\(\int \frac {A+B x}{x (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [719]

Optimal result

Integrand size = 29, antiderivative size = 140 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {A}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A (a+b x) \log (x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {784, 78} \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {A b-a B}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A \log (x) (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {a \left (3 a A b-a^2 B+2 A b^2 x\right )+2 A b (a+b x)^2 \log (x)-2 A b (a+b x)^2 \log (a+b x)}{2 a^3 b (a+b x) \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {A b x}{a^{2}}+\frac {3 A b -B a}{2 b a}\right )}{\left (b x +a \right )^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, A \ln \left (-x \right )}{\left (b x +a \right ) a^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, A \ln \left (b x +a \right )}{\left (b x +a \right ) a^{3}}\)	\(97\)
default	\(\frac {\left (2 A \ln \left (x \right ) x^{2} b^{3}-2 A \ln \left (b x +a \right ) x^{2} b^{3}+4 A \ln \left (x \right ) x a \,b^{2}-4 A \ln \left (b x +a \right ) x a \,b^{2}+2 A \ln \left (x \right ) a^{2} b -2 A \ln \left (b x +a \right ) a^{2} b +2 A x a \,b^{2}+3 A \,a^{2} b -B \,a^{3}\right ) \left (b x +a \right )}{2 b \,a^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(117\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, A a b^{2} x - B a^{3} + 3 \, A a^{2} b - 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \left (b x + a\right ) + 2 \, {\left (A b^{3} x^{2} + 2 \, A a b^{2} x + A a^{2} b\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{3} x^{2} + 2 \, a^{4} b^{2} x + a^{5} b\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} - \frac {B}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {A}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.59 \[ \int \frac {A+B x}{x \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {A \log \left ({\left | b x + a \right |}\right )}{a^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {A \log \left ({\left | x \right |}\right )}{a^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, A a b^{2} x - B a^{3} + 3 \, A a^{2} b}{2 \, {\left (b x + a\right )}^{2} a^{3} b \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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