\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx\) [663]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {(A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

-1/5*a*A*((b*x+a)^2)^(1/2)/x^5/(b*x+a)-1/4*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^4/(b*x+a)-1/3*b*B*((b*x+a)^2)^(1/2)/x
^3/(b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 114, 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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

-1/5*(a*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x^5*(a + b*x)) - ((A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*
(a + b*x)) - (b*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*x^3*(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^6} \, 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^6}+\frac {b (A b+a B)}{x^5}+\frac {b^2 B}{x^4}\right ) \, dx}{a b+b^2 x} \\ & = -\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {(A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.43 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {\sqrt {(a+b x)^2} (5 b x (3 A+4 B x)+3 a (4 A+5 B x))}{60 x^5 (a+b x)} \]

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(5*b*x*(3*A + 4*B*x) + 3*a*(4*A + 5*B*x)))/(x^5*(a + b*x))

Maple [C] (warning: unable to verify)

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

Time = 0.43 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.30

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (20 B b \,x^{2}+15 A b x +15 a B x +12 a A \right )}{60 x^{5}}\)	\(34\)
gosper	\(-\frac {\left (20 B b \,x^{2}+15 A b x +15 a B x +12 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{60 x^{5} \left (b x +a \right )}\)	\(44\)
risch	\(\frac {\left (-\frac {B b \,x^{2}}{3}+\left (-\frac {A b}{4}-\frac {B a}{4}\right ) x -\frac {a A}{5}\right ) \sqrt {\left (b x +a \right )^{2}}}{x^{5} \left (b x +a \right )}\)	\(44\)

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/60*csgn(b*x+a)*(20*B*b*x^2+15*A*b*x+15*B*a*x+12*A*a)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.24 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {20 \, B b x^{2} + 12 \, A a + 15 \, {\left (B a + A b\right )} x}{60 \, x^{5}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="fricas")

[Out]

-1/60*(20*B*b*x^2 + 12*A*a + 15*(B*a + A*b)*x)/x^5

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=\int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{x^{6}}\, dx \]

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**6,x)

[Out]

Integral((A + B*x)*sqrt((a + b*x)**2)/x**6, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (75) = 150\).

Time = 0.22 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.76 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4}}{2 \, a^{4}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{5}}{2 \, a^{5}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3}}{2 \, a^{3} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4}}{2 \, a^{4} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2}}{2 \, a^{4} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{2 \, a^{5} x^{2}} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b}{12 \, a^{3} x^{3}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{20 \, a^{4} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{4 \, a^{2} x^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{5 \, a^{2} x^{5}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="maxima")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^4/a^4 - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^5/a^5 + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2)*B*b^3/(a^3*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^4/(a^4*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2
)^(3/2)*B*b^2/(a^4*x^2) + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/(a^5*x^2) + 5/12*(b^2*x^2 + 2*a*b*x + a^2)
^(3/2)*B*b/(a^3*x^3) - 9/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^2/(a^4*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*B/(a^2*x^4) + 7/20*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b/(a^3*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A/(
a^2*x^5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {{\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} \mathrm {sgn}\left (b x + a\right )}{60 \, a^{4}} - \frac {20 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 15 \, B a x \mathrm {sgn}\left (b x + a\right ) + 15 \, A b x \mathrm {sgn}\left (b x + a\right ) + 12 \, A a \mathrm {sgn}\left (b x + a\right )}{60 \, x^{5}} \]

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/60*(5*B*a*b^4 - 3*A*b^5)*sgn(b*x + a)/a^4 - 1/60*(20*B*b*x^2*sgn(b*x + a) + 15*B*a*x*sgn(b*x + a) + 15*A*b*
x*sgn(b*x + a) + 12*A*a*sgn(b*x + a))/x^5

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^6} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (12\,A\,a+15\,A\,b\,x+15\,B\,a\,x+20\,B\,b\,x^2\right )}{60\,x^5\,\left (a+b\,x\right )} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^6,x)

[Out]

-(((a + b*x)^2)^(1/2)*(12*A*a + 15*A*b*x + 15*B*a*x + 20*B*b*x^2))/(60*x^5*(a + b*x))