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

Optimal. Leaf size=25 \[ a^2 x+\frac{2}{3} a b x^3+\frac{b^2 x^5}{5} \]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

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Rubi [A]  time = 0.011589, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1584, 194} \[ a^2 x+\frac{2}{3} a b x^3+\frac{b^2 x^5}{5} \]

Antiderivative was successfully verified.

[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 1584

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]

Rule 194

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]

Rubi steps

\begin{align*} \int \frac{\left (a x+b x^3\right )^2}{x^2} \, dx &=\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]  time = 0.0009442, size = 25, normalized size = 1. \[ a^2 x+\frac{2}{3} a b x^3+\frac{b^2 x^5}{5} \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.002, size = 22, normalized size = 0.9 \begin{align*}{a}^{2}x+{\frac{2\,ab{x}^{3}}{3}}+{\frac{{b}^{2}{x}^{5}}{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.995564, size = 28, normalized size = 1.12 \begin{align*} \frac{1}{5} \, b^{2} x^{5} + \frac{2}{3} \, a b x^{3} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.31853, size = 47, normalized size = 1.88 \begin{align*} \frac{1}{5} \, b^{2} x^{5} + \frac{2}{3} \, a b x^{3} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.059626, size = 22, normalized size = 0.88 \begin{align*} a^{2} x + \frac{2 a b x^{3}}{3} + \frac{b^{2} x^{5}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.22034, size = 28, normalized size = 1.12 \begin{align*} \frac{1}{5} \, b^{2} x^{5} + \frac{2}{3} \, a b x^{3} + a^{2} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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