\(\int \frac {(a x+b x^3)^2}{x^2} \, dx\) [10]

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[Out]

a^2*x+2/3*a*b*x^3+1/5*b^2*x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1598, 200} \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[In]

Int[(a*x + b*x^3)^2/x^2,x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

Rule 200

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[In]

Integrate[(a*x + b*x^3)^2/x^2,x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
default	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
risch	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
parallelrisch	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
gosper	\(\frac {x \left (3 b^{2} x^{4}+10 a b \,x^{2}+15 a^{2}\right )}{15}\)	\(25\)
norman	\(\frac {a^{2} x^{2}+\frac {1}{5} b^{2} x^{6}+\frac {2}{3} a b \,x^{4}}{x}\)	\(28\)

[In]

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

[Out]

a^2*x+2/3*a*b*x^3+1/5*b^2*x^5

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5} \]

[In]

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

[Out]

a**2*x + 2*a*b*x**3/3 + b**2*x**5/5

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[In]

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

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a x+b x^3\right )^2}{x^2} \, dx=a^2\,x+\frac {2\,a\,b\,x^3}{3}+\frac {b^2\,x^5}{5} \]

[In]

int((a*x + b*x^3)^2/x^2,x)

[Out]

a^2*x + (b^2*x^5)/5 + (2*a*b*x^3)/3