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

Optimal. Leaf size=62 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{b^2}-\frac{a (a+b x) \log (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0172623, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {640, 608, 31} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2}}{b^2}-\frac{a (a+b x) \log (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 640

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

Rule 608

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0103491, size = 33, normalized size = 0.53 \[ \frac{(a+b x) (b x-a \log (a+b x))}{b^2 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.22, size = 33, normalized size = 0.5 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( a\ln \left ( bx+a \right ) -bx \right ) }{{b}^{2}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.24071, size = 57, normalized size = 0.92 \begin{align*} -\frac{a \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right )}{b} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.63242, size = 38, normalized size = 0.61 \begin{align*} \frac{b x - a \log \left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x - a*log(b*x + a))/b^2

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Sympy [A]  time = 0.482535, size = 14, normalized size = 0.23 \begin{align*} - \frac{a \log{\left (a + b x \right )}}{b^{2}} + \frac{x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(a + b*x)/b**2 + x/b

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Giac [A]  time = 1.34215, size = 42, normalized size = 0.68 \begin{align*} \frac{x \mathrm{sgn}\left (b x + a\right )}{b} - \frac{a \log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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