\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx\) [660]

Optimal result

Integrand size = 29, antiderivative size = 108 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {(A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b B \sqrt {a^2+2 a b x+b^2 x^2} \log (x)}{a+b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {784, 77} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{x (a+b x)}-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}+\frac {b B \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \]

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^3,x]

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right ) (A+B x)}{x^3} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a A b}{x^3}+\frac {b (A b+a B)}{x^2}+\frac {b^2 B}{x}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {(A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b B \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. \(225\) vs. \(2(108)=216\).

Time = 0.34 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.08 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=-\frac {-a \sqrt {a^2} A-2 \sqrt {a^2} A b x-2 a \sqrt {a^2} B x+a A \sqrt {(a+b x)^2}+A b x \sqrt {(a+b x)^2}+2 a B x \sqrt {(a+b x)^2}+4 a b B x^2 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )+4 \sqrt {a^2} b B x^2 \log (x)-2 \sqrt {a^2} b B x^2 \log \left (a \left (\sqrt {a^2}-b x-\sqrt {(a+b x)^2}\right )\right )-2 \sqrt {a^2} b B x^2 \log \left (a \left (\sqrt {a^2}+b x-\sqrt {(a+b x)^2}\right )\right )}{4 a x^2} \]

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^3,x]

[Out]

-1/4*(-(a*Sqrt[a^2]*A) - 2*Sqrt[a^2]*A*b*x - 2*a*Sqrt[a^2]*B*x + a*A*Sqrt[(a + b*x)^2] + A*b*x*Sqrt[(a + b*x)^
2] + 2*a*B*x*Sqrt[(a + b*x)^2] + 4*a*b*B*x^2*ArcTanh[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])] + 4*Sqrt[a^2]*b*B*
x^2*Log[x] - 2*Sqrt[a^2]*b*B*x^2*Log[a*(Sqrt[a^2] - b*x - Sqrt[(a + b*x)^2])] - 2*Sqrt[a^2]*b*B*x^2*Log[a*(Sqr
t[a^2] + b*x - Sqrt[(a + b*x)^2])])/(a*x^2)

Maple [C] (warning: unable to verify)

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

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.35

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (-2 B \ln \left (-b x \right ) b \,x^{2}+2 A b x +2 a B x +a A \right )}{2 x^{2}}\)	\(38\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-A b -B a \right ) x -\frac {a A}{2}\right )}{\left (b x +a \right ) x^{2}}+\frac {b B \ln \left (x \right ) \sqrt {\left (b x +a \right )^{2}}}{b x +a}\)	\(59\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.27 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=\frac {2 \, B b x^{2} \log \left (x\right ) - A a - 2 \, {\left (B a + A b\right )} x}{2 \, x^{2}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{x^{3}}\, dx \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**3,x)

[Out]

Integral((A + B*x)*sqrt((a + b*x)**2)/x**3, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (73) = 146\).

Time = 0.23 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.59 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=\left (-1\right )^{2 \, b^{2} x + 2 \, a b} B b \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{2 \, a^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{x} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{2 \, a x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{2 \, a^{2} x^{2}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

(-1)^(2*b^2*x + 2*a*b)*B*b*log(2*b^2*x + 2*a*b) - (-1)^(2*a*b*x + 2*a^2)*B*b*log(2*a*b*x/abs(x) + 2*a^2/abs(x)
) + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^2/a^2 - sqrt(b^2*x^2 + 2*a*b*x + a^2)*B/x + 1/2*sqrt(b^2*x^2 + 2*a*b
*x + a^2)*A*b/(a*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/(a^2*x^2)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.46 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=B b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {A a \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (B a \mathrm {sgn}\left (b x + a\right ) + A b \mathrm {sgn}\left (b x + a\right )\right )} x}{2 \, x^{2}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

B*b*log(abs(x))*sgn(b*x + a) - 1/2*(A*a*sgn(b*x + a) + 2*(B*a*sgn(b*x + a) + A*b*sgn(b*x + a))*x)/x^2

Mupad [B] (verification not implemented)

Time = 10.25 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^3} \, dx=B\,\ln \left (a\,b+\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {b^2}+b^2\,x\right )\,\sqrt {b^2}-\frac {B\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}-\frac {A\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+2\,b\,x\right )}{2\,x^2\,\left (a+b\,x\right )}-\frac {B\,a\,b\,\ln \left (a\,b+\frac {a^2}{x}+\frac {\sqrt {a^2}\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x}\right )}{\sqrt {a^2}} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^3,x)

[Out]

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