\(\int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx\) [1260]

Optimal result

Integrand size = 29, antiderivative size = 27 \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\text {arctanh}\left (\sqrt {1+a^2+2 a b x+b^2 x^2}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {702, 214} \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\text {arctanh}\left (\sqrt {a^2+2 a b x+b^2 x^2+1}\right )}{b} \]

[In]

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

[Out]

-(ArcTanh[Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]]/b)

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 702

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(294\) vs. \(2(27)=54\).

Time = 0.40 (sec) , antiderivative size = 294, normalized size of antiderivative = 10.89 \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {-\frac {a^2}{\sqrt {1+a^2} b^2}-\frac {a^4}{\sqrt {1+a^2} b^2}-\frac {a^3 x}{\sqrt {1+a^2} b}+\frac {x^2}{\sqrt {1+a^2}}+\frac {a^2 \sqrt {1+a^2+2 a b x+b^2 x^2}}{b^2}}{x^2}\right )}{b}+\frac {\log \left (a+a^3+a^2 b x-\sqrt {1+a^2} b x-a \sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}+b x \sqrt {1+a^2+2 a b x+b^2 x^2}\right )}{2 b}-\frac {\log \left (a b+a^3 b+a^2 b^2 x+\sqrt {1+a^2} b^2 x-a \sqrt {1+a^2} b \sqrt {1+a^2+2 a b x+b^2 x^2}-b^2 x \sqrt {1+a^2+2 a b x+b^2 x^2}\right )}{2 b} \]

[In]

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

[Out]

-(ArcTanh[(-(a^2/(Sqrt[1 + a^2]*b^2)) - a^4/(Sqrt[1 + a^2]*b^2) - (a^3*x)/(Sqrt[1 + a^2]*b) + x^2/Sqrt[1 + a^2
] + (a^2*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/b^2)/x^2]/b) + Log[a + a^3 + a^2*b*x - Sqrt[1 + a^2]*b*x - a*Sqrt[
1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] + b*x*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]]/(2*b) - Log[a*b + a^3*b +
a^2*b^2*x + Sqrt[1 + a^2]*b^2*x - a*Sqrt[1 + a^2]*b*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] - b^2*x*Sqrt[1 + a^2 + 2
*a*b*x + b^2*x^2]]/(2*b)

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {b^{2} \left (x +\frac {a}{b}\right )^{2}+1}}\right )}{b}\)	\(24\)
pseudoelliptic	\(-\frac {\operatorname {arctanh}\left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{b}\)	\(26\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44 \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right ) - \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 1\right )}{b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=-\frac {\operatorname {arsinh}\left (\frac {1}{{\left | b x + a \right |}}\right )}{b} \]

[In]

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

[Out]

-arcsinh(1/abs(b*x + a))/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (25) = 50\).

Time = 0.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.30 \[ \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx=\frac {\log \left (\frac {{\left | -2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} b - 2 \, a {\left | b \right |} - 2 \, {\left | b \right |} \right |}}{{\left | -2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} b - 2 \, a {\left | b \right |} + 2 \, {\left | b \right |} \right |}}\right )}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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