\(\int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [2423]

Optimal result

Integrand size = 24, antiderivative size = 68 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \log (x)}{a \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 36, 29, 31} \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log (x) (a+b x)}{a \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 \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 \frac {1}{x} \, dx}{a b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (b \left (a b+b^2 x\right )\right ) \int \frac {1}{a b+b^2 x} \, dx}{a \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {(a+b x) \log (x)}{a \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.76 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {-2 a \log (x)+\left (a-\sqrt {a^2}\right ) \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+a \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )+\sqrt {a^2} \log \left (a \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{2 a \sqrt {a^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.46

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (\ln \left (b x +a \right )-\ln \left (x \right )\right )}{\sqrt {\left (b x +a \right )^{2}}\, a}\)	\(31\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (-x \right )}{\left (b x +a \right ) a}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) a}\)	\(53\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.24 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\log \left (b x + a\right ) - \log \left (x\right )}{a} \]

[In]

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

[Out]

-(log(b*x + a) - log(x))/a

Sympy [F]

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

[In]

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

[Out]

Integral(1/(x*sqrt((a + b*x)**2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-{\left (\frac {\log \left ({\left | b x + a \right |}\right )}{a} - \frac {\log \left ({\left | x \right |}\right )}{a}\right )} \mathrm {sgn}\left (b x + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {1}{x \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )}{\sqrt {a^2}} \]

[In]

int(1/(x*((a + b*x)^2)^(1/2)),x)

[Out]

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