bx+Ab{a}^{2}+3\,B{a}^{3} \right ) }{12\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/12*(b*x+a)*(12*B*b^3*x^3+6*A*b^3*x^2+18*B*a*b^2*x^2+4*A*a*b^2*x+12*B*a^2*b*x+A*a^2*b+3*B*a^3)/b^4/((b*x+a)^
2)^(5/2)

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Maxima [B]  time = 1.12303, size = 269, normalized size = 3.49 \begin{align*} -\frac{B x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{2 \, B a^{2}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{4}} - \frac{B a^{3} b}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} - \frac{A a^{2} b^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, B a^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} + \frac{2 \, A a b}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{A}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{B a}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} + \frac{B a^{3}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b^{3}{\left (x + \frac{a}{b}\right )}^{4}} \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)^(5/2),x, algorithm="maxima")

[Out]

-B*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*B*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) - 1/4*B*a^3*b/(
(b^2)^(9/2)*(x + a/b)^4) - 1/4*A*a^2*b^2/((b^2)^(9/2)*(x + a/b)^4) + 2/3*B*a^2/((b^2)^(7/2)*(x + a/b)^3) + 2/3
*A*a*b/((b^2)^(7/2)*(x + a/b)^3) - 1/2*A/((b^2)^(5/2)*(x + a/b)^2) - 1/2*B*a/((b^2)^(5/2)*b*(x + a/b)^2) + 1/2
*B*a^3/((b^2)^(5/2)*b^3*(x + a/b)^4)

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Fricas [A]  time = 1.52193, size = 219, normalized size = 2.84 \begin{align*} -\frac{12 \, B b^{3} x^{3} + 3 \, B a^{3} + A a^{2} b + 6 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 4 \,{\left (3 \, B a^{2} b + A a b^{2}\right )} x}{12 \,{\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(12*B*b^3*x^3 + 3*B*a^3 + A*a^2*b + 6*(3*B*a*b^2 + A*b^3)*x^2 + 4*(3*B*a^2*b + A*a*b^2)*x)/(b^8*x^4 + 4*
a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)

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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{5}{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)**(5/2),x)

[Out]

Integral(x**2*(A + B*x)/((a + b*x)**2)**(5/2), x)

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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)^(5/2),x, algorithm="giac")

[Out]

sage0*x