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

Optimal. Leaf size=121 \[ -\frac{A b-2 a B}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0724546, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {770, 77} \[ -\frac{A b-2 a B}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A*b - a*B))/(4*b^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (A*b - 2*a*B)/(3*b^3*(a + b*x)^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]) - B/(2*b^3*(a + b*x)*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 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{x (A+B x)}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac{a (-A b+a B)}{b^7 (a+b x)^5}+\frac{A b-2 a B}{b^7 (a+b x)^4}+\frac{B}{b^7 (a+b x)^3}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-2 a B}{3 b^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0240356, size = 56, normalized size = 0.46 \[ \frac{a^2 (-B)-a b (A+4 B x)-2 b^2 x (2 A+3 B x)}{12 b^3 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 52, normalized size = 0.4 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 6\,{b}^{2}B{x}^{2}+4\,A{b}^{2}x+4\,abBx+Aab+B{a}^{2} \right ) }{12\,{b}^{3}} \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*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.20019, size = 142, normalized size = 1.17 \begin{align*} -\frac{A}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{B a^{2} b^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, B a b}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{B}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{A a}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57307, size = 166, normalized size = 1.37 \begin{align*} -\frac{6 \, B b^{2} x^{2} + B a^{2} + A a b + 4 \,{\left (B a b + A b^{2}\right )} x}{12 \,{\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(x*(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*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x