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

Optimal. Leaf size=23 \[ a^2 \log (x)+a b x^2+\frac{b^2 x^4}{4} \]

[Out]

a*b*x^2 + (b^2*x^4)/4 + a^2*Log[x]

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Rubi [A]  time = 0.0127247, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ a^2 \log (x)+a b x^2+\frac{b^2 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^2/x,x]

[Out]

a*b*x^2 + (b^2*x^4)/4 + a^2*Log[x]

Rule 266

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

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (2 a b+\frac{a^2}{x}+b^2 x\right ) \, dx,x,x^2\right )\\ &=a b x^2+\frac{b^2 x^4}{4}+a^2 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0007734, size = 23, normalized size = 1. \[ a^2 \log (x)+a b x^2+\frac{b^2 x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^2/x,x]

[Out]

a*b*x^2 + (b^2*x^4)/4 + a^2*Log[x]

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Maple [A]  time = 0.002, size = 22, normalized size = 1. \begin{align*} ab{x}^{2}+{\frac{{b}^{2}{x}^{4}}{4}}+{a}^{2}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*x^2+1/4*b^2*x^4+a^2*ln(x)

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Maxima [A]  time = 1.90964, size = 32, normalized size = 1.39 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + a b x^{2} + \frac{1}{2} \, a^{2} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + a*b*x^2 + 1/2*a^2*log(x^2)

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Fricas [A]  time = 1.48527, size = 49, normalized size = 2.13 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + a b x^{2} + a^{2} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + a*b*x^2 + a^2*log(x)

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Sympy [A]  time = 0.244421, size = 20, normalized size = 0.87 \begin{align*} a^{2} \log{\left (x \right )} + a b x^{2} + \frac{b^{2} x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) + a*b*x**2 + b**2*x**4/4

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Giac [A]  time = 1.50345, size = 32, normalized size = 1.39 \begin{align*} \frac{1}{4} \, b^{2} x^{4} + a b x^{2} + \frac{1}{2} \, a^{2} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*x^4 + a*b*x^2 + 1/2*a^2*log(x^2)