\(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx\) [662]

Optimal result

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

[Out]

-1/4*a*A*((b*x+a)^2)^(1/2)/x^4/(b*x+a)-1/3*(A*b+B*a)*((b*x+a)^2)^(1/2)/x^3/(b*x+a)-1/2*b*B*((b*x+a)^2)^(1/2)/x
^2/(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} \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{3 x^3 (a+b x)}-\frac {a A \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {b B \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^5,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.41 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\sqrt {(a+b x)^2} \left (3 a A+4 A b x+4 a B x+6 b B x^2\right )}{12 x^4 (a+b x)} \]

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^5,x]

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(3*a*A + 4*A*b*x + 4*a*B*x + 6*b*B*x^2))/(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 = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.30

method	result	size
default	\(-\frac {\operatorname {csgn}\left (b x +a \right ) \left (6 B b \,x^{2}+4 A b x +4 a B x +3 a A \right )}{12 x^{4}}\)	\(34\)
gosper	\(-\frac {\left (6 B b \,x^{2}+4 A b x +4 a B x +3 a A \right ) \sqrt {\left (b x +a \right )^{2}}}{12 x^{4} \left (b x +a \right )}\)	\(44\)
risch	\(\frac {\left (-\frac {B b \,x^{2}}{2}+\left (-\frac {A b}{3}-\frac {B a}{3}\right ) x -\frac {a A}{4}\right ) \sqrt {\left (b x +a \right )^{2}}}{x^{4} \left (b x +a \right )}\)	\(44\)

[In]

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

[Out]

-1/12*csgn(b*x+a)*(6*B*b*x^2+4*A*b*x+4*B*a*x+3*A*a)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.24 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {6 \, B b x^{2} + 3 \, A a + 4 \, {\left (B a + A b\right )} x}{12 \, x^{4}} \]

[In]

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

[Out]

-1/12*(6*B*b*x^2 + 3*A*a + 4*(B*a + A*b)*x)/x^4

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*x)*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. 255 vs. \(2 (75) = 150\).

Time = 0.22 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{3}}{2 \, a^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4}}{2 \, a^{4}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2}}{2 \, a^{2} x} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{2 \, a^{3} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b}{2 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{2}}{2 \, a^{4} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{3 \, a^{2} x^{3}} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b}{12 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A}{4 \, a^{2} x^{4}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.68 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=\frac {{\left (2 \, B a b^{3} - A b^{4}\right )} \mathrm {sgn}\left (b x + a\right )}{12 \, a^{3}} - \frac {6 \, B b x^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, B a x \mathrm {sgn}\left (b x + a\right ) + 4 \, A b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a \mathrm {sgn}\left (b x + a\right )}{12 \, x^{4}} \]

[In]

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

[Out]

1/12*(2*B*a*b^3 - A*b^4)*sgn(b*x + a)/a^3 - 1/12*(6*B*b*x^2*sgn(b*x + a) + 4*B*a*x*sgn(b*x + a) + 4*A*b*x*sgn(
b*x + a) + 3*A*a*sgn(b*x + a))/x^4

Mupad [B] (verification not implemented)

Time = 10.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.38 \[ \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{x^5} \, dx=-\frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (3\,A\,a+4\,A\,b\,x+4\,B\,a\,x+6\,B\,b\,x^2\right )}{12\,x^4\,\left (a+b\,x\right )} \]

[In]

int((((a + b*x)^2)^(1/2)*(A + B*x))/x^5,x)

[Out]

-(((a + b*x)^2)^(1/2)*(3*A*a + 4*A*b*x + 4*B*a*x + 6*B*b*x^2))/(12*x^4*(a + b*x))