\(\int \frac {x (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [717]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {A b-2 a B}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (A b-a B)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 113, 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} \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {A b-2 a B}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (A b-a B)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[(x*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.65 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\frac {b x \left (2 a^5 B-a^3 A b^2 x+3 a^4 b B x-a A b^4 x^3+a^2 b^3 B x^3-\sqrt {a^2} \sqrt {(a+b x)^2} \left (2 a^3 B+a^2 b B x+A b^3 x^2-a b^2 x (A+B x)\right )\right )}{a^3 (a+b x) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}-4 B \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{2 b^3} \]

[In]

Integrate[(x*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

((b*x*(2*a^5*B - a^3*A*b^2*x + 3*a^4*b*B*x - a*A*b^4*x^3 + a^2*b^3*B*x^3 - Sqrt[a^2]*Sqrt[(a + b*x)^2]*(2*a^3*
B + a^2*b*B*x + A*b^3*x^2 - a*b^2*x*(A + B*x))))/(a^3*(a + b*x)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)
^2]))) - 4*B*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])])/(2*b^3)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {\left (A b -2 B a \right ) x}{b^{2}}-\frac {a \left (A b -3 B a \right )}{2 b^{3}}\right )}{\left (b x +a \right )^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, B \ln \left (b x +a \right )}{\left (b x +a \right ) b^{3}}\)	\(75\)
default	\(-\frac {\left (-2 B \ln \left (b x +a \right ) b^{2} x^{2}-4 B \ln \left (b x +a \right ) x a b +2 A \,b^{2} x -2 B \ln \left (b x +a \right ) a^{2}-4 B a b x +A b a -3 B \,a^{2}\right ) \left (b x +a \right )}{2 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(83\)

[In]

int(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x + 2 \, {\left (B b^{2} x^{2} + 2 \, B a b x + B a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x \left (A + B x\right )}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(x*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(x*(A + B*x)/((a + b*x)**2)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.79 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, B a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, B a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {A a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.62 \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (2 \, B a - A b\right )} x + \frac {3 \, B a^{2} - A a b}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(x*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

B*log(abs(b*x + a))/(b^3*sgn(b*x + a)) + 1/2*(2*(2*B*a - A*b)*x + (3*B*a^2 - A*a*b)/b)/((b*x + a)^2*b^2*sgn(b*
x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {x (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {x\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

int((x*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((x*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)