3.2.84 \(\int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [184]

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]

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

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Rubi [A]
time = 0.01, 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 = {654, 622, 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 622

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 654

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*}

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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])

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Maple [A]
time = 0.41, size = 33, normalized size = 0.53

method	result	size
default	\(-\frac {\left (b x +a \right ) \left (a \ln \left (b x +a \right )-b x \right )}{\sqrt {\left (b x +a \right )^{2}}\, b^{2}}\)	\(33\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, x}{\left (b x +a \right ) b}-\frac {\sqrt {\left (b x +a \right )^{2}}\, a \ln \left (b x +a \right )}{\left (b x +a \right ) b^{2}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, 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

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Fricas [A]
time = 1.79, 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

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Sympy [A]
time = 0.03, 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

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Giac [A]
time = 1.63, 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

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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)

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