\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx\) [142]

Optimal result

Integrand size = 24, antiderivative size = 62 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=\frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {a \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]

[Out]

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

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 = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=\frac {b x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {a \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

[In]

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

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 660

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]

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(398\) vs. \(2(62)=124\).

Time = 0.88 (sec) , antiderivative size = 398, normalized size of antiderivative = 6.42 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=\frac {-2 a \sqrt {a^2} b x-2 \sqrt {a^2} b^2 x^2+2 a b x \sqrt {(a+b x)^2}-2 a \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right ) \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )-2 \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x-a^2 \sqrt {(a+b x)^2}\right ) \log (x)+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+a \sqrt {a^2} b x \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )-a^2 \sqrt {(a+b x)^2} \log \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )+\left (a^2\right )^{3/2} \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )+a \sqrt {a^2} b x \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )-a^2 \sqrt {(a+b x)^2} \log \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )}{2 \left (a^2+a b x-\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.32

method	result	size
default	\(\operatorname {csgn}\left (b x +a \right ) \left (b x +a +a \ln \left (-b x \right )\right )\)	\(20\)
risch	\(\frac {b x \sqrt {\left (b x +a \right )^{2}}}{b x +a}+\frac {a \ln \left (x \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(41\)

[In]

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

[Out]

csgn(b*x+a)*(b*x+a+a*ln(-b*x))

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.13 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=b x + a \log \left (x\right ) \]

[In]

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

[Out]

b*x + a*log(x)

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt((a + b*x)**2)/x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (40) = 80\).

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} a \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} a \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.34 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=b x \mathrm {sgn}\left (b x + a\right ) + a \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) \]

[In]

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

[Out]

b*x*sgn(b*x + a) + a*log(abs(x))*sgn(b*x + a)

Mupad [B] (verification not implemented)

Time = 9.19 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.58 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x} \, dx=\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}-\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )\,\sqrt {a^2}+\frac {a\,b\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )}{\sqrt {b^2}} \]

[In]

int(((a + b*x)^2)^(1/2)/x,x)

[Out]

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