\(\int \frac {x (A+B x)}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [725]

Optimal result

Integrand size = 27, antiderivative size = 121 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-2 a B}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {784, 78} \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {A b-2 a B}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (A b-a B)}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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 78

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 784

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*} \text {integral}& = \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 [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(121)=242\).

Time = 0.84 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.64 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x^2 \left (-3 a^2 b^5 B x^6-3 a^3 \sqrt {a^2} b^2 x^2 \sqrt {(a+b x)^2} (A+B x)+a^6 b x (4 A+B x)+a^4 \sqrt {a^2} b x \sqrt {(a+b x)^2} (2 A+3 B x)+a^7 (6 A+4 B x)-3 a b^4 x^4 \left (-A b^2 x^2+\sqrt {a^2} A \sqrt {(a+b x)^2}+\sqrt {a^2} B x \sqrt {(a+b x)^2}\right )-a^5 \left (-A b^2 x^2+6 \sqrt {a^2} A \sqrt {(a+b x)^2}+4 \sqrt {a^2} B x \sqrt {(a+b x)^2}\right )+3 \sqrt {a^2} b^3 x^3 \sqrt {(a+b x)^2} \left (A b^2 x^2+a^2 (A+B x)\right )\right )}{12 a^7 (a+b x)^3 \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (6 b^{2} B \,x^{2}+4 A \,b^{2} x +4 B a b x +A b a +B \,a^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(52\)
default	\(-\frac {\left (b x +a \right ) \left (6 b^{2} B \,x^{2}+4 A \,b^{2} x +4 B a b x +A b a +B \,a^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(52\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,x^{2}}{2 b}-\frac {\left (A b +B a \right ) x}{3 b^{2}}-\frac {a \left (A b +B a \right )}{12 b^{3}}\right )}{\left (b x +a \right )^{5}}\)	\(54\)

[In]

int(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} x^{2} + B a^{2} + A a b + 4 \, {\left (B a b + A b^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {x \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {A}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {B}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, B a}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {B a^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {A a}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} \]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.43 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, B b^{2} x^{2} + 4 \, B a b x + 4 \, A b^{2} x + B a^{2} + A a b}{12 \, {\left (b x + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.51 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (B\,a^2+4\,B\,a\,b\,x+A\,a\,b+6\,B\,b^2\,x^2+4\,A\,b^2\,x\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \]

[In]

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

[Out]

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