\(\int \frac {(a+\frac {b}{x})^3}{x^3} \, dx\) [1579]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {(b+a x)^4}{5 b x^5}+\frac {a (b+a x)^4}{20 b^2 x^4} \]

[Out]

-1/5*(a*x+b)^4/b/x^5+1/20*a*(a*x+b)^4/b^2/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {269, 47, 37} \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=\frac {a (a x+b)^4}{20 b^2 x^4}-\frac {(a x+b)^4}{5 b x^5} \]

[In]

Int[(a + b/x)^3/x^3,x]

[Out]

-1/5*(b + a*x)^4/(b*x^5) + (a*(b + a*x)^4)/(20*b^2*x^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a x)^3}{x^6} \, dx \\ & = -\frac {(b+a x)^4}{5 b x^5}-\frac {a \int \frac {(b+a x)^3}{x^5} \, dx}{5 b} \\ & = -\frac {(b+a x)^4}{5 b x^5}+\frac {a (b+a x)^4}{20 b^2 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {b^3}{5 x^5}-\frac {3 a b^2}{4 x^4}-\frac {a^2 b}{x^3}-\frac {a^3}{2 x^2} \]

[In]

Integrate[(a + b/x)^3/x^3,x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {-\frac {1}{2} a^{3} x^{3}-a^{2} b \,x^{2}-\frac {3}{4} a \,b^{2} x -\frac {1}{5} b^{3}}{x^{5}}\)	\(35\)
risch	\(\frac {-\frac {1}{2} a^{3} x^{3}-a^{2} b \,x^{2}-\frac {3}{4} a \,b^{2} x -\frac {1}{5} b^{3}}{x^{5}}\)	\(35\)
gosper	\(-\frac {10 a^{3} x^{3}+20 a^{2} b \,x^{2}+15 a \,b^{2} x +4 b^{3}}{20 x^{5}}\)	\(36\)
default	\(-\frac {a^{2} b}{x^{3}}-\frac {a^{3}}{2 x^{2}}-\frac {3 a \,b^{2}}{4 x^{4}}-\frac {b^{3}}{5 x^{5}}\)	\(36\)
parallelrisch	\(\frac {-10 a^{3} x^{3}-20 a^{2} b \,x^{2}-15 a \,b^{2} x -4 b^{3}}{20 x^{5}}\)	\(36\)

[In]

int((a+b/x)^3/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {10 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x + 4 \, b^{3}}{20 \, x^{5}} \]

[In]

integrate((a+b/x)^3/x^3,x, algorithm="fricas")

[Out]

-1/20*(10*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x + 4*b^3)/x^5

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=\frac {- 10 a^{3} x^{3} - 20 a^{2} b x^{2} - 15 a b^{2} x - 4 b^{3}}{20 x^{5}} \]

[In]

integrate((a+b/x)**3/x**3,x)

[Out]

(-10*a**3*x**3 - 20*a**2*b*x**2 - 15*a*b**2*x - 4*b**3)/(20*x**5)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {10 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x + 4 \, b^{3}}{20 \, x^{5}} \]

[In]

integrate((a+b/x)^3/x^3,x, algorithm="maxima")

[Out]

-1/20*(10*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x + 4*b^3)/x^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {10 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x + 4 \, b^{3}}{20 \, x^{5}} \]

[In]

integrate((a+b/x)^3/x^3,x, algorithm="giac")

[Out]

-1/20*(10*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x + 4*b^3)/x^5

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^3} \, dx=-\frac {\frac {a^3\,x^3}{2}+a^2\,b\,x^2+\frac {3\,a\,b^2\,x}{4}+\frac {b^3}{5}}{x^5} \]

[In]

int((a + b/x)^3/x^3,x)

[Out]

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