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]

-((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])

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Rubi [A]  time = 0.0350412, antiderivative size = 103, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.0131405, size = 41, normalized size = 0.4 \[ -\frac{(a+b x) (-b x \log (a+b x)+a+b x \log (x))}{a^2 x \sqrt{(a+b x)^2}} \]

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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Maple [A]  time = 0.177, size = 40, normalized size = 0.4 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( b\ln \left ( x \right ) x-b\ln \left ( bx+a \right ) x+a \right ) }{{a}^{2}x}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.66927, size = 61, normalized size = 0.59 \begin{align*} \frac{b x \log \left (b x + a\right ) - b x \log \left (x\right ) - a}{a^{2} x} \end{align*}

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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Sympy [A]  time = 1.08252, size = 19, normalized size = 0.18 \begin{align*} - \frac{1}{a x} + \frac{b \left (- \log{\left (x \right )} + \log{\left (\frac{a}{b} + x \right )}\right )}{a^{2}} \end{align*}

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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Giac [A]  time = 1.3248, size = 50, normalized size = 0.49 \begin{align*}{\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 ) \end{align*}

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)