3.2426 \(\int \frac {1}{x \sqrt {-a^2-2 a b x-b^2 x^2}} \, dx\)

Optimal. Leaf size=74 \[ \frac {\log (x) (a+b x)}{a \sqrt {-a^2-2 a b x-b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a \sqrt {-a^2-2 a b x-b^2 x^2}} \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {646, 36, 29, 31} \[ \frac {\log (x) (a+b x)}{a \sqrt {-a^2-2 a b x-b^2 x^2}}-\frac {(a+b x) \log (a+b x)}{a \sqrt {-a^2-2 a b x-b^2 x^2}} \]

Antiderivative was successfully verified.

[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 646

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.45 \[ \frac {(a+b x) (\log (x)-\log (a+b x))}{a \sqrt {-(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.86, size = 66, normalized size = 0.89 \[ -\sqrt {-\frac {1}{a^{2}}} \log \left (\frac {i \, a^{2} \sqrt {-\frac {1}{a^{2}}} + 2 \, b x + a}{2 \, b}\right ) + \sqrt {-\frac {1}{a^{2}}} \log \left (\frac {-i \, a^{2} \sqrt {-\frac {1}{a^{2}}} + 2 \, b x + a}{2 \, b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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maple [A]  time = 0.06, size = 33, normalized size = 0.45 \[ -\frac {\left (b x +a \right ) \left (-\ln \relax (x )+\ln \left (b x +a \right )\right )}{\sqrt {-\left (b x +a \right )^{2}}\, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.92, size = 38, normalized size = 0.51 \[ -\frac {i \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.17, size = 55, normalized size = 0.74 \[ -\frac {\ln \left (\frac {\sqrt {-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2}}{x}-\frac {a^2}{x}-a\,b\right )}{\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(-(a + b*x)^2)^(1/2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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