\(\int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [187]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) \log (x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 103, 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} \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b \log (x) (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

-((a + b*x)/(a*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (b*(a + b*x)*Log[x])/(a^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) +
(b*(a + b*x)*Log[a + b*x])/(a^2*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 (a b+b^2 x\right ) \int \frac {1}{x^2 \left (a b+b^2 x\right )} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (a b+b^2 x\right ) \int \left (\frac {1}{a b x^2}-\frac {1}{a^2 x}+\frac {b}{a^2 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) \log (x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a^2-\sqrt {a^2} \sqrt {(a+b x)^2}+2 a b x \log (x)+\left (-a+\sqrt {a^2}\right ) b x \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-a b x \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )-\sqrt {a^2} b x \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{2 \left (a^2\right )^{3/2} x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.39

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (b \ln \left (x \right ) x -b \ln \left (b x +a \right ) x +a \right )}{\sqrt {\left (b x +a \right )^{2}}\, a^{2} x}\)	\(40\)
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right ) a x}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{2}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, b \ln \left (x \right )}{\left (b x +a \right ) a^{2}}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.25 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{2} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{a^{2} x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.36 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx={\left (\frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x}\right )} \mathrm {sgn}\left (b x + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.59 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}\right )}{{\left (a^2\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{a^2\,x} \]

[In]

int(1/(x^2*((a + b*x)^2)^(1/2)),x)

[Out]

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