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

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 31

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

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 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]

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 = 0.01, size = 33, normalized size = 0.53 \begin {gather*} \frac {(a+b x) (b x-a \log (a+b x))}{b^2 \sqrt {(a+b x)^2}} \end {gather*}

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])

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 0.19, size = 144, normalized size = 2.32 \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{2 b^2}+\frac {a \left (\sqrt {b^2}+b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b^3}+\frac {a \left (\sqrt {b^2}-b\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b^3}-\frac {x}{2 \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*x/Sqrt[b^2] + Sqrt[a^2 + 2*a*b*x + b^2*x^2]/(2*b^2) + (a*(b + Sqrt[b^2])*Log[-a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(2*b^3) + (a*(-b + Sqrt[b^2])*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*
b^3)

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 17, normalized size = 0.27 \begin {gather*} \frac {b x - a \log \left (b x + a\right )}{b^{2}} \end {gather*}

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

________________________________________________________________________________________

giac [A]  time = 0.19, size = 31, normalized size = 0.50 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 33, normalized size = 0.53 \begin {gather*} -\frac {\left (b x +a \right ) \left (a \ln \left (b x +a \right )-b x \right )}{\sqrt {\left (b x +a \right )^{2}}\, b^{2}} \end {gather*}

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

________________________________________________________________________________________

maxima [A]  time = 1.31, size = 37, normalized size = 0.60 \begin {gather*} -\frac {a \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.28, size = 57, normalized size = 0.92 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^2}-\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )}{{\left (b^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 14, normalized size = 0.23 \begin {gather*} - \frac {a \log {\left (a + b x \right )}}{b^{2}} + \frac {x}{b} \end {gather*}

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

________________________________________________________________________________________