\(\int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [185]

Optimal result

Integrand size = 20, antiderivative size = 35 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {622, 31} \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

Rule 31

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

Rule 622

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\left (b x +a \right ) \ln \left (b x +a \right )}{b \sqrt {\left (b x +a \right )^{2}}}\)	\(25\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \ln \left (b x +a \right )}{\left (b x +a \right ) b}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log \left (b x + a\right )}{b} \]

[In]

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

[Out]

log(b*x + a)/b

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} \]

[In]

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

[Out]

(a/b + x)*log(a/b + x)/sqrt(b**2*(a/b + x)**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log \left (x + \frac {a}{b}\right )}{b} \]

[In]

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

[Out]

log(x + a/b)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.49 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b} \]

[In]

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

[Out]

log(abs(b*x + a))*sgn(b*x + a)/b

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\ln \left (a+b\,x+\sqrt {{\left (a+b\,x\right )}^2}\right )}{b} \]

[In]

int(1/((a + b*x)^2)^(1/2),x)

[Out]

log(a + b*x + ((a + b*x)^2)^(1/2))/b