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

Optimal. Leaf size=103 \[ -\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b \log (x) (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 44} \[ -\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b \log (x) (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{x^2 \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 \left (\frac {1}{a b x^2}-\frac {1}{a^2 x}+\frac {b}{a^2 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {a+b x}{a x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) \log (x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (a+b x) \log (a+b x)}{a^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.40 \[ -\frac {(a+b x) (-b x \log (a+b x)+a+b x \log (x))}{a^2 x \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 26, normalized size = 0.25 \[ \frac {b x \log \left (b x + a\right ) - b x \log \relax (x) - a}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 37, normalized size = 0.36 \[ {\left (\frac {b \log \left ({\left | b x + a \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} - \frac {1}{a x}\right )} \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 41, normalized size = 0.40 \[ \frac {\left (b x +a \right ) \left (-b x \ln \relax (x )+b x \ln \left (b x +a \right )-a \right )}{\sqrt {\left (b x +a \right )^{2}}\, a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 65, normalized size = 0.63 \[ \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.25, size = 68, normalized size = 0.66 \[ \frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a}{\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}\right )}{{\left (a^2\right )}^{3/2}}-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{a^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*((a + b*x)^2)^(1/2)),x)

[Out]

(a*b*atanh((a^2 + a*b*x)/((a^2)^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))))/(a^2)^(3/2) - (a^2 + b^2*x^2 + 2*a*b*
x)^(1/2)/(a^2*x)

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sympy [A]  time = 0.21, size = 19, normalized size = 0.18 \[ - \frac {1}{a x} + \frac {b \left (- \log {\relax (x )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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