\(\int (\frac {a}{x}+b x)^2 \, dx\) [344]

Optimal result

Integrand size = 11, antiderivative size = 24 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=-\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \]

[Out]

-a^2/x+2*a*b*x+1/3*b^2*x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1607, 276} \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=-\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \]

[In]

Int[(a/x + b*x)^2,x]

[Out]

-(a^2/x) + 2*a*b*x + (b^2*x^3)/3

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=-\frac {a^2}{x}+2 a b x+\frac {b^2 x^3}{3} \]

[In]

Integrate[(a/x + b*x)^2,x]

[Out]

-(a^2/x) + 2*a*b*x + (b^2*x^3)/3

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {a^{2}}{x}+2 a b x +\frac {b^{2} x^{3}}{3}\)	\(23\)
risch	\(-\frac {a^{2}}{x}+2 a b x +\frac {b^{2} x^{3}}{3}\)	\(23\)
norman	\(\frac {\frac {1}{3} b^{2} x^{4}+2 a b \,x^{2}-a^{2}}{x}\)	\(26\)
parallelrisch	\(\frac {b^{2} x^{4}+6 a b \,x^{2}-3 a^{2}}{3 x}\)	\(26\)
gosper	\(-\frac {-b^{2} x^{4}-6 a b \,x^{2}+3 a^{2}}{3 x}\)	\(27\)

[In]

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

[Out]

-a^2/x+2*a*b*x+1/3*b^2*x^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=\frac {b^{2} x^{4} + 6 \, a b x^{2} - 3 \, a^{2}}{3 \, x} \]

[In]

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

[Out]

1/3*(b^2*x^4 + 6*a*b*x^2 - 3*a^2)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=- \frac {a^{2}}{x} + 2 a b x + \frac {b^{2} x^{3}}{3} \]

[In]

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

[Out]

-a**2/x + 2*a*b*x + b**2*x**3/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=\frac {1}{3} \, b^{2} x^{3} + 2 \, a b x - \frac {a^{2}}{x} \]

[In]

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

[Out]

1/3*b^2*x^3 + 2*a*b*x - a^2/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=\frac {1}{3} \, b^{2} x^{3} + 2 \, a b x - \frac {a^{2}}{x} \]

[In]

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

[Out]

1/3*b^2*x^3 + 2*a*b*x - a^2/x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \left (\frac {a}{x}+b x\right )^2 \, dx=\frac {b^2\,x^3}{3}-\frac {a^2}{x}+2\,a\,b\,x \]

[In]

int((b*x + a/x)^2,x)

[Out]

(b^2*x^3)/3 - a^2/x + 2*a*b*x