\(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx\) [146]

Optimal result

Integrand size = 24, antiderivative size = 71 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {a \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

-1/4*a*((b*x+a)^2)^(1/2)/x^4/(b*x+a)-1/3*b*((b*x+a)^2)^(1/2)/x^3/(b*x+a)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 71, 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} \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {a \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[In]

Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x^5,x]

[Out]

-1/4*(a*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x^4*(a + b*x)) - (b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(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^5} \, dx}{a b+b^2 x} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a b}{x^5}+\frac {b^2}{x^4}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {a \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\sqrt {(a+b x)^2} (3 a+4 b x)}{12 x^4 (a+b x)} \]

[In]

Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]/x^5,x]

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(3*a + 4*b*x))/(x^4*(a + b*x))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.28

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (4 b x +3 a \right )}{12 x^{4}}\)	\(20\)
risch	\(\frac {\left (-\frac {b x}{3}-\frac {a}{4}\right ) \sqrt {\left (b x +a \right )^{2}}}{x^{4} \left (b x +a \right )}\)	\(29\)
gosper	\(-\frac {\left (4 b x +3 a \right ) \sqrt {\left (b x +a \right )^{2}}}{12 x^{4} \left (b x +a \right )}\)	\(30\)

[In]

int(((b*x+a)^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/12*csgn(b*x+a)*(4*b*x+3*a)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {4 \, b x + 3 \, a}{12 \, x^{4}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

-1/12*(4*b*x + 3*a)/x^4

Sympy [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=\int \frac {\sqrt {\left (a + b x\right )^{2}}}{x^{5}}\, dx \]

[In]

integrate(((b*x+a)**2)**(1/2)/x**5,x)

[Out]

Integral(sqrt((a + b*x)**2)/x**5, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (45) = 90\).

Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.94 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}}{2 \, a^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}}{2 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}}{2 \, a^{4} x^{2}} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b}{12 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}{4 \, a^{2} x^{4}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4/a^4 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3/(a^3*x) - 1/2*(b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*b^2/(a^4*x^2) + 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b/(a^3*x^3) - 1/4*(b^2*x^2 + 2*a*b*x +
a^2)^(3/2)/(a^2*x^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {b^{4} \mathrm {sgn}\left (b x + a\right )}{12 \, a^{3}} - \frac {4 \, b x \mathrm {sgn}\left (b x + a\right ) + 3 \, a \mathrm {sgn}\left (b x + a\right )}{12 \, x^{4}} \]

[In]

integrate(((b*x+a)^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/12*b^4*sgn(b*x + a)/a^3 - 1/12*(4*b*x*sgn(b*x + a) + 3*a*sgn(b*x + a))/x^4

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\left (3\,a+4\,b\,x\right )\,\sqrt {{\left (a+b\,x\right )}^2}}{12\,x^4\,\left (a+b\,x\right )} \]

[In]

int(((a + b*x)^2)^(1/2)/x^5,x)

[Out]

-((3*a + 4*b*x)*((a + b*x)^2)^(1/2))/(12*x^4*(a + b*x))