3.428 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^2}{x} \, dx\)

Optimal. Leaf size=50 \[ \frac{3}{2} a^2 b^2 x^4+2 a^3 b x^2+a^4 \log (x)+\frac{2}{3} a b^3 x^6+\frac{b^4 x^8}{8} \]

[Out]

2*a^3*b*x^2 + (3*a^2*b^2*x^4)/2 + (2*a*b^3*x^6)/3 + (b^4*x^8)/8 + a^4*Log[x]

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Rubi [A]  time = 0.0337603, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {28, 266, 43} \[ \frac{3}{2} a^2 b^2 x^4+2 a^3 b x^2+a^4 \log (x)+\frac{2}{3} a b^3 x^6+\frac{b^4 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x,x]

[Out]

2*a^3*b*x^2 + (3*a^2*b^2*x^4)/2 + (2*a*b^3*x^6)/3 + (b^4*x^8)/8 + a^4*Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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^2+2 a b x^2+b^2 x^4\right )^2}{x} \, dx &=\frac{\int \frac{\left (a b+b^2 x^2\right )^4}{x} \, dx}{b^4}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a b+b^2 x\right )^4}{x} \, dx,x,x^2\right )}{2 b^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (4 a^3 b^5+\frac{a^4 b^4}{x}+6 a^2 b^6 x+4 a b^7 x^2+b^8 x^3\right ) \, dx,x,x^2\right )}{2 b^4}\\ &=2 a^3 b x^2+\frac{3}{2} a^2 b^2 x^4+\frac{2}{3} a b^3 x^6+\frac{b^4 x^8}{8}+a^4 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0041856, size = 50, normalized size = 1. \[ \frac{3}{2} a^2 b^2 x^4+2 a^3 b x^2+a^4 \log (x)+\frac{2}{3} a b^3 x^6+\frac{b^4 x^8}{8} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/x,x]

[Out]

2*a^3*b*x^2 + (3*a^2*b^2*x^4)/2 + (2*a*b^3*x^6)/3 + (b^4*x^8)/8 + a^4*Log[x]

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Maple [A]  time = 0.041, size = 45, normalized size = 0.9 \begin{align*} 2\,{a}^{3}b{x}^{2}+{\frac{3\,{a}^{2}{b}^{2}{x}^{4}}{2}}+{\frac{2\,a{b}^{3}{x}^{6}}{3}}+{\frac{{b}^{4}{x}^{8}}{8}}+{a}^{4}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^2/x,x)

[Out]

2*a^3*b*x^2+3/2*a^2*b^2*x^4+2/3*a*b^3*x^6+1/8*b^4*x^8+a^4*ln(x)

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Maxima [A]  time = 0.976765, size = 63, normalized size = 1.26 \begin{align*} \frac{1}{8} \, b^{4} x^{8} + \frac{2}{3} \, a b^{3} x^{6} + \frac{3}{2} \, a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + \frac{1}{2} \, a^{4} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x,x, algorithm="maxima")

[Out]

1/8*b^4*x^8 + 2/3*a*b^3*x^6 + 3/2*a^2*b^2*x^4 + 2*a^3*b*x^2 + 1/2*a^4*log(x^2)

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Fricas [A]  time = 1.67549, size = 100, normalized size = 2. \begin{align*} \frac{1}{8} \, b^{4} x^{8} + \frac{2}{3} \, a b^{3} x^{6} + \frac{3}{2} \, a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x,x, algorithm="fricas")

[Out]

1/8*b^4*x^8 + 2/3*a*b^3*x^6 + 3/2*a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4*log(x)

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Sympy [A]  time = 0.274536, size = 49, normalized size = 0.98 \begin{align*} a^{4} \log{\left (x \right )} + 2 a^{3} b x^{2} + \frac{3 a^{2} b^{2} x^{4}}{2} + \frac{2 a b^{3} x^{6}}{3} + \frac{b^{4} x^{8}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**2/x,x)

[Out]

a**4*log(x) + 2*a**3*b*x**2 + 3*a**2*b**2*x**4/2 + 2*a*b**3*x**6/3 + b**4*x**8/8

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Giac [A]  time = 1.11139, size = 63, normalized size = 1.26 \begin{align*} \frac{1}{8} \, b^{4} x^{8} + \frac{2}{3} \, a b^{3} x^{6} + \frac{3}{2} \, a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + \frac{1}{2} \, a^{4} \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^2/x,x, algorithm="giac")

[Out]

1/8*b^4*x^8 + 2/3*a*b^3*x^6 + 3/2*a^2*b^2*x^4 + 2*a^3*b*x^2 + 1/2*a^4*log(x^2)