3.639 \(\int \frac{(a+b x^4)^3}{x} \, dx\)

Optimal. Leaf size=39 \[ \frac{3}{4} a^2 b x^4+a^3 \log (x)+\frac{3}{8} a b^2 x^8+\frac{b^3 x^{12}}{12} \]

[Out]

(3*a^2*b*x^4)/4 + (3*a*b^2*x^8)/8 + (b^3*x^12)/12 + a^3*Log[x]

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Rubi [A]  time = 0.0182281, antiderivative size = 39, 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} \[ \frac{3}{4} a^2 b x^4+a^3 \log (x)+\frac{3}{8} a b^2 x^8+\frac{b^3 x^{12}}{12} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^3/x,x]

[Out]

(3*a^2*b*x^4)/4 + (3*a*b^2*x^8)/8 + (b^3*x^12)/12 + a^3*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^4\right )^3}{x} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{(a+b x)^3}{x} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (3 a^2 b+\frac{a^3}{x}+3 a b^2 x+b^3 x^2\right ) \, dx,x,x^4\right )\\ &=\frac{3}{4} a^2 b x^4+\frac{3}{8} a b^2 x^8+\frac{b^3 x^{12}}{12}+a^3 \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0036939, size = 39, normalized size = 1. \[ \frac{3}{4} a^2 b x^4+a^3 \log (x)+\frac{3}{8} a b^2 x^8+\frac{b^3 x^{12}}{12} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^3/x,x]

[Out]

(3*a^2*b*x^4)/4 + (3*a*b^2*x^8)/8 + (b^3*x^12)/12 + a^3*Log[x]

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Maple [A]  time = 0.001, size = 34, normalized size = 0.9 \begin{align*}{\frac{3\,{a}^{2}b{x}^{4}}{4}}+{\frac{3\,a{b}^{2}{x}^{8}}{8}}+{\frac{{b}^{3}{x}^{12}}{12}}+{a}^{3}\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^3/x,x)

[Out]

3/4*a^2*b*x^4+3/8*a*b^2*x^8+1/12*b^3*x^12+a^3*ln(x)

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Maxima [A]  time = 0.974596, size = 49, normalized size = 1.26 \begin{align*} \frac{1}{12} \, b^{3} x^{12} + \frac{3}{8} \, a b^{2} x^{8} + \frac{3}{4} \, a^{2} b x^{4} + \frac{1}{4} \, a^{3} \log \left (x^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x,x, algorithm="maxima")

[Out]

1/12*b^3*x^12 + 3/8*a*b^2*x^8 + 3/4*a^2*b*x^4 + 1/4*a^3*log(x^4)

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Fricas [A]  time = 1.44322, size = 81, normalized size = 2.08 \begin{align*} \frac{1}{12} \, b^{3} x^{12} + \frac{3}{8} \, a b^{2} x^{8} + \frac{3}{4} \, a^{2} b x^{4} + a^{3} \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x,x, algorithm="fricas")

[Out]

1/12*b^3*x^12 + 3/8*a*b^2*x^8 + 3/4*a^2*b*x^4 + a^3*log(x)

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Sympy [A]  time = 0.323616, size = 37, normalized size = 0.95 \begin{align*} a^{3} \log{\left (x \right )} + \frac{3 a^{2} b x^{4}}{4} + \frac{3 a b^{2} x^{8}}{8} + \frac{b^{3} x^{12}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**3/x,x)

[Out]

a**3*log(x) + 3*a**2*b*x**4/4 + 3*a*b**2*x**8/8 + b**3*x**12/12

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Giac [A]  time = 1.12822, size = 49, normalized size = 1.26 \begin{align*} \frac{1}{12} \, b^{3} x^{12} + \frac{3}{8} \, a b^{2} x^{8} + \frac{3}{4} \, a^{2} b x^{4} + \frac{1}{4} \, a^{3} \log \left (x^{4}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^3/x,x, algorithm="giac")

[Out]

1/12*b^3*x^12 + 3/8*a*b^2*x^8 + 3/4*a^2*b*x^4 + 1/4*a^3*log(x^4)