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

Optimal. Leaf size=27 \[ -\frac {\tanh ^{-1}\left (\sqrt {a^2+2 a b x+b^2 x^2+1}\right )}{b} \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {688, 208} \begin {gather*} -\frac {\tanh ^{-1}\left (\sqrt {a^2+2 a b x+b^2 x^2+1}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[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 208

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

Rule 688

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*} \int \frac {1}{(a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2}} \, dx &=\left (4 b^2\right ) \operatorname {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*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.70 \begin {gather*} -\frac {\tanh ^{-1}\left (\sqrt {(a+b x)^2+1}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[1 + (a + b*x)^2]]/b)

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IntegrateAlgebraic [B]  time = 0.49, size = 112, normalized size = 4.15 \begin {gather*} \frac {\left (\sqrt {b^2}+b\right ) \tanh ^{-1}\left (-\sqrt {a^2+2 a b x+b^2 x^2+1}+a+\sqrt {b^2} x\right )}{b \sqrt {b^2}}+\frac {\left (b-\sqrt {b^2}\right ) \tanh ^{-1}\left (\sqrt {a^2+2 a b x+b^2 x^2+1}+a-\sqrt {b^2} x\right )}{b \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 66, normalized size = 2.44 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.33, size = 89, normalized size = 3.30 \begin {gather*} \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 |}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 24, normalized size = 0.89 \begin {gather*} -\frac {\arctanh \left (\frac {1}{\sqrt {\left (x +\frac {a}{b}\right )^{2} b^{2}+1}}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.93, size = 14, normalized size = 0.52 \begin {gather*} -\frac {\operatorname {arsinh}\left (\frac {1}{{\left | b x + a \right |}}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\left (a+b\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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