\(\int \frac {1}{x^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [196]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 b}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 46} \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b \log (x) (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-2*b)/(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(2*a^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a + b*x)/(a^
3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b*(a + b*x)*Log[x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b*(a + b*
x)*Log[a + b*x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{x^2 \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 {1}{a^3 b^3 x^2}-\frac {3}{a^4 b^2 x}+\frac {1}{a^2 b (a+b x)^3}+\frac {2}{a^3 b (a+b x)^2}+\frac {3}{a^4 b (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {2 b}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a+b x}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {-a \left (2 a^2+9 a b x+6 b^2 x^2\right )-6 b x (a+b x)^2 \log (x)+6 b x (a+b x)^2 \log (a+b x)}{2 a^4 x (a+b x) \sqrt {(a+b x)^2}} \]

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(-(a*(2*a^2 + 9*a*b*x + 6*b^2*x^2)) - 6*b*x*(a + b*x)^2*Log[x] + 6*b*x*(a + b*x)^2*Log[a + b*x])/(2*a^4*x*(a +
b*x)*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 2.31 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {3 b^{2} x^{2}}{a^{3}}-\frac {9 b x}{2 a^{2}}-\frac {1}{a}\right )}{\left (b x +a \right )^{3} x}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{4}}-\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \ln \left (x \right )}{\left (b x +a \right ) a^{4}}\)	\(101\)
default	\(-\frac {\left (6 b^{3} \ln \left (x \right ) x^{3}-6 b^{3} \ln \left (b x +a \right ) x^{3}+12 a \,b^{2} \ln \left (x \right ) x^{2}-12 \ln \left (b x +a \right ) x^{2} a \,b^{2}+6 a^{2} b \ln \left (x \right ) x -6 \ln \left (b x +a \right ) a^{2} b x +6 a \,b^{2} x^{2}+9 a^{2} b x +2 a^{3}\right ) \left (b x +a \right )}{2 x \,a^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(117\)

[In]

int(1/x^2/(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*(-3*b^2/a^3*x^2-9/2*b/a^2*x-1/a)/x+3*((b*x+a)^2)^(1/2)/(b*x+a)/a^4*b*ln(-b*x-a)-3*
((b*x+a)^2)^(1/2)/(b*x+a)*b*ln(x)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3} - 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (b x + a\right ) + 6 \, {\left (b^{3} x^{3} + 2 \, a b^{2} x^{2} + a^{2} b x\right )} \log \left (x\right )}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

[In]

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

[Out]

-1/2*(6*a*b^2*x^2 + 9*a^2*b*x + 2*a^3 - 6*(b^3*x^3 + 2*a*b^2*x^2 + a^2*b*x)*log(b*x + a) + 6*(b^3*x^3 + 2*a*b^
2*x^2 + a^2*b*x)*log(x))/(a^4*b^2*x^3 + 2*a^5*b*x^2 + a^6*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {3 \, b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}} - \frac {1}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x} - \frac {1}{2 \, a^{2} b {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {3 \, b \log \left ({\left | b x + a \right |}\right )}{a^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {6 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{2 \, {\left (b x + a\right )}^{2} a^{4} x \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

3*b*log(abs(b*x + a))/(a^4*sgn(b*x + a)) - 3*b*log(abs(x))/(a^4*sgn(b*x + a)) - 1/2*(6*a*b^2*x^2 + 9*a^2*b*x +
2*a^3)/((b*x + a)^2*a^4*x*sgn(b*x + a))

Mupad [F(-1)]

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

[In]

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

[Out]

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