\(\int \frac {1}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1610]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[Out]

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

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} \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \]

[In]

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

[Out]

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

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}{4 b (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {a+b x}{4 b \left ((a+b x)^2\right )^{5/2}} \]

[In]

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

[Out]

-1/4*(a + b*x)/(b*((a + b*x)^2)^(5/2))

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59

method	result	size
gosper	\(-\frac {b x +a}{4 b \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(20\)
default	\(-\frac {b x +a}{4 b \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(20\)
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{4 \left (b x +a \right )^{5} b}\)	\(22\)

[In]

int(1/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*(b*x+a)/b/((b*x+a)^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 \, {\left (b^{5} x^{4} + 4 \, a b^{4} x^{3} + 6 \, a^{2} b^{3} x^{2} + 4 \, a^{3} b^{2} x + a^{4} b\right )}} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-1/4/(b^5*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {1}{4 \, {\left (b x + a\right )}^{4} b \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate(1/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

-1/4/((b*x + a)^4*b*sgn(b*x + a))

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b\,{\left (a+b\,x\right )}^5} \]

[In]

int(1/(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)/(4*b*(a + b*x)^5)