\(\int \frac {1}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1600]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-1/2/b/(b*x+a)/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {621} \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

-1/2*1/(b*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 621

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[2*((a + b*x + c*x^2)^(p + 1)/((2*p + 1)*(b + 2*
c*x))), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {a+b x}{2 b \left ((a+b x)^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {b x +a}{2 b \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(20\)
default	\(-\frac {b x +a}{2 b \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(20\)
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{2 \left (b x +a \right )^{3} b}\)	\(22\)

[In]

int(1/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/2/(b^3*x^2 + 2*a*b^2*x + a^2*b)

Sympy [F]

\[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/2/(b^3*(x + a/b)^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 \, {\left (b x + a\right )}^{2} b \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

-1/2/((b*x + a)^2*b*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 9.74 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b\,{\left (a+b\,x\right )}^3} \]

[In]

int(1/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

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