\(\int \frac {(a x+b x^3)^2}{x} \, dx\) [9]

Optimal result

Integrand size = 15, antiderivative size = 16 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {\left (a+b x^2\right )^3}{6 b} \]

[Out]

1/6*(b*x^2+a)^3/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1598, 267} \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {\left (a+b x^2\right )^3}{6 b} \]

[In]

Int[(a*x + b*x^3)^2/x,x]

[Out]

(a + b*x^2)^3/(6*b)

Rule 267

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

Rule 1598

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {\left (a+b x^2\right )^3}{6 b} \]

[In]

Integrate[(a*x + b*x^3)^2/x,x]

[Out]

(a + b*x^2)^3/(6*b)

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (b \,x^{2}+a \right )^{3}}{6 b}\)	\(15\)
norman	\(\frac {1}{6} b^{2} x^{6}+\frac {1}{2} a b \,x^{4}+\frac {1}{2} a^{2} x^{2}\)	\(25\)
parallelrisch	\(\frac {1}{6} b^{2} x^{6}+\frac {1}{2} a b \,x^{4}+\frac {1}{2} a^{2} x^{2}\)	\(25\)
gosper	\(\frac {x^{2} \left (b^{2} x^{4}+3 a b \,x^{2}+3 a^{2}\right )}{6}\)	\(26\)
risch	\(\frac {b^{2} x^{6}}{6}+\frac {a b \,x^{4}}{2}+\frac {a^{2} x^{2}}{2}+\frac {a^{3}}{6 b}\)	\(33\)

[In]

int((b*x^3+a*x)^2/x,x,method=_RETURNVERBOSE)

[Out]

1/6*(b*x^2+a)^3/b

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {1}{2} \, a b x^{4} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

integrate((b*x^3+a*x)^2/x,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {a^{2} x^{2}}{2} + \frac {a b x^{4}}{2} + \frac {b^{2} x^{6}}{6} \]

[In]

integrate((b*x**3+a*x)**2/x,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {1}{2} \, a b x^{4} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

integrate((b*x^3+a*x)^2/x,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {1}{6} \, b^{2} x^{6} + \frac {1}{2} \, a b x^{4} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

integrate((b*x^3+a*x)^2/x,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a x+b x^3\right )^2}{x} \, dx=\frac {a^2\,x^2}{2}+\frac {a\,b\,x^4}{2}+\frac {b^2\,x^6}{6} \]

[In]

int((a*x + b*x^3)^2/x,x)

[Out]

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