\(\int (a+\frac {b}{x})^2 x^5 \, dx\) [1559]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {b^2 x^4}{4}+\frac {2}{5} a b x^5+\frac {a^2 x^6}{6} \]

[Out]

1/4*b^2*x^4+2/5*a*b*x^5+1/6*a^2*x^6

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {269, 45} \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {a^2 x^6}{6}+\frac {2}{5} a b x^5+\frac {b^2 x^4}{4} \]

[In]

Int[(a + b/x)^2*x^5,x]

[Out]

(b^2*x^4)/4 + (2*a*b*x^5)/5 + (a^2*x^6)/6

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 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 x^3 (b+a x)^2 \, dx \\ & = \int \left (b^2 x^3+2 a b x^4+a^2 x^5\right ) \, dx \\ & = \frac {b^2 x^4}{4}+\frac {2}{5} a b x^5+\frac {a^2 x^6}{6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {b^2 x^4}{4}+\frac {2}{5} a b x^5+\frac {a^2 x^6}{6} \]

[In]

Integrate[(a + b/x)^2*x^5,x]

[Out]

(b^2*x^4)/4 + (2*a*b*x^5)/5 + (a^2*x^6)/6

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
gosper	\(\frac {x^{4} \left (10 a^{2} x^{2}+24 a b x +15 b^{2}\right )}{60}\)	\(25\)
default	\(\frac {1}{4} b^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} a^{2} x^{6}\)	\(25\)
risch	\(\frac {1}{4} b^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} a^{2} x^{6}\)	\(25\)
parallelrisch	\(\frac {1}{4} b^{2} x^{4}+\frac {2}{5} a b \,x^{5}+\frac {1}{6} a^{2} x^{6}\)	\(25\)
norman	\(\frac {\frac {1}{6} a^{2} x^{7}+\frac {1}{4} b^{2} x^{5}+\frac {2}{5} a b \,x^{6}}{x}\)	\(29\)

[In]

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

[Out]

1/60*x^4*(10*a^2*x^2+24*a*b*x+15*b^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {1}{6} \, a^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, b^{2} x^{4} \]

[In]

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

[Out]

1/6*a^2*x^6 + 2/5*a*b*x^5 + 1/4*b^2*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {a^{2} x^{6}}{6} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{4}}{4} \]

[In]

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

[Out]

a**2*x**6/6 + 2*a*b*x**5/5 + b**2*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {1}{6} \, a^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, b^{2} x^{4} \]

[In]

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

[Out]

1/6*a^2*x^6 + 2/5*a*b*x^5 + 1/4*b^2*x^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {1}{6} \, a^{2} x^{6} + \frac {2}{5} \, a b x^{5} + \frac {1}{4} \, b^{2} x^{4} \]

[In]

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

[Out]

1/6*a^2*x^6 + 2/5*a*b*x^5 + 1/4*b^2*x^4

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (a+\frac {b}{x}\right )^2 x^5 \, dx=\frac {a^2\,x^6}{6}+\frac {2\,a\,b\,x^5}{5}+\frac {b^2\,x^4}{4} \]

[In]

int(x^5*(a + b/x)^2,x)

[Out]

(a^2*x^6)/6 + (b^2*x^4)/4 + (2*a*b*x^5)/5