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

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{b^2}-\frac {a (a+b x) \log (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {654, 622, 31} \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{b^2}-\frac {a (a+b x) \log (a+b x)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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*} \text {integral}& = \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] (verified)

Time = 0.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.56 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\frac {b \left (-\sqrt {a^2} x (a+b x)+a x \sqrt {(a+b x)^2}\right )}{a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}}+2 a \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{b^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 33, normalized size of antiderivative = 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	\(\frac {x \sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right ) b}-\frac {\sqrt {\left (b x +a \right )^{2}}\, a \ln \left (b x +a \right )}{\left (b x +a \right ) b^{2}}\)	\(51\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.27 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {b x - a \log \left (b x + a\right )}{b^{2}} \]

[In]

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

[Out]

(b*x - a*log(b*x + a))/b^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (39) = 78\).

Time = 0.67 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.69 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} - \frac {a \left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{b \sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} + \frac {\sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{b^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {- a^{2} \sqrt {a^{2} + 2 a b x} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2}}{2 \sqrt {a^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a*(a/b + x)*log(a/b + x)/(b*sqrt(b**2*(a/b + x)**2)) + sqrt(a**2 + 2*a*b*x + b**2*x**2)/b**2, Ne(b
**2, 0)), ((-a**2*sqrt(a**2 + 2*a*b*x) + (a**2 + 2*a*b*x)**(3/2)/3)/(2*a**2*b**2), Ne(a*b, 0)), (x**2/(2*sqrt(
a**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.60 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=-\frac {a \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}}{b^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.50 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {x \mathrm {sgn}\left (b x + a\right )}{b} - \frac {a \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^2}-\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )}{{\left (b^2\right )}^{3/2}} \]

[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)